# Unitary canonical forms over Clifford algebras, and an observed unification
of some real-matrix decompositions

Ran Gutin
Department of Computer Science, Imperial College London

###### Abstract

We show that the spectral theorem – which we understand to be a statement that
every self-adjoint matrix admits a certain type of canonical form under
unitary similarity – admits analogues over other $*$-algebras distinct from
the complex numbers. If these $*$-algebras contain nilpotents, then it is
shown that there is a consistent way in which many classic matrix
decompositions – such as the Singular Value Decomposition, the Takagi
decomposition, the skew-Takagi decomposition, and the Jordan decomposition,
among others – are immediate consequences of these. If producing the relevant
canonical form of a self-adjoint matrix were a subroutine in some programming
language, then the corresponding classic matrix decomposition would be a
1-line invocation with no additional steps. We also suggest that by employing
operator overloading in a programming language, a numerical algorithm for
computing a unitary diagonalisation of a complex self-adjoint matrix would
generalise immediately to solving problems like SVD or Takagi. While algebras
without nilpotents (like the quaternions) allow for similar unifying
behaviour, the classic matrix decompositions which they unify are never
obtained as easily. In the process of doing this, we develop some spectral
theory over Clifford algebras of the form $Cl_{p,q,0}(\mathbb{R})$ and
$Cl_{p,q,1}(\mathbb{R})$ where the former is admittedly quite easy. We propose
a broad conjecture about spectral theorems.

##  1 Introduction

In this paper, we prove “spectral theorems” for all Clifford $*$-algebras with
a limited amount of degeneracy (up to 1 nilpotent generator) and suggest that
a “unification” between some classic matrix decompositions results from this.
The unification is interesting because: If producing the canonical form were a
subroutine, then these classical matrix decompositions would be obtained from
1 application of the subroutine, and nothing more. Other reductions between
matrix decompositions are usually less “efficient” (in the sense of not being
just 1 application with no additional steps). Some connections with the theory
of quiver representations are also obtained.

The classic matrix decompositions we consider are the:

  * •

The unitary diagonalisation of a self-adjoint matrix.

  * •

The Takagi decomposition [1] of a complex-symmetric matrix.

  * •

The skew-Takagi decomposition [6] of a complex skew-symmetric matrix.

  * •

The SVD of a real matrix.

  * •

The Jordan decomposition of a real matrix.

In some sense, the first of these cases is equivalent to the rest, if one
varies the $*$-algebra. We will now discuss the notion of a $*$-algebra, and
its motivation.

We are intending to generalise notions like _self-adjoint matrix_ , _unitary
matrix_ and _singular value decomposition_ (among others) to various “number
systems” for various reasons. These notions were originally developed over the
complex numbers $\mathbb{C}$. If we focus attention to the notion of, let’s
say, a unitary matrix, we see that to generalise this to novel “number
systems” it is not sufficient to simply redefine the operations
$\\{+,-,\times,\div\\}$, but also the _complex conjugation_ operation, which
we will denote $*$. A bit of experience with similar “number systems” (like
the quaternions, or the $2\times 2$ real matrices) suggests that a promising
generalisation of complex conjugation over a ring would be an arbitrary ring
anti-automorphism of order up to $2$. Such an operation is called an
_involution_ , and we must include it in our list of operations to redefine:
$\\{+,-,\times,\div,*\\}$.

The above discussion suggests that the claim in some linear algebra courses
that linear algebra takes place over _fields_ is incorrect, because notions
like unitary matrices are defined in terms of involutions which are not field
operations. It is thus interesting to suggest that linear algebra might be
done instead over $*$-fields, which are fields equipped with involutions.
Unfortunately, this is not sufficient for our paper, where our “number
systems” may contain zero divisors.

When we refer to an algebra over a field, we understand this to be something
unital, associative and finite-dimensional. The notion of an algebra is not
sufficient because there are many involutions which an algebra can be equipped
with. The notion of a $*$-algebra [5] is clearly a better notion of “number
system” than just an algebra when generalising spectral theory. Our
$*$-algebras will be over $*$-fields in the expected way.

Most of the $*$-algebras we’ll consider will be Clifford $*$-algebras over a
$*$-field $\mathbb{F}$ which will be either the real number $\mathbb{R}$ or
the complex numbers $\mathbb{C}$ equipped with their standard involutions. We
will in fact consider _two_ different involutions for each algebra.

To give a very quick example of why passing from $\mathbb{C}$ (equipped with
its standard involution) to a larger $*$-algebra can unify matrix
decompositions, consider the Takagi decomposition (though the SVD would
provide a very similar example): This states that given a $\mathbb{C}$-matrix
$M$ which satisfies $M=M^{T}$, there is a unitary matrix $U$ and a
$\mathbb{R}$-diagonal matrix $D$ such that $M=UDU^{T}$. Notice that while this
_looks like_ a diagonalisation of a linear map, it actually isn’t one because
$U^{T}\neq U^{-1}$. The columns of $U$ are not eigenvectors. But now imagine
introducing a new imaginary number $\delta$ which satisfies $\delta
i=-i\delta$ and $\delta^{2}=0$. We then have that
$M\delta=UDU^{T}\delta=U(D\delta)U^{*}=U(D\delta)U^{-1}$. We see that while
the columns of $U$ are not eigenvectors of $M$, they are instead eigenvectors
of $M\delta$. Additionally, if we extend the involution $*$ so that
$\delta^{*}=\delta$, we get that $M\delta$ is self-adjoint, which $M$ wasn’t.
If we brazenly assume that $M\delta$ can be unitarily diagonalised (by an
optimistic extension of the spectral theorem), then it’s immediate that the
unitary diagonalisation will take the form $M\delta=U(D\delta)U^{*}$ where $U$
is immediately a $\mathbb{C}$-matrix and $D$ is immediately an
$\mathbb{R}$-matrix. This is more efficient than introducing the quaternion
$j$ which satisfies $ji=ij$ but not $j^{2}=0$, because we cannot immediately
conclude that the unitary diagonalisation of $Mj$ will yield the Takagi
decomposition.

###  1.1 Preliminary definitions

We assume the reader knows what a field is. A $*$-field is a pair
$(\mathbb{F},*_{\mathbb{F}})$ where $\mathbb{F}$ is a field and
$*_{\mathbb{F}}:\mathbb{F}\to\mathbb{F}$ is a function called the involution.
The involution satisfies $(x+y)^{*}=x^{*}+y^{*}$, $(xy)^{*}=x^{*}y^{*}$,
$(-x)^{*}=-x^{*}$, $1^{*}=1$, $0^{*}=0$.

A $*$-algebra $(\mathcal{A},*:\mathcal{A}\to\mathcal{A})$ over a $*$-field
$(\mathbb{F},*_{\mathbb{F}})$ is an algebra over $\mathbb{F}$ equipped with a
map $*:\mathcal{A}\to\mathcal{A}$ which we call the involution, satisfying
$(x^{*})^{*}=x$, $(xy)^{*}=y^{*}x^{*}$ and $(x+\lambda
y)^{*}=x^{*}+\lambda^{*_{\mathbb{F}}}y^{*}$. In this paper, we assume that our
$*$-algebras are both associative and unital. Associativity means
$(xy)z=x(yz)$. Unital means that there exists an element $1\in\mathcal{A}$
such that $x1=1x=x$, and $1^{*}=1$.

We hope that it is clear what a matrix should be over a $*$-algebra
$(\mathcal{A},*:\mathcal{A}\to\mathcal{A})$. We define a map which we _also
call $*$_ over matrices over a $*$-algebra, which we define to satisfy
$(M^{*})_{ij}=(M_{ji})^{*}$ (where the $*$ on the right hand side is the
involution over the $*$-algebra, but the $*$ on the LHS is a map over
matrices), which we call either the _adjoint_ or the _conjugate-transpose_. A
self-adjoint matrix is one which satisfies $M^{*}=M$, and a unitary matrix is
one which satisfies $M^{*}=M^{-1}$.

##  2 The double number example, and Jordan decomposition

We will illustrate how matrix-decomposition-unification and “spectral
theorems” over “exotic” $*$-algebras relate to each other by considering the
Clifford algebra $Cl_{1,0,0}(\mathbb{R})$ (which is merely an algebra until
equipped with an involution). This algebra is called by different names:
Sometimes it’s called the “double numbers” or the “split-complex numbers”. We
will call it the double numbers [3]. This algebra is sometimes defined as
being “like the complex numbers” but with the role of $i$ being replaced with
a number $j$ for which $j^{2}=1$. This results in the number $1$ having 4
different square roots. This algebra is isomorphic to
$\mathbb{R}\oplus\mathbb{R}$. This implies that the algebra is far more
convenient to work with when its elements are written as $(a,b)$, and all
arithmetic operations $\\{+,-,\times,\div\\}$ are understood to happen
componentwise. We now consider two possible involutions.

###  2.1 The sterile $*_{1}$ involution

The first involution will be denoted $*_{1}$. This will be defined by
$(a,b)^{*_{1}}=(a,b)$. The corresponding $*$-algebra over $\mathbb{R}$ will be
denoted $(\mathbb{R}\oplus\mathbb{R},*_{1})$. This is a trivial definition,
and will result in a rather sterile “spectral theorem”. How should we write an
$(\mathbb{R}\oplus\mathbb{R},*_{1})$-matrix? We will write it in the form
$(M,K)$ where $M$ and $K$ are real matrices of equal dimensions. What we would
be the adjoint operation (sometimes called conjugate-transpose)? It would
simply be $(M,K)^{*}=(M^{T},K^{T})$. Based on this, we see that $(M,K)$ is
unitary (respectively, self-adjoint) whenever $M$ and $K$ are individually
unitary (respectively, self-adjoint). Trivially we obtain a spectral theorem:
Given a self-adjoint $(\mathbb{R}\oplus\mathbb{R},*_{1})$-matrix $(H,K)$, we
see that there exists an $(\mathbb{R}\oplus\mathbb{R},*_{1})$-unitary matrix
$(U,V)$ and an $(\mathbb{R}\oplus\mathbb{R},*_{1})$-diagonal matrix $(D,E)$
such that $(H,K)=(U,V)(D,E)(U,V)^{*}$. But this is trivial and uninteresting.

###  2.2 The intriguing $*_{-1}$ involution

More interesting would be to consider the other possible involution. This
involution will be denoted $*_{-1}$, but otherwise just $*$ if ambiguity won’t
arise. This will be defined by $(a,b)^{*_{-1}}=(b,a)$. The corresponding
$*$-algebra over $\mathbb{R}$ will be denoted
$(\mathbb{R}\oplus\mathbb{R},*_{-1})$. We argue that an
$(\mathbb{R}\oplus\mathbb{R},*_{-1})$-matrix should now be written as $[M,K]$,
and define this to mean $(1,0)M+(0,1)K^{T}$. We use square brackets instead of
round brackets because of the transpose, which ultimately serves to simplify
things.111We think it would be fairer to reserve round bracket notation
$(M,K)$ for when it stands for $(1,0)M+(0,1)K$. We observe the following
identities:

$\displaystyle[C,D]$ $\displaystyle=[AC,DB]$ $\displaystyle[A,B]+[C,D]$
$\displaystyle=[A+C,B+D]$ $\displaystyle[A,B]^{*}$ $\displaystyle=[B,A]$

By defining the square bracket notation the way we did, we made multiplication
slightly more complicated while greatly simplifying the adjoint operation
(which is the third one in our list). The multiplication operation is made
more complicated because the second component multiplies in the opposite order
to the first component.

What is now a self-adjoint matrix over $(\mathbb{R}\oplus\mathbb{R},*_{-1})$?
It needs to satisfy $[M,K]^{*}=[M,K]$. This simplifies to $[K,M]=[M,K]$, so it
is precisely of the form $[M,M]$ where $M$ is an arbitrary square real matrix.
Note that the embedding $M\mapsto[M,M]$ of square real matrices into
$(\mathbb{R}\oplus\mathbb{R},*_{-1})$-matrices does not preserve
multiplication, and so is not a ring homomorphism.

What is now an $(\mathbb{R}\oplus\mathbb{R},*_{-1})$-unitary matrix? It needs
to satisfy $[M,K]^{*}=[M,K]^{-1}$. This simplifies to $[K,M]=[M^{-1},K^{-1}]$,
so it is precisely of the form $[P,P^{-1}]$ where $P$ is an arbitrary
invertible real matrix.

Given a self-adjoint $(\mathbb{R}\oplus\mathbb{R},*_{-1})$-matrix $[M,M]$, the
matrices unitarily similar to it are all of the form
$[P,P^{-1}][M,M][P,P^{-1}]^{*}$, which simplifies to $[PMP^{-1},PMP^{-1}]$.
The “spectral theorem” for $(\mathbb{R}\oplus\mathbb{R},*_{-1})$ is thus
essentially equivalent to the Jordan decomposition for $\mathbb{R}$-matrices:
$[M,M]=[P,P^{-1}][J,J][P^{-1},P]$ where $J$ is a unique Jordan matrix. Thus,
the spectral theorem for $(\mathbb{R}\oplus\mathbb{R},*_{-1})$ is (somehow)
the same as the $\mathbb{R}$-Jordan decomposition. This achieves a
unification.

In summary:

######  Definition 2.1.

The $*$-algebra $(\mathbb{R}\oplus\mathbb{R},*_{-1})$ has as its involution
$(a,b)^{*_{-1}}=(b,a)$. A matrix over $(\mathbb{R}\oplus\mathbb{R},*_{-1})$ is
written in the form $[M,K]$, defined to mean $(1,0)M+(0,1)K^{T}$. The square
brackets hint at the transpose.

######  Proposition 2.1.

A self-adjoint $(\mathbb{R}\oplus\mathbb{R},*_{-1})$-matrix is of the form
$[M,M]$ where $M$ is an arbitrary square $\mathbb{R}$-matrix. A unitary
$(\mathbb{R}\oplus\mathbb{R},*_{-1})$-matrix is of the form $[P,P^{-1}]$ where
$P$ is an invertible $\mathbb{R}$-matrix.

######  Remark 2.1.

The mapping $M\mapsto[M,M]$ preserves the additive, but not the
multiplicative, structure of $\mathbb{R}$-matrices.

######  Theorem 2.1.

Every self-adjoint $(\mathbb{R}\oplus\mathbb{R},*_{-1})$-matrix is unitarily
similar to a unique matrix of the form $[J,J]$, where $J$ is a Jordan matrix,
and the uniqueness is up to permutation of the Jordan blocks of $J$.

The proofs are immediate.

##  3 What we propose a spectrum theorem is for general $*$-algebras, and a
conjecture

In this section, we define what we think a spectral theorem ought to be, and
pose a conjecture. A proof of this would be like a (sometimes purely
qualitative and non-constructive) generalisation of many matrix
decompositions. We begin by defining terms:

We recall some definitions related to monoids:

######  Definition 3.1.

A _monoid_ is the same notion as a group, but without the requirement of
existence of inverses. An _abelian monoid_ is a monoid where the product is
commutative. The product in an abelian monoid is usually written using the
additive symbol $+$. A _free abelian monoid generated by a set $S$_ is the
monoid whose underlying set is the set of functions of the form
$f:S\to\mathbb{N}$ with finite support (i.e. where the set $\\{x\in S\mid
f(x)\neq 0\\}$ is finite), where the monoid operation is $(f+g)(x)=f(x)+g(x)$.
We define a _subfree abelian monoid_ to be a submonoid of a free abelian
monoid.

Let $(\mathcal{A},*)$ be a $*$-algebra over $\mathbb{R}$.

######  Definition 3.2.

By the _spectral monoid_ of $(\mathcal{A},*)$, which we denote
$\overline{\operatorname{Her}}(\mathcal{A},*)$, we mean the monoid
${\operatorname{Her}}(\mathcal{A},*)/{\sim}$ where
${\operatorname{Her}}(\mathcal{A},*)$ is the monoid formed from the set of
Hermitian matrices over $(\mathcal{A},*)$ with the monoid operation being
direct sum of matrices $\oplus$, and $\sim$ denotes the equivalence relation
_unitary similarity_ : $M\sim K\iff\exists\,\text{unitary
matrix}\,U:UMU^{*}=K$.

######  Conjecture 3.1 (The spectral conjecture).

Let $(\mathcal{A},*)$ be an arbitrary associative, unital, finite-dimensional
$*$-algebra over $\mathbb{R}$. We conjecture that the spectral monoid
$\overline{\operatorname{Her}}(\mathcal{A},*)$ is a subfree abelian monoid.

######  Example 3.1.

We refer to only the examples in section 2. For
$(\mathcal{A},*)=(\mathbb{R}\oplus\mathbb{R},*_{-1})$,
$\overline{\operatorname{Her}}(\mathcal{A},*)$ is a free abelian monoid with
the generators corresponding to real Jordan blocks. This captures the
uniqueness of the Jordan Normal Form of an $\mathbb{R}$-matrix, up to
permutation of the Jordan blocks. For
$(\mathcal{A},*)=(\mathbb{R}\oplus\mathbb{R},*_{1})$,
$\overline{\operatorname{Her}}(\mathcal{A},*)$ is only a subfree abelian
monoid; it is in fact a submonoid of
$\overline{\operatorname{Her}}(\mathbb{R},\operatorname{id}_{\mathbb{R}})$.

######  Remark 3.1.

We will later show that the Singular Value Decomposition for the $*$-algebra
$(\mathcal{A},*)$ is implied by the spectral theorem for the $*$-algebra
$(\mathcal{B},\dagger)$ (called the “SVD algebra” for $(\mathcal{A},*)$) where
$\mathcal{B}$ is the result of adjoining an element $\delta$ to
$\mathcal{A}\oplus\mathcal{A}$ (where $\oplus$ denotes direct sum of
algebras), such that a general element of $\mathcal{B}$ is of the form
$(x,y)+(y^{\prime},x^{\prime})\delta$, with: $\delta^{2}=0$,
$\delta(x^{\prime},y^{\prime})=(y^{\prime},x^{\prime})\delta$,
$\delta^{\dagger}=\delta$, $(x,y)^{\dagger}=(x^{*},y^{*})$. The spectral
theorem for the enlarged $*$-algebra is more general than the SVD for the
original $*$-algebra, but nevertheless provides insight into its SVD.

##  4 Introducing three $*$-algebras corresponding to the SVD, Takagi and
skew-Takagi decompositions respectively

###  4.1 The “SVD $*$-algebra”

Consider the Clifford algebra $Cl_{1,0,1}(\mathbb{R})$, which we will equip
with a certain involution. To make the reader’s life easier, we will describe
this algebra explicitly. The elements of this algebra are all of the form

$(a,b)+(a^{\prime},b^{\prime})\delta$

where $a,b,a^{\prime},b^{\prime}$ are all real numbers. The two pairs $(a,b)$
and $(a^{\prime},b^{\prime})$ are double numbers, or numbers belonging to the
algebra $\mathbb{R}\oplus\mathbb{R}$. The number $\delta$ on the other hand is
quite exotic. First of all, $\delta$ satisfies $\delta^{2}=0$. Additionally, a
number of the form $(a^{\prime},b^{\prime})\delta$ is essentially in its
simplest form, but a number of the form $\delta(a^{\prime},b^{\prime})$
simplifies to $(b^{\prime},a^{\prime})\delta$. $\delta$ therefore acts as a
swapping operator. The swapping operation here is essentially identical to the
operation we defined as $*_{-1}$ over $\mathbb{R}\oplus\mathbb{R}$.

The algebra $Cl_{1,0,1}(\mathbb{R})$ is still not a $*$-algebra because we
have not equipped it with an involution. We will define an involution $*_{1}$
by
$((a,b)+(a^{\prime},b^{\prime})\delta)^{*_{1}}=(a,b)+(b^{\prime},a^{\prime})\delta$.
We will denote this more simply as $*$ unless this results in ambiguity. We
will denote the corresponding $*$-algebra as $(Cl_{1,0,1}(\mathbb{R}),*_{1})$.

How should we write a matrix over $(Cl_{1,0,1}(\mathbb{R}),*_{1})$? We can
write it as $(M,K)+(M^{\prime},K^{\prime})\delta$ with the obvious meaning.
What is the adjoint operation over $(Cl_{1,0,1}(\mathbb{R}),*_{1})$? It is
$((M,K)+(M^{\prime},K^{\prime})\delta)^{*}=(M^{T},K^{T})+((K^{\prime})^{T},(M^{\prime})^{T})\delta$.
What then is a $(Cl_{1,0,1}(\mathbb{R}),*_{1})$-unitary matrix? It is of the
form $(U,V)(I+(K,-K^{T})\delta)$ where $U$ and $V$ are orthogonal matrices.
What is then a self-adjoint matrix? It is of the form $(H,K)+(M,M^{T})\delta$
where $H=H^{T}$ and $K=K^{T}$. We can consider an _infinitesimal_ self-adjoint
matrix to be of the form $(M,M^{T})\delta$ because $\delta$ behaves like an
infinitesimal.

We now consider the special case of the “spectral theorem” for only
_infinitesimal_ self-adjoint matrices. This would be a canonical form for an
infinitesimal self-adjoint matrix under unitary similarity. Consider an
infinitesimal self-adjoint matrix $(M,M^{T})\delta$ conjugated by a unitary
matrix $(U,V)(I+(K,-K^{T})\delta)$; this is written as
$(U,V)(I+(K,-K^{T})\delta)\times(M,M^{T})\delta\times((U,V)(I+(K,-K^{T})\delta))^{*}$,
and simplifies to $(UMV^{T},VM^{T}U^{T})\delta$. The canonical form must
therefore be the same as the $\mathbb{R}$-singular value decomposition. A
unification is thus achieved between a special case of the spectral theorem
over $(Cl_{1,0,1}(\mathbb{R}),*_{1})$ (over only the infinitesimal self-
adjoint matrices) and the singular value decomposition over $\mathbb{R}$. Note
that the reasoning is valid with respect to any $*$-field in place of
$\mathbb{R}$ (including $\mathbb{C}$).

Appendix remark: Note that we could have defined another involution $*_{-1}$
by
$((a,b)+(a^{\prime},b^{\prime})\delta)^{*_{-1}}=(a,b)-(b^{\prime},a^{\prime})\delta$
and considered the $*$-algebra $(Cl_{1,0,1}(\mathbb{R}),*_{-1})$ instead. But
this is in fact redundant because $(Cl_{1,0,1}(\mathbb{R}),*_{-1})$ is
isomorphic to $(Cl_{1,0,1}(\mathbb{R}),*_{1})$. The isomorphism is given by
$(a,b)+(a^{\prime},b^{\prime})\delta\mapsto(a,b)+(a^{\prime},-b^{\prime})\delta$,
and works in either direction.

######  Proposition 4.1.

$(Cl_{1,0,1}(\mathbb{R}),*_{1})$ is isomorphic to
$(Cl_{1,0,1}(\mathbb{R}),*_{-1})$.

###  4.2 The “Takagi $*$-algebra”

Consider the Clifford algebra $Cl_{0,1,1}(\mathbb{R})$, which we will equip
with a certain involution. To make the reader’s life easier, we will describe
this algebra explicitly. The elements of this algebra are all of the form

$(a+bi)+(a^{\prime}+b^{\prime}i)\delta$

where $a,b,a^{\prime},b^{\prime}$ are all real numbers. The two components
$a+bi$ and $a^{\prime}+b^{\prime}i$ are complex numbers. The number $\delta$
on the other hand is quite exotic. First of all, $\delta$ satisfies
$\delta^{2}=0$. Additionally, a number of the form
$(a^{\prime}+b^{\prime}i)\delta$ is essentially in its simplest form, but a
number of the form $\delta(a^{\prime}+b^{\prime}i)$ simplifies to
$(a^{\prime}-b^{\prime}i)\delta$. $\delta$ therefore acts as a complex
conjugation operator.

The algebra $Cl_{0,1,1}(\mathbb{R})$ is still not a $*$-algebra because we
have not equipped it with an involution. We will define an involution $*_{1}$
by
$((a+bi)+(a^{\prime}+b^{\prime}i)\delta)^{*_{1}}=(a-bi)+(a^{\prime}+b^{\prime}i)\delta$.
We will denote this more simply as $*$ unless this results in ambiguity. We
will denote the corresponding $*$-algebra as $(Cl_{1,0,1}(\mathbb{R}),*_{1})$.

How should we write a matrix over $(Cl_{0,1,1}(\mathbb{R}),*_{1})$? We can
write it as $M+K\delta$ with the obvious meaning. What is the adjoint
operation over $(Cl_{0,1,1}(\mathbb{R}),*_{1})$? It is
$(M+K\delta)^{*}=M^{*}+K^{T}\delta$ where the $M^{*}$ denotes conjugate-
transpose of a complex matrix. What then is a
$(Cl_{0,1,1}(\mathbb{R}),*_{1})$-unitary matrix? It is of the form
$U(I+K\delta)$ where $U$ is complex unitary and $K$ is skew complex-symmetric.
What is then a self-adjoint matrix over $(Cl_{0,1,1}(\mathbb{R}),*_{1})$? It
is of the form $H+S\delta$ where $H=H^{*}$ and $S=S^{T}$. We can consider an
_infinitesimal_ self-adjoint matrix to be of the form $S\delta$ where $S$ is
complex-symmetric because $\delta$ behaves like an infinitesimal.

We now consider the special case of the “spectral theorem” for only
_infinitesimal_ self-adjoint $(Cl_{0,1,1}(\mathbb{R}),*_{1})$-matrices. This
would be a canonical form for an infinitesimal self-adjoint
$(Cl_{0,1,1}(\mathbb{R}),*_{1})$-matrix under unitary similarity. Consider an
infinitesimal self-adjoint $(Cl_{0,1,1}(\mathbb{R}),*_{1})$-matrix $S\delta$
(where $S=S^{T}$) conjugated by a $(Cl_{0,1,1}(\mathbb{R}),*_{1})$-unitary
matrix $U(I+K\delta)$; this is written as $U(I+K\delta)\times
S\delta\times(U(I+K\delta))^{*}$, and simplifies to $USU^{T}\delta$. The
canonical form must therefore be the same as the Takagi decomposition. A
unification is thus achieved between a special case of the spectral theorem
over $(Cl_{0,1,1}(\mathbb{R}),*_{1})$ (over only the infinitesimal self-
adjoint matrices) and the Takagi decomposition for symmetric matrices over
$\mathbb{C}$.

###  4.3 The “skew-Takagi $*$-algebra”

Consider the same algebra $Cl_{0,1,1}(\mathbb{R})$ as in the previous
subsection, but with a different involution. Define
$*_{-1}:Cl_{0,1,1}(\mathbb{R})\to Cl_{0,1,1}(\mathbb{R})$ by
$(w+z\delta)^{*{-1}}=w^{*}-z^{*}\delta$, where in the previous definition we
instead had $w^{*}+z^{*}\delta$. An infinitesimal self-adjoint matrix is now
of the form $S\delta$ where $S$ is _skew complex-symmetric_ instead of
_complex-symmetric_ as before (i.e. we need $S=-S^{T}$). Applying a unitary
similarity to it simplifes to $USU^{T}\delta$ where $U$ is some complex
unitary matrix. The spectral theorem for infinitesimal self-adjoint matrices
over $(Cl_{0,1,1}(\mathbb{R}),*_{-1})$ is thus the same as the skew-Takagi
decomposition.

###  4.4 Remark about dualities between matrix decompositions

The above is surprising because the Takagi decomposition is usually seen as a
special case of the SVD. What we’ve uncovered though is instead a duality
between them. Takagi is not a special case of SVD, but is its dual. This
duality suggests that somehow, whatever “works” (let’s say in numerical
computing) for the SVD should also work over Takagi. Since the SVD is more
thoroughly studied, this suggests that one can transfer the numerical theory
of the SVD wholesale onto the Takagi and skew-Takagi decompositions.

The skew-Takagi decomposition is obviously dual to the Takagi decomposition in
a different way to how it is dual to the SVD.

Notice that when diagonalising a self-adjoint matrix $H$ over $\mathbb{C}$,
the following steps are taken:

  1. 1.

Find an eigenvector $v$ of $H$.

  2. 2.

Find a basis $B$ for the orthogonal complement of $v$, written $v^{\perp}$.

  3. 3.

Restrict $H$ to $v^{\perp}$ by using the basis $B$.

  4. 4.

Return to the first step.

The trick is to realise that the SVD and Takagi follow the same plan. In the
case of the SVD, given an arbitrary $\mathbb{R}$-matrix $M$, the matrix
$(M,M^{T})\delta$ is a self-adjoint matrix in
$(Cl_{1,0,1}(\mathbb{R}),*_{1})$. A left and right singular vector of $M$,
when paired together as $(u,v)$, is actually an eigenvector of
$(M,M^{T})\delta$. Conversely, an eigenvector of $(M,M^{T})\delta$ is
immediately reducible to the $(Cl_{1,0,1}(\mathbb{R}),*_{1})$-vector $(u,v)$
where $u$ and $v$ are singular vectors of $M$. Steps 2 and 3 are simple. In
step 2, the orthogonal complement of $(u,v)$ is the Cartesian product of the
orthogonal complements of $u$ and $v$. In symbols:
$(u,v)^{\perp}=u^{\perp}\times v^{\perp}$. Thus, given an ordered basis
$(b_{1},b_{2},\dotsc,b_{n})$ of $u^{\perp}$ and an ordered basis
$(c_{1},c_{2},\dotsc,c_{n})$ of $v^{\perp}$, we can obtain an ordered basis
$((b_{1},c_{1}),(b_{2},c_{2}),\dotsc,(b_{n},c_{n}))$ of $(u,v)^{\perp}$.

Step 1 is usually the most complicated, but there are approaches which
sometimes work, depending on the $*$-algebra. A good approach can be called
the _unpack-and-unwind_ method (for lack of a better name). Given a
$*$-algebra $(\mathcal{A},*)$, and a sub-$*$-algebra $\mathcal{B}$, we define
a function “$\operatorname{unpack}$” that sends a $(\mathcal{A},*)$-matrix to
a $(\mathcal{B},*)$-matrix. For example, consider the matrix $(M,M^{T})\delta$
over $(Cl_{1,0,1}(\mathbb{R}),*_{1})$, and let $\mathcal{B}$ be the dual
numbers. We have that
$\operatorname{unpack}((M,M^{T})\delta)=\begin{pmatrix}0&M\varepsilon\\\
M^{T}\varepsilon&0\end{pmatrix}$. Ignoring the $\varepsilon$, we can easily
obtain an eigenvector $v$ of $\operatorname{unpack}((M,M^{T})\delta)$ using
the $\mathbb{R}$-spectral theorem. Notice though that an eigenvector of
$\operatorname{unpack}((M,M^{T})\delta)$ is not an eigenvector of
$(M,M^{T})\delta$. We define another function called $\operatorname{unwind}$.
$\operatorname{unwind}$ sends an $\mathcal{A}$-vector to a
$\mathcal{B}$-vector in such a way that
$\operatorname{unwind}(Kv)=\operatorname{unpack}(K)\operatorname{unwind}(v)$.
For our eigenvector $v$ of $\operatorname{unpack}((M,M^{T})\delta)$, we can
verify that $v=\operatorname{unwind}(v^{\prime})$ for some
$(Cl_{1,0,1}(\mathbb{R}),*_{1})$-vector $v^{\prime}$. We then have that
$\operatorname{unwind}(\lambda
v^{\prime})=\lambda\operatorname{unwind}(v^{\prime})=\operatorname{unpack}((M,M^{T})\delta)\operatorname{unwind}(v^{\prime})=\operatorname{unwind}((M,M^{T})\delta
v^{\prime})$. By the injectivity of $\operatorname{unwind}$, we cancel to get
that $v^{\prime}$ is an eigenvector of $(M,M^{T})\delta$.

Note that all of this is _not_ the same thing as reducing SVD or Takagi to the
diagonalisation of some suitable self-adjoint $\mathbb{C}$-matrix. These
tricks are well-known, suboptimal, and they require additional post-processing
steps which the method we’re proposing doesn’t.

##  5 Infinitesimal spectral theorems, and the resulting equivalence relations
which give rise to numerous matrix decompositions

###  5.1 Understanding $\delta$ in general

The algebras above – that is, $Cl_{0,1,1}(\mathbb{R})$ and
$Cl_{1,0,1}(\mathbb{R})$, and without considering involutions – could be
obtained from $\mathbb{C}(\cong Cl_{0,1,0}(\mathbb{R}))$ or
$\mathbb{R}\oplus\mathbb{R}(\cong Cl_{1,0,0}(\mathbb{R}))$ respectively in
multiple ways. On the one hand, they could be obtained by introducing a
nilsquare generator $\delta$ into the Clifford algebra to obtain a larger
Clifford algebra.

A broader direction of generalisation is also apparent: We introduced an
element $\delta$ into some algebra $A$ over some field $\mathbb{F}$ such that
$\delta z=\phi(z)\delta$ where $\phi:A\to A$ is some function. This is a lot
like the Cayley-Dickson construction. What properties should $\phi$ satisfy?

To have associativity, we will need to have $\phi(wz)=\phi(w)\phi(z)$. Why?
Because by associativity, we have that $\phi(wz)\delta=\delta(wz)=(\delta
w)z=(\phi(w)\delta)z=\phi(w)(\delta
z)=\phi(w)(\phi(z)\delta)=(\phi(w)\phi(z))\delta$. This already rules out the
standard quaternion involution, or the matrix transpose, as possible
instantiations of $\phi$.

$\phi$ will need to be linear over the underlying field. Why? Because
$\phi(w+\lambda z)\delta=\delta(w+\lambda z)=\delta w+\delta\lambda
z=\phi(w)\delta+\lambda\phi(z)\delta=(\phi(w)+\lambda\phi(z))\delta$ where
$\lambda$ is in the field.

Notice though that it would not suffice for $\phi$ to be an antiautomorphism.
The usual quaternion involution is actually an antiautomorphism and not an
automorphism – as is the matrix transpose – and so could not be used as a
$\phi$.

We can now state:

######  Proposition 5.1.

Let $A$ be an algebra over a field $\mathbb{F}$. Let $A[[\delta]]$ denote the
algebra obtained by adjoining to $A$ an element $\delta$ such that
$\delta^{2}=0$ and $\delta z=\phi(z)\delta$ for some function $\phi:A\to A$
and all $z\in A$. The result is an associative algebra over $\mathbb{F}$ if
and only if $\phi$ is an algebra automorphism of $A$.

###  5.2 The unifying list

Each matrix decomposition we consider presents a canonical form for a matrix
under some equivalence relation. In our statement of conjecture 3.1, we
speculate that this canonical form has a particularly nice structure for the
general type of equivalence relation we consider here.

Some matrix decompositions are precisely equivalent to spectral theorems over
$*$-algebras. We’ve seen this with the spectral theorem for
$(\mathbb{R}\oplus\mathbb{R},*_{-1})$, which is precisely equivalent to the
Jordan decomposition. But oftentimes, the equivalence only holds for so-called
_infinitesimal matrices_ , with the non-infinitesimal spectrum theorem being
strictly more general than is needed to describe a classic matrix
decomposition.

Given a $*$-algebra $(R,*)$, an endomorphism $\phi:R\to R$, and a value
$s\in\\{-1,+1\\}$, we can define the $*$-algebra $R[[\delta]]$ whose elements
are of the form $w+z\delta$ ($w,z\in R$), where $\delta z=\phi(z)\delta$, and
where $(w+z\delta)^{*}=w^{*}+s\phi(z^{*})\delta$.

A matrix $M$ over the $*$-algebra $R[[\delta]]$ is called _infinitesimal_ if
it is equal to $K\delta$ for some $(R,*)$-matrix $K$. An _infinitesimal self-
adjoint matrix_ $H$ is one which is infinitesimal and self-adjoint.

The unitary-similarity relation specialised to infinitesimal self-adjoint
matrices is equivalent (depending on $R,*,\phi,s$) to numerous equivalence
relations on $\mathbb{R}$-matrices, with a likely spectral theorem for each of
them:

  * •

Ordinary matrix similarity, whose canonical form is the Jordan decomposition,
if $R=\mathbb{R}\oplus\mathbb{R}$, $(a,b)^{*}=(b,a)$, $s=1$ and
$\phi(a,b)=(a,b)$.

  * •

The equivalence relation $M\sim UMV^{T}$ (for $M$ real, and $U$ and $V$ real-
unitary), whose canonical form is the SVD, if $R=\mathbb{R}\oplus\mathbb{R}$,
$(a,b)^{*}=(a,b)$, $s=1$ and $\phi(a,b)=(b,a)$.

  * •

The equivalence relation $M\sim VMV^{T}$ for $M$ complex-symmetric and $V$
complex-unitary, whose canonical form is the Takagi decomposition, if
$R=\mathbb{C}$, $(a+bi)^{*}=a-bi$, $s=1$ and $\phi(a+bi)=a-bi$.

  * •

The equivalence relation $M\sim VMV^{T}$ for $M$ complex skew-symmetric and
$V$ complex-unitary, whose canonical form is the skew-Takagi decomposition if
$R=\mathbb{C}$, $(a+bi)^{*}=a-bi$, $s=-1$ and $\phi(a+bi)=a-bi$.

  * •

Consider four symmetric bilinear forms
$B_{1},B^{\prime}_{1},B_{2},B^{\prime}_{2}:V\otimes V\to\mathbb{R}$, the first
two being non-degenerate, over an $\mathbb{R}$-vector space $V$. We say that
$(B_{1},B_{2})$ is equivalent to $(B^{\prime}_{1},B^{\prime}_{2})$ if there is
an invertible linear operator $P:V\to V$ such that for all $i\in\\{1,2\\}$ we
have $B_{i}(P(u),P(v))=B^{\prime}_{i}(u,v)$. This equivalence relation occurs
when $R=\mathbb{R}\oplus\mathbb{R}$, $(a,b)^{*}=(b,a)$, $s=1$ and
$\phi(a,b)=(b,a)$. We don’t consider this one further.

  * •

The equivalence relation $M\sim VMV^{T}$ for $M$ dual-number symmetric and $V$
dual-number unitary, whose canonical form is a dual-Takagi decomposition, if
$R=\mathbb{D}$, $(a+b\varepsilon)^{*}=a+b\varepsilon$, $s=1$ and
$\phi(a+b\varepsilon)=a-b\varepsilon$.

There are other special cases of spectral theorems over $*$-algebras that are
equivalent to other matrix decompositions.

##  6 The unpack-and-unwind method for computing and verifying existence of
some classic matrix decompositions

The following theorems aren’t new, but are proved using the same technique.
These are special cases of spectral theorems which we will prove fully later.
Note that by $\mathbb{C}$, we will mean the $*$-algebra of complex numbers
equipped with their standard involution: $(a+bi)^{*}=a-bi$. We won’t write
this $(\mathbb{C},*_{-1})$ for the sake of readability, but be aware that the
complex numbers may (outside of this paper) be equipped with a different
involution.

Below, we piecemeal define operations we call $\operatorname{unpack}$ and
$\operatorname{unwind}$. The $\operatorname{unwind}$ operation acts on column
vectors, and is at least _partially_ a mere change of scalars. The
$\operatorname{unpack}$ operation is then defined such that
$\operatorname{unpack}(M)\operatorname{unwind}(v)=\operatorname{unwind}(Mv)$
for all $v$. We say that $\operatorname{unwind}$ is only partially a change of
scalars, because it changes the scalar $*$-algebra to one of its subalgebras,
but in such a way that for instance the “length squared” of a vector might be
preserved.

For example, the following would be a _bad way_ to define
$\operatorname{unwind}$ from $\mathbb{C}$-vectors to $\mathbb{R}$-vectors:
$\operatorname{unwind}(u+vi)=(u+v,u)^{T}$. The problem with this definition is
that if $z=u+vi$, then
$z^{*}z\neq\operatorname{unwind}(z)^{*}\operatorname{unwind}(z)$ according to
this proposal.

In some instances below, we unfortunately cannot define
$\operatorname{unwind}$ such that
$v^{*}v=\operatorname{unwind}(v)^{*}\operatorname{unwind}(v)$, but a seemingly
natural definition of $\operatorname{unwind}$ is still possible, and we still
give one. Such difficulties arise in any ring which contains nontrivial
idempotents (i.e. solutions to $x^{2}=x$ which are not either $0$ or $1$): For
example, in the ring $\mathbb{R}\oplus\mathbb{R}$.

In general, the $\operatorname{unwind}$ and $\operatorname{unpack}$ operations
ought to satisfy:

$\displaystyle\operatorname{unwind}(u+v)$
$\displaystyle=\operatorname{unwind}(u)+\operatorname{unwind}(v)$
$\displaystyle\operatorname{unwind}(Av)$
$\displaystyle=\operatorname{unpack}(A)\operatorname{unwind}(v)$
$\displaystyle\operatorname{unpack}(AB)$
$\displaystyle=\operatorname{unpack}(A)\operatorname{unpack}(B)$
$\displaystyle\operatorname{unpack}(A^{*})$
$\displaystyle=\operatorname{unpack}(A)^{*}$ $\operatorname{unwind}$ is
bijective

######  Proposition 6.1 (Takagi decomposition).

Given a complex-symmetric matrix $M$ (i.e. one which satisfies $M=M^{T}$, and
not to be confused with a self-adjoint $\mathbb{C}$-matrix) there exists a
$\mathbb{C}$-unitary matrix $U$ and an $\mathbb{R}$-diagonal matrix $D$ such
that $M=UDU^{T}$. (Note that $U^{T}\neq U^{-1}$).

###### Proof.

Observe that while $M$ is not Hermitian, we may introduce a new scalar
$\delta$ such that $\delta^{2}=0,\delta i=-i\delta$ and $\delta^{*}=\delta$.
We see that $M\delta$ is now a $(Cl_{0,1,1}(\mathbb{R}),*_{1})$-matrix, and is
indeed self-adjoint (over $(Cl_{0,1,1}(\mathbb{R}),*_{1})$) because
$(M\delta)^{*}=\delta^{*}M^{*}=\delta\overline{M}^{T}=M^{T}\delta=M\delta$. We
might then hope to unitarily diagonalise $M\delta$. We seek to show that a
$(Cl_{0,1,1}(\mathbb{R}),*_{1})$-unitary diagonalisation of $M\delta$ – should
it exist – gives rise to a Takagi decomposition of $M$. To see this, we begin
with assuming that a $(Cl_{0,1,1}(\mathbb{R}),*_{1})$-unitary diagonalisation
exists: $M\delta=UDU^{*}$ for $(Cl_{0,1,1}(\mathbb{R}),*_{1})$-matrices $U$
and $D$ where $U$ is $(Cl_{0,1,1}(\mathbb{R}),*_{1})$-unitary and $D$ is
$(Cl_{0,1,1}(\mathbb{R}),*_{1})$-diagonal. We write each of $U$ and $D$ as:

$\displaystyle U=U_{0}+U^{\prime}\delta$ $\displaystyle
D=D_{0}+D^{\prime}\delta$

where the four matrices $U_{0},U^{\prime},D_{0},D^{\prime}$ are all complex
matrices, and $U_{0}$ is $\mathbb{C}$-unitary. Substituting these into
$M\delta=UDU^{*}$ and simplifying gives
$M\delta=U_{0}D^{\prime}U_{0}^{T}\delta$. Taking the $\delta$ component
finally gives $M=U_{0}D^{\prime}U_{0}^{T}$. This is a Takagi decomposition of
$M$. Therefore, our problem is reduced to finding the unitary diagonalisation
of $M\delta$.

We define
$\operatorname{unpack}(A+Bi+C\delta+Di\delta):=\begin{pmatrix}A-C\varepsilon&-B+D\varepsilon\\\
B+D\varepsilon&A+C\varepsilon\end{pmatrix}$ (where $A,B,C,D$ are arbitrary
$\mathbb{R}$-matrices of equal dimensions) and
$\operatorname{unwind}(a+bi+c\delta+di\delta):=\begin{pmatrix}a+c\epsilon\\\
b+d\varepsilon\end{pmatrix}$ (where $a,b,c,d$ are $\mathbb{R}$-vectors of
equal dimensions). We observe that
$\operatorname{unpack}(MK)=\operatorname{unpack}(M)\operatorname{unpack}(K)$
and
$\operatorname{unwind}(Mv)=\operatorname{unpack}(M)\operatorname{unwind}(v)$.

We have that $\operatorname{unpack}(M\delta)=\begin{pmatrix}-\Re(M)&\Im(M)\\\
\Im(M)&\Re(M)\end{pmatrix}\varepsilon$. Ignoring the $\varepsilon$, we have a
self-adjoint $\mathbb{R}$-matrix. By the $\mathbb{R}$-spectral theorem, we
obtain an $\mathbb{R}$-eigenvector $v$ of $\operatorname{unpack}(M\delta)$.
Unfortunately, $v$ is not an eigenvector of $M\delta$, but only of
$\operatorname{unpack}(M\delta)$. By the expression for
$\operatorname{unwind}$ above, we have that there is a $\mathbb{C}$-vector
$v^{\prime}$ such that $\operatorname{unwind}(v^{\prime})=v$. We see that
$v^{\prime}$ is an eigenvector of $M\delta$.

Since $M\delta$ is self-adjoint, it maps the orthogonal complement of
$v^{\prime}$ to itself. We can define a matrix representation of $M\delta$
over this subspace. Since $v^{\prime}$ is a $\mathbb{C}$-vector, a
$(Cl_{0,1,1}(\mathbb{R}),*_{1})$-orthonormal basis $B$ for
$(v^{\prime})^{\perp}$ is obtained simply from the $\mathbb{C}$-orthogonal
complement of $v^{\prime}$. The matrix representation of $M\delta$ over
$(v^{\prime})^{\perp}$ is obtained using the basis $B$. The result is a self-
adjoint $(Cl_{0,1,1}(\mathbb{R}),*_{1})$-matrix $M^{\prime}\delta$ with one
less dimension than $M\delta$. We repeat by finding another eigenvector,
restricting to the orthogonal complement of _that_ , etc. Formally, this proof
is by induction. ∎

######  Proposition 6.2 (Singular Value Decomposition).

Given a square $\mathbb{R}$-matrix $M$ there exists a pair of
$\mathbb{R}$-unitary matrices $U$ and $V$, and an $\mathbb{R}$-diagonal matrix
$D$, such that $UDV^{T}=M$.

######  Remark 6.1.

Before we begin the proof, we provide a justification for providing a new
proof of the singular value decomposition, other than that we can use the same
method for other decompositions. The approach usually taught to students for
proving or computing the SVD, which revolves around diagonalising $M^{T}M$
does _not work over the ring of dual numbers_. This is because the eigenvalues
of $M^{T}M$ are the squares of the singular values of $M$. The singular values
of a dual number matrix may square to zero without actually being zero. This
provides a simplified model of numerical underflow in inexact floating point
or fixed point computation. While our method below uses a well-known block
matrix $\begin{pmatrix}0&M\\\ M^{T}&0\end{pmatrix}$ whose singular values are
plus-and-minus the singular values of $M$, the argument below succeeds in
building an existence-of-SVD proof around it, where we ensure that the unitary
diagonalisation we obtain has the block structure we need. For the dual-number
spectral theorem(s) and initial proofs of the corresponding SVD(s), see [2].
Note that we say “(s)” because there is a separate dual-number spectral
theorem (and SVD) for each of the two possible involutions over the dual
numbers.

###### Proof.

Observe that $(M,M^{T})\delta$ is self-adjoint over
$(Cl_{1,0,1}(\mathbb{R}),*_{1})$. We might then reasonably hope to obtain a
unitary diagonalisation of $(M,M^{T})\delta$. We seek to show that a
$(Cl_{1,0,1}(\mathbb{R}),*_{1})$-unitary diagonalisation of $(M,M^{T})\delta$
– should it exist – gives rise to a singular value decomposition of $M$. To
see this, we begin with assuming that a
$(Cl_{1,0,1}(\mathbb{R}),*_{1})$-unitary diagonalisation exists:
$(M,M^{T})\delta=WDW^{*}$ for $(Cl_{1,0,1}(\mathbb{R}),*_{1})$-matrices $W$
and $D$ where $W$ is $(Cl_{1,0,1}(\mathbb{R}),*_{1})$-unitary and $D$ is
$(Cl_{1,0,1}(\mathbb{R}),*_{1})$-diagonal. We write each of $W$ and $D$ as:

$\displaystyle W=(U,V)+(U^{\prime},V^{\prime})\delta$ $\displaystyle
D=(E,F)+(\Sigma,\Sigma^{\prime})\delta$

where the eight matrices
$U,V,U^{\prime},V^{\prime},E,F,\Sigma,\Sigma^{\prime}$ are all real matrices,
and the two matrices $U$ and $V$ are $\mathbb{R}$-unitary. Substituting these
into $(M,M^{T})\delta=WDW^{*}$ and simplifying gives
$\Sigma^{\prime}=\Sigma^{T}$ and $(M,M^{T})\delta=(U\Sigma V^{T},V\Sigma
U^{T})\delta$. Taking the $\delta$ component finally gives $(M,M^{T})=(U\Sigma
V^{T},V\Sigma U^{T})$, which implies $M=U\Sigma V^{T}$. This is an SVD of $M$.
Therefore, our problem is reduced to finding the unitary diagonalisation of
$(M,M^{T})\delta$.

We define
$\operatorname{unpack}((A,B)+(C,D)\delta):=\begin{pmatrix}A&C\varepsilon\\\
D\varepsilon&B\end{pmatrix}$ (where $A,B,C,D$ are arbitrary
$\mathbb{R}$-matrices of equal dimensions) and
$\operatorname{unwind}((a,b)+(c,d)\delta):=\begin{pmatrix}a+d\varepsilon\\\
b+c\varepsilon\end{pmatrix}$ (where $a,b,c,d$ are $\mathbb{R}$-vectors of
equal dimensions). We observe that
$\operatorname{unpack}(MK)=\operatorname{unpack}(M)\operatorname{unpack}(K)$
and
$\operatorname{unwind}(Mv)=\operatorname{unpack}(M)\operatorname{unwind}(v)$.

We have that $\operatorname{unpack}((M,M^{T})\delta)=\begin{pmatrix}0&M\\\
M^{T}&0\end{pmatrix}\varepsilon$. Ignoring the $\varepsilon$, we have a self-
adjoint $\mathbb{R}$-matrix. By the $\mathbb{R}$-spectral theorem, we obtain
an $\mathbb{R}$-eigenvector $w$ of $\operatorname{unpack}((M,M^{T})\delta)$.
Unfortunately, $w$ is not an eigenvector of $(M,M^{T})\delta$, but only of
$\operatorname{unpack}((M,M^{T})\delta)$. By the expression for
$\operatorname{unwind}$ above, we have that there is a $\mathbb{R}$-vector
$(u,v)$ such that $\operatorname{unwind}((u,v))=w$. We see that $(u,v)$ is an
eigenvector of $(M,M^{T})\delta$.

Since $(M,M^{T})\delta$ is self-adjoint, it maps the orthogonal complement of
$(u,v)$ to itself. We can define a matrix representation of $(M,M^{T})\delta$
over this subspace. Since $(u,v)$ is a $\mathbb{R}\oplus\mathbb{R}$-vector, a
$(Cl_{1,0,1}(\mathbb{R}),*_{1})$-orthonormal basis $B$ for $(u,v)^{\perp}$ is
obtained simply as the $\mathbb{R}$-orthogonal complement of $v^{\prime}$. The
matrix representation of $(M,M^{T})\delta$ over $(v^{\prime})^{\perp}$ is
obtained using the basis $B$. The result is a self-adjoint
$(Cl_{1,0,1}(\mathbb{R}),*_{1})$-matrix $(M^{\prime},(M^{\prime})^{T})\delta$
with one less dimension than $(M,M^{T})\delta$. We repeat by finding another
eigenvector, restricting to the orthogonal complement of _that_ , etc.
Formally, this proof is by induction. ∎

######  Proposition 6.3 (skew-Takagi decomposition).

Given a complex skew-symmetric matrix $M$ (i.e. one which satisfies
$M=-M^{T}$, and not to be confused with a skew-Hermitian $\mathbb{C}$-matrix)
there exists a $\mathbb{C}$-unitary matrix $U$ and an $\mathbb{R}$-matrix
$D=D_{1}\oplus D_{2}\oplus\dotsb\oplus D_{k}$ such that $D_{i}$ is either of
the form $\begin{pmatrix}0&-\mu\\\ \mu&0\end{pmatrix}$ or $(0)$, and
$M=UDU^{T}$. (Note that $U^{T}\neq U^{-1}$).

###### Proof.

Observe that while $M$ is not Hermitian, we may introduce a new scalar
$\delta$ such that $\delta^{2}=0,\delta i=-i\delta$ and $\delta^{*}=-\delta$.
We see that $M\delta$ is now a $(Cl_{0,1,1}(\mathbb{R}),*_{-1})$-matrix, and
is indeed self-adjoint (over $(Cl_{0,1,1}(\mathbb{R}),*_{-1})$) because
$(M\delta)^{*}=\delta^{*}M^{*}=-\delta\overline{M}^{T}=-M^{T}\delta=M\delta$.
We can now hope to employ some analogue of the spectral theorem. It’s easily
seen that if for a matrix $K$ we have that $K\delta$ is unitarily similar to
$M\delta$, then there is a $U$ such that $UMU^{T}=K$. We intend to find a
canonical $K$.

We define
$\operatorname{unpack}(A+Bi+C\delta+Di\delta):=\begin{pmatrix}A+C\varepsilon&-B-D\varepsilon\\\
B-D\varepsilon&A+C\varepsilon\end{pmatrix}$ (where $A,B,C,D$ are arbitrary
$\mathbb{R}$-matrices of equal dimensions) and
$\operatorname{unwind}(a+bi+c\delta+di\delta):=\begin{pmatrix}a+c\epsilon\\\
b+d\varepsilon\end{pmatrix}$ (where $a,b,c,d$ are $\mathbb{R}$-vectors of
equal dimensions). We observe that
$\operatorname{unpack}(MK)=\operatorname{unpack}(M)\operatorname{unpack}(K)$
and
$\operatorname{unwind}(Mv)=\operatorname{unpack}(M)\operatorname{unwind}(v)$.

We have that $\operatorname{unpack}(M\delta)=\begin{pmatrix}\Re(M)&-\Im(M)\\\
-\Im(M)&\Re(M)\end{pmatrix}\varepsilon$. Ignoring the $\varepsilon$, we have a
skew-symmetric $\mathbb{R}$-matrix. By the skew-symmetric
$\mathbb{R}$-spectral theorem, we obtain a pair of $\mathbb{R}$-eigenvectors
$u$ and $v$ such that there exists a $\mu\in\mathbb{R}$ for which
$\operatorname{unpack}(M)u=-\mu v$ and $\operatorname{unpack}(M)v=\mu u$.
Unfortunately, $u$ and $v$ are not vectors over the same algebra as $M\delta$.
By the expression for $\operatorname{unwind}$ above, we have that there are
$\mathbb{C}$-vectors $u^{\prime}$ and $v^{\prime}$ such that
$\operatorname{unwind}(u^{\prime})=u$ and
$\operatorname{unwind}(v^{\prime})=v$. We see that $M\delta v^{\prime}=-\mu
v^{\prime}$ and $M\delta u^{\prime}=\mu v^{\prime}$.

Since $M\delta$ is self-adjoint, it maps the orthogonal complement of
$\operatorname{span}\\{u^{\prime},v^{\prime}\\}$ to itself. We can define a
matrix representation of $M\delta$ over this subspace. Since $u^{\prime}$ and
$v^{\prime}$ are $\mathbb{C}$-vectors, a
$(Cl_{0,1,1}(\mathbb{R}),*_{-1})$-orthonormal basis $B$ for
$\operatorname{span}\\{u^{\prime},v^{\prime}\\}^{\perp}$ is obtained simply
from the $\mathbb{C}$-orthogonal complement of
$\operatorname{span}\\{u^{\prime},v^{\prime}\\}$. The matrix representation of
$M\delta$ over $\operatorname{span}\\{u^{\prime},v^{\prime}\\}^{\perp}$ is
obtained using the basis $B$. The result is a self-adjoint
$(Cl_{0,1,1}(\mathbb{R}),*_{-1})$-matrix $M^{\prime}\delta$ with one less
dimension than $M\delta$. We repeat the above with $M^{\prime}\delta$.
Formally, this proof is by induction. ∎

######  Proposition 6.4 (quaternion skew-spectral theorem).

Given a skew-Hermitian quaternion matrix $M$ (i.e. one which satisfies
$M=-M^{*}$, where we use the standard quaternion involution) there exists a
$\mathbb{H}$-unitary matrix $U$ and an $\mathbb{R}$-matrix $D=D_{1}\oplus
D_{2}\oplus\dotsb\oplus D_{k}$ such that $D_{i}$ is either of the form
$\begin{pmatrix}0&-\mu\\\ \mu&0\end{pmatrix}$ or $(0)$, and $M=UDU^{*}$.

###### Proof.

We are not able to employ the same trick as for a skew-Hermitian matrix over
the complex numbers (with their usual involution). We employ unpack-and-unwind
instead.

We define $\operatorname{unpack}(A+Bi+Cj+Dk):=\begin{pmatrix}A+Bi&-C+Di\\\
C+Di&A-Bi\end{pmatrix}$ (where $A,B,C,D$ are arbitrary $\mathbb{R}$-matrices
of equal dimensions) and
$\operatorname{unwind}(a+bi+cj+dk):=\begin{pmatrix}a+bi\\\ c+di\end{pmatrix}$
(where $a,b,c,d$ are $\mathbb{R}$-vectors of equal dimensions). We observe
that
$\operatorname{unpack}(MK)=\operatorname{unpack}(M)\operatorname{unpack}(K)$
and
$\operatorname{unwind}(Mv)=\operatorname{unpack}(M)\operatorname{unwind}(v)$.

We have that $\operatorname{unpack}(M)$ is a skew-symmetric
$\mathbb{R}$-matrix. By the skew-symmetric $\mathbb{R}$-spectral theorem, we
obtain a pair of $\mathbb{R}$-eigenvectors $u$ and $v$ such that there exists
a $\mu\in\mathbb{R}$ for which $\operatorname{unpack}(M)u=-\mu v$ and
$\operatorname{unpack}(M)v=\mu u$. Unfortunately, $u$ and $v$ are not vectors
over the same algebra as $M$. By the expression for $\operatorname{unwind}$
above, we have that there are $\mathbb{C}$-vectors $u^{\prime}$ and
$v^{\prime}$ such that $\operatorname{unwind}(u^{\prime})=u$ and
$\operatorname{unwind}(v^{\prime})=v$. We see that $M\delta v^{\prime}=-\mu
v^{\prime}$ and $M\delta u^{\prime}=\mu v^{\prime}$. Since $M$ is a skew-
Hermitian, it maps the orthogonal complement of
$\operatorname{span}\\{u^{\prime},v^{\prime}\\}$ to itself. We can define a
matrix representation of $M$ over this subspace. We know that all modules over
the quaternions admit a basis, and this basis can always be orthonormalised.
We restrict $M$ to the orthogonal complement of
$\operatorname{span}\\{u^{\prime},v^{\prime}\\}$ by use of a basis, and obtain
a matrix $M^{\prime}$. Formally, this proof is by induction. ∎

There are many more examples of this trick, for instance involving some matrix
decompositions over the dual numbers, reducing them to the dual-number
spectral theorem [2].

##  7 Proving the spectral theorem for the “SVD $*$-algebra”
$(Cl_{1,0,1}(\mathbb{R}),*_{1})$

In this section, we prove a spectral theorem for
$(Cl_{1,0,1}(\mathbb{R}),*_{1})$.

First though, we must define the $\operatorname{unpack}$ and
$\operatorname{unwind}$ operations:

$\displaystyle\operatorname{unpack}\left((A,B)+(B^{\prime},A^{\prime})\delta\right)$
$\displaystyle=\begin{bmatrix}A&A^{\prime}\varepsilon\\\
B^{\prime}\varepsilon&B\end{bmatrix},$
$\displaystyle\operatorname{unwind}\left((u,v)+(v^{\prime},u^{\prime})\delta\right)$
$\displaystyle=\begin{bmatrix}u+\varepsilon u^{\prime}\\\
v+v^{\prime}\varepsilon\end{bmatrix}.$

We see that

$\displaystyle\operatorname{unpack}(XY)$
$\displaystyle=\operatorname{unpack}(X)\operatorname{unpack}(Y),$
$\displaystyle\operatorname{unpack}(X+Y)$
$\displaystyle=\operatorname{unpack}(X)+\operatorname{unpack}(Y),$
$\displaystyle\operatorname{unwind}(Xv)$
$\displaystyle=\operatorname{unpack}(X)\operatorname{unwind}(v),$
$\displaystyle\operatorname{unwind}(u+v)$
$\displaystyle=\operatorname{unwind}(u)+\operatorname{unwind}(v),$
$\displaystyle\operatorname{unpack}(X^{*})$
$\displaystyle=\operatorname{unpack}(X)^{*}.$

#### Some preliminaries on notation:

In the following arguments, we sometimes conflate a dual number
$a+b\varepsilon$ with a member of $(Cl_{1,0,1}(\mathbb{R}),*_{1})$ of the form
$a+b\delta$. We define $\operatorname{st}(a+b\varepsilon)=a$ (the “standard
part”) and $\operatorname{nst}(a+b\varepsilon)=b$ (the “non-standard part”).
Be aware that when we write $(a,b)$, we mean a member of
$(Cl_{1,0,1}(\mathbb{R}),*_{1})$, and not a row vector – we don’t presently
expect this to cause confusion. When we write $(M,K)$ (or some other capital
letters), we mean a matrix over $(Cl_{1,0,1}(\mathbb{R}),*_{1})$ of the form
$M(1,0)+K(0,1)$.

#### Some facts we’re assuming:

We make extensive use of the fact that every dual number matrix satisfying
$S=S^{T}$ also satisfies a form of the spectral theorem: We have that
$S=UDU^{T}$ is true for a _unique_ diagonal matrix $D$ over the dual numbers,
and some orthogonal matrix $U$ over the dual numbers (i.e. satisfying
$U^{T}=U^{-1}$).

######  Lemma 7.1.

If a self-adjoint $(Cl_{1,0,1}(\mathbb{R}),*_{1})$-matrix $H$ admits a pair of
vectors $w_{L}$ and $w_{R}$ such that:

  1. 1.

$w_{L}=w_{L}(1,0)$ and $w_{R}=w_{R}(0,1)$,

  2. 2.

$w_{L}^{*}w_{L}=(1,0)$ and $w_{R}^{*}w_{R}=(0,1)$,

  3. 3.

There exist real numbers $\lambda_{L}$ and $\lambda_{R}$ for which
$Hw_{L}=w_{L}\lambda_{L}$ and $Hw_{R}=w_{R}\lambda_{R}$,

then $w=w_{L}+w_{R}$ is a unit eigenvector of $H$ with eigenvalue
$(\lambda_{L},\lambda_{R})$.

###### Proof.

From 1, we may expand $w_{L}$ to $w_{L}=(u,0)+\delta(u^{\prime},0)$ and
$w_{R}$ to $w_{R}=(0,v)+\delta(0,v^{\prime})$. Then
$w_{L}+w_{R}=(u,v)+\delta(u^{\prime},v^{\prime})$. Let $w=w_{L}+w_{R}$. It’s
easy to see that $w$ is an eigenvector of $H$ of eigenvalue
$(\lambda_{L},\lambda_{R})$. It remains to consider $w^{*}w$.

We have that $w^{*}w=A+B\delta$ for some $A$ and $B$. We know that $A=1$
because $u^{*}u=v^{*}v=1$ (from item 2). We also get that $B$ is real because
$B\delta=w^{*}w-1=(w^{*}w-1)^{*}=(B\delta)^{*}$.

In fact, $B=0$, as we will now show. Let $\mu$ be the eigenvalue of $H$
corresponding to $w$. We have
$(1+B\delta)\mu=w^{*}w\mu=w^{*}Hw=(Hw)^{*}w=(w\mu)^{*}w=\mu^{*}w^{*}w=\mu(1+B\delta)$.
Since $\mu=(\lambda_{L},\lambda_{R})$ and $(1+B\delta)\mu=\mu(1+B\delta)$, we
get $B(\lambda_{L}-\lambda_{R})=0$. Since furthermore
$\lambda_{L}\neq\lambda_{R}$, we have that $B=0$. ∎

######  Lemma 7.2.

Let $\operatorname{unwind}(v)$ be an eigenvector of $\operatorname{unpack}(H)$
(for a self-adjoint $(Cl_{1,0,1}(\mathbb{R}),*_{1})$ matrix $H$) with
eigenvalue $\lambda$ that is not real. We have that the normalised vector
$w=v(v^{*}v)^{-1/2}$ satisfies $w^{*}w=1$ and $Hw=w\lambda$.

###### Proof.

Observe that $Hv=v\lambda$. From this, we get that
$H(v(1,-1))=(v(1,-1))\overline{\lambda}$. Expand $v$ to $v=(a,b)+\delta(c,d)$.
Expand $\lambda$ to $\lambda_{0}+\lambda^{\prime}\varepsilon$.

#### We seek to show that $a^{*}a=b^{*}b$:

We have that
$v^{*}v(1,-1)\overline{\lambda}=v^{*}Hv(1,-1)=(Hv)^{*}v(1,-1)=(v\lambda)^{*}v(1,-1)=\lambda
v^{*}v(1,-1)$. We therefore have that $v^{*}v(1,-1)\overline{\lambda}=\lambda
v^{*}v(1,-1)$. Substituting our expansion of $\lambda$ into this gives
$v^{*}v(1,-1)(\lambda_{0}-\lambda^{\prime}\delta)=(\lambda_{0}+\lambda^{\prime}\delta)v^{*}v(1,-1)$.
After some cancellation, this simplifies to $-v^{*}v(1,-1)\delta=\delta
v^{*}v(1,-1)$. Further substituting our expansion of $v$ and rearranging
$\delta$ gives $\delta(b^{*}b,-a^{*}a)=\delta(a^{*}a,-b^{*}b)$. We thus get
that $a^{*}a=b^{*}b$.

#### We seek to show that $a^{*}a=b^{*}b=1/2$:

We know from the previous paragraph that $a^{*}a=b^{*}b$. We also know that
$\operatorname{unwind}(v)=\begin{bmatrix}a+c\varepsilon\\\
b+d\varepsilon\end{bmatrix}$ is a unit vector by the
$(\mathbb{D},\operatorname{id})$ spectral theorem. So we conclude that
$a^{*}a=b^{*}b=1/2$.

#### We seek to show that $v^{*}v$ is in $\mathbb{D}$:

From the previous paragraph, we get that $v^{*}v=1/2+X\delta$ for some $X$.
Clearly, $(v^{*}v)^{*}=v^{*}v$, so we conclude $X$ is real.

#### We prove the main claim:

Now let $w=v(v^{*}v)^{-1/2}$. Clearly, $w^{*}w=1$, as we sought. We also
observe that
$Hw=Hv(v^{*}v)^{-1/2}=v\lambda(v^{*}v)^{-1/2}=v(v^{*}v)^{-1/2}\lambda$, with
the last equality holding because dual numbers commute. So $Hw=w\lambda$ as
desired. ∎

######  Lemma 7.3.

If all eigenvalues of $\operatorname{unpack}(H)$ (for self-adjoint
$(Cl_{1,0,1}(\mathbb{R}),*_{1})$ matrix $H$) are real, then there exists an
eigenvector $v$ of $H$ for which $v^{*}v=1$, with eigenvalue of the form
$(x,y)\in\mathbb{R}^{2}$.

###### Proof.

Let
$\\{\operatorname{unwind}(u_{1}),\operatorname{unwind}(u_{2}),\dotsc,\operatorname{unwind}(u_{2n})\\}$
be an orthonormal eigenbasis of $\operatorname{unpack}(H)$ with corresponding
real eigenvalues $\\{\lambda_{1},\lambda_{2},\dotsc,\lambda_{2n}\\}$.

Observe that from
$\operatorname{unpack}(H)\operatorname{unwind}(u_{i})=\operatorname{unwind}(u_{i})\lambda_{i}$,
we get $Hu_{i}=u_{i}\lambda_{i}$. Furthermore, we may scale $u_{i}$ by the
scalars $(1,0)$ and $(0,1)$ and the analogous identity remains true: More
explicitly, we have that $Hu_{i}(1,0)=u_{i}(1,0)\lambda_{i}$ and
$Hu_{i}(0,1)=u_{i}(0,1)\lambda_{i}$.

#### We seek to show that for each $i$, at least one of $u_{i}(1,0)$ or
$u_{i}(0,1)$ is not a multiple of $\delta$:

If they are both multiples of $\delta$, then $u_{i}$ is also a multiple of
$\delta$ because $u_{i}=u_{i}(1,0)+u_{i}(0,1)$. But then
$\operatorname{unwind}(u_{i})$ is a multiple of $\varepsilon$. This is clearly
impossible because $\operatorname{unwind}(u_{i})$ has unit length.

#### We seek to show that it is _not_ the case that $u_{i}(1,0)$ is a multiple
of $\delta$ for every $i$:

Assume otherwise. We have that the $\operatorname{unwind}$ of a multiple of
$\delta$ is a multiple of $\varepsilon$. We thus get that the following is a
multiple of $\varepsilon$:
$\operatorname{unwind}(u_{i}(1,0))=\begin{pmatrix}I_{n}&0_{n}\\\
0_{n}&0_{n}\end{pmatrix}\operatorname{unwind}(u_{i})$. This means that the
first half of the components of each vector $\operatorname{unwind}(u_{i})$ is
infinitesimal. But then
$\\{\operatorname{unwind}(u_{i}):i\in\\{1,\dotsc,2n\\}\\}$ cannot be linearly
independent.

#### We seek to show that it is _not_ the case that $u_{i}(0,1)$ is a multiple
of $\delta$ for every $i$:

Same argument as above.

#### We seek to show that there exists a unit eigenvector of $H$:

Pick a $u_{i}$ such that $u_{i}(1,0)$ is not a multiple of $\delta$. Either
there exists a $u_{j}$ such that $u_{j}(0,1)$ is not a multiple of $\delta$
_and_ $\lambda_{i}\neq\lambda_{j}$, or there doesn’t:

  * •

If there doesn’t, then we conclude that all eigenvalues of
$\operatorname{unpack}(H)$ are the same real number $\lambda$. But then
$\operatorname{unpack}(H)$ is a real multiple of the identity matrix.
Therefore so is $H$. Any vector is now an eigenvector of $H$, and so we are
done.

  * •

If there does, then pick this $i$ and $j$. Let $w=u_{i}(1,0)+u_{j}(0,1)$. We
see that $Hw=w(\lambda_{i},\lambda_{j})$. It remains to show that $w^{*}w=1$.
But this follows from lemma 7.1, so we are done. ∎

The following lemma is necessary to be able to take orthogonal complements. In
general, this can be quite complicated if we _only_ assume the conditions
$P^{2}=P$ and $P^{*}=P$. We therefore need to add the condition that $P$
should be displaced from a real orthogonal projection $Q$ by only a multiple
of $\delta$.

######  Lemma 7.4.

If $P^{2}=P$, $P^{*}=P$ and $P=Q+\delta(K,K^{T})$ for some real projection
matrix $Q$ and some real matrix $K$, then $P$ is unitarily diagonalisable with
eigenvalues either $0$ or $1$.

###### Proof.

The real matrix $Q$ can be unitarily diagonalised using a real matrix $U$. We
therefore have that
$UPU^{*}=\operatorname{diag}(1,\dotsc,1,0,\dotsc,0)+\delta(L,L^{T})$ for some
real matrix $L$. We write $UPU^{*}$ as a block matrix for clarity:
$UPU^{*}=\begin{pmatrix}I+\delta(L_{11},L_{11}^{T})&\delta(L_{12},L_{12}^{T})\\\
\delta(L_{12},L_{12}^{T})&\delta(L_{22},L_{22}^{T})\end{pmatrix}$. If we
square this block matrix and recall that $(UPU^{*})^{2}=UPU^{*}$, we get that
$L_{11}=L_{22}=0$. We can finally diagonalise $UPU^{*}$ using the block matrix
$V=\begin{pmatrix}I&\delta(L_{12},L_{12}^{T})\\\
-\delta(L_{12},L_{12}^{T})&I\end{pmatrix}$. ∎

######  Theorem 7.1.

Every self-adjoint $(Cl_{1,0,1}(\mathbb{R}),*_{1})$-matrix $H$ is unitarily
diagonalisable with each eigenvalue either a dual number or of the form
$(x,y)\in\mathbb{R}^{2}$.

###### Proof.

We prove this by induction on $k$ where $H$ is $k\times k$.

The claim is clearly true for $k=0$.

Assume that the claim is true for $k=n-1$. Now $\operatorname{unpack}(H)$ is a
self-adjoint $(\mathbb{D},\operatorname{id})$ matrix. Therefore, we may obtain
a unitarily diagonalisation of it $UDU^{*}=\operatorname{unpack}(H)$. This
gives us a unitary eigenbasis $\\{u_{1},u_{2},\dotsc,u_{2n}\\}$ of
$\operatorname{unpack}(H)$ with corresponding eigenvalues
$\\{\lambda_{1},\lambda_{2},\dotsc,\lambda_{2n}\\}$.

Either all the eigenvalues are real or they’re not.

Consider the case when some eigenvalue $\lambda_{i}$ is not real, but dual.
There is a $u^{\prime}_{i}$ such that
$\operatorname{unwind}(u^{\prime}_{i})=u_{i}$. Lemma 7.2 implies that for $w$
obtained by normalising $u^{\prime}_{i}$, we have that $Hw=w\lambda_{i}$ and
$w^{*}w=1$. We have that $\lambda$ in this case is a dual number. By lemma
7.4, we may take the orthogonal complement of $w$, restrict $H$ to
$w^{\perp}$, and apply the induction hypothesis. We are done.

Consider the case when all the eigenvalues are real: By lemma 7.3, we may find
a unit eigenvector $v$. $\lambda$ in this case is a pair
$(x,y)\in\mathbb{R}^{2}$. By lemma 7.4, we may take the orthogonal complement
of $w$. We restrict $H$ to $w^{\perp}$ and apply the induction hypothesis. We
are done. ∎

######  Definition 7.1.

We define the spectrum of a self-adjoint
$(Cl_{1,0,1}(\mathbb{R}),*_{1})$-matrix $H$ (which we will later show is
unique) by a triple of finite multisets $(C,L,R)$ where:

  * •

$C$ consists of dual numbers, while $L$ and $R$ consist of real numbers,

  * •

$(\forall x\in C)\operatorname{nst}(x)>0$,

  * •

$|L|=|R|$,

  * •

$H$ is unitarily similar to $c\oplus(l,r)$ where $c$, $l$ and $r$ are diagonal
matrices whose entries belong to $C$, $L$ and $R$ respectively.

######  Theorem 7.2.

Given a self-adjoint $(Cl_{1,0,1}(\mathbb{R}),*_{1})$ matrix $H$, any two
spectra $(C,L,R)$ and $(C^{\prime},L^{\prime},R^{\prime})$ of $H$ are equal.

###### Proof.

Consider two spectral decompositions of $H$:

  * •

One where the unitary eigenbasis is $\\{u_{1},u_{2},\dotsc,u_{n}\\}$ with the
corresponding spectrum being $(C,L,R)$. We say that the vectors in
$\\{u_{1},\dotsc,u_{k}\\}$ have eigenvalues in $C$.

  * •

One where the unitary eigenbasis is
$\\{u^{\prime}_{1},u^{\prime}_{2},\dotsc,u^{\prime}_{n}\\}$ with the
eigenvalues coming from the spectrum $(C^{\prime},L^{\prime},R^{\prime})$. We
say that the vectors in $\\{u^{\prime}_{1},\dotsc,u^{\prime}_{k^{\prime}}\\}$
have eigenvalues in $C^{\prime}$.

Observe that
$\\{u_{1},u_{1}(1,-1),u_{2},u_{2}(1,-1),\dotsc,u_{k},u_{k}(1,-1)\\}$ $\cup$
$\\{u_{k+1}(1,0),u_{k+1}(0,1)\dotsc,u_{n}(1,0),u_{n}(0,1)\\}$ form a spanning
set. Furthermore, applying $\operatorname{unwind}$ retains the spanning
property. We get that for each dual $\lambda_{i}$, we get that both
$\lambda_{i}$ and $\overline{\lambda_{i}}$ are eigenvalues of
$\operatorname{unpack}(H)$. Since the eigenvalues of
$\operatorname{unpack}(H)$ are unique, this establishes that $C=C^{\prime}$
and $L+R=L^{\prime}+R^{\prime}$. By projecting $\operatorname{st}(H)$ on its
two components, and applying the uniqueness of real eigenspectra, we get that
$L=L^{\prime}$ and $R=R^{\prime}$ . ∎

######  Corollary 7.1.

$\overline{\operatorname{Herm}}(Cl_{1,0,1}(\mathbb{R}),*_{1})$ is a subfree
abelian monoid, with the monoid operation being
$(C,L,R)+(C^{\prime},L^{\prime},R^{\prime})=(C+C^{\prime},L+L^{\prime},R+R^{\prime})$.
This is not free because $|L|=|R|$.

######  Corollary 7.2.

A self-adjoint $(Cl_{1,0,1}(\mathbb{R}),*_{-1})$-matrix is unitarily
diagonalisable with a unique spectrum.

###### Proof.

Proposition 4.1 says that $(Cl_{1,0,1}(\mathbb{R}),*_{-1})$ is isomorphic to
$(Cl_{1,0,1}(\mathbb{R}),*_{1})$, so this is trivial. ∎

##  8 Proving the spectral theorem for the “Takagi $*$-algebra”
$(Cl_{0,1,1}(\mathbb{R}),*_{1})$

I have discovered this result independently. It is also proved in a paper by
Qi et al.222It’s called “Low Rank Approximation of Dual Complex Matrices” and
remains unpublished. I have not added it to the bibliography because of an
ongoing dispute.

######  Theorem 8.1.

Every self-adjoint $(Cl_{0,1,1}(\mathbb{R}),*_{1})$-matrix $H$ is unitarily
diagonalisable.

###### Proof.

Let $M$ be a self-adjoint $(Cl_{0,1,1}(\mathbb{R}),*_{1})$-matrix. We find a
complex unitary matrix $S$ such that $S\operatorname{st}(A)S^{*}$ is diagonal.
We let $M^{\prime}=SMS^{*}$, which we write as a block matrix

$M^{\prime}=\begin{pmatrix}\lambda_{1}I+B_{11}\delta&B_{12}\delta&\dotsb&B_{1n}\delta\\\
B_{12}^{T}\delta&\lambda_{2}I+B_{22}\delta&\ddots&\vdots\\\
\vdots&\ddots&\ddots&B_{n-1,n}\delta\\\
B_{1n}^{T}\delta&\dotsb&B_{n-1,n}^{T}\delta&\lambda_{n}I+B_{nn}\delta\end{pmatrix}.$

where each $B_{ii}$ is complex symmetric. We let

$P=\begin{pmatrix}I&\frac{B_{12}\delta}{\lambda_{1}-\lambda_{2}}&\dotsb&\frac{B_{1n}\delta}{\lambda_{1}-\lambda_{n}}\\\
-\frac{B_{12}^{T}\delta}{\lambda_{1}-\lambda_{2}}&I&\ddots&\vdots\\\
\vdots&\ddots&\ddots&\frac{B_{n-1,n}\delta}{\lambda_{n-1}-\lambda_{n}}\\\
-\frac{B_{1n}^{T}\delta}{\lambda_{1}-\lambda_{n}}&\dotsb&-\frac{B_{n-1,n}^{T}\delta}{\lambda_{n-1}-\lambda_{n}}&I\end{pmatrix},$

and let $M^{\prime\prime}=PM^{\prime}P^{*}$. We end up with $M^{\prime\prime}$
being equal to a direct sum of matrices:
$M^{\prime\prime}=(\lambda_{1}I+B_{11}\delta)\oplus(\lambda_{2}I+B_{22}\delta)\oplus\dotsb\oplus(\lambda_{n}I+B_{nn}\delta)$.
We finally use the Takagi decomposition (whose existence we proved for a
general complex-symmetric matrix using the unpack-and-unwind method in
proposition 6.1) to find matrices $Q_{i}$ such that $Q_{i}B_{ii}Q_{i}^{T}$ is
equal to a real diagonal matrix. We thus get that $(Q_{1}\oplus
Q_{2}\oplus\dotsb\oplus Q_{n})M^{\prime\prime}(Q_{1}\oplus
Q_{2}\oplus\dotsb\oplus Q_{n})^{*}$ is a diagonal matrix. ∎

##  9 Proving the spectral theorem for the “skew-Takagi $*$-algebra”
$(Cl_{0,1,1}(\mathbb{R}),*_{-1})$

Presently, this result has been proved in a paper in Arxiv where I have been
promised coauthorship.333It’s called “Eigenvalues and Singular Value
Decomposition of Dual Complex Matrices”. I have not added the paper to the
bibliography because of an ongoing dispute.

######  Theorem 9.1.

Every self-adjoint $(Cl_{0,1,1}(\mathbb{R}),*_{-1})$-matrix $H$ is unitarily
similar to a direct sum of matrices of the form
$\begin{pmatrix}\lambda_{i}&-\lambda^{\prime}_{i}\delta\\\
\lambda^{\prime}_{i}\delta&\lambda_{i}\end{pmatrix}$ and $(\lambda_{i})$
(where $\lambda_{i}\in\mathbb{R}$ and $\lambda^{\prime}_{i}>0$).

###### Proof.

Let $M$ be a self-adjoint $(Cl_{0,1,1}(\mathbb{R}),*_{-1})$-matrix. We find a
complex unitary matrix $S$ such that $S\operatorname{st}(A)S^{*}$ is diagonal.
We let $M^{\prime}=SMS^{*}$, which we write as a block matrix

$M^{\prime}=\begin{pmatrix}\lambda_{1}I+B_{11}\delta&B_{12}\delta&\dotsb&B_{1n}\delta\\\
-B_{12}^{T}\delta&\lambda_{2}I+B_{22}\delta&\ddots&\vdots\\\
\vdots&\ddots&\ddots&B_{n-1,n}\delta\\\
-B_{1n}^{T}\delta&\dotsb&-B_{n-1,n}^{T}\delta&\lambda_{n}I+B_{nn}\delta\end{pmatrix}.$

where each $B_{ii}$ is complex skew-symmetric. We let

$P=\begin{pmatrix}I&\frac{B_{12}\delta}{\lambda_{1}-\lambda_{2}}&\dotsb&\frac{B_{1n}\delta}{\lambda_{1}-\lambda_{n}}\\\
\frac{B_{12}^{T}\delta}{\lambda_{1}-\lambda_{2}}&I&\ddots&\vdots\\\
\vdots&\ddots&\ddots&\frac{B_{n-1,n}\delta}{\lambda_{n-1}-\lambda_{n}}\\\
\frac{B_{1n}^{T}\delta}{\lambda_{1}-\lambda_{n}}&\dotsb&\frac{B_{n-1,n}^{T}\delta}{\lambda_{n-1}-\lambda_{n}}&I\end{pmatrix},$

and let $M^{\prime\prime}=PM^{\prime}P^{*}$. We end up with $M^{\prime\prime}$
being equal to a direct sum of matrices:
$M^{\prime\prime}=(\lambda_{1}I+B_{11}\delta)\oplus(\lambda_{2}I+B_{22}\delta)\oplus\dotsb\oplus(\lambda_{n}I+B_{nn}\delta)$.
We finally use the skew-Takagi decomposition (whose existence is proved in 6.3
using unpack-and-unwind) to find matrices $Q_{i}$ such that
$Q_{i}B_{ii}Q_{i}^{T}$ is equal to a direct sum of matrices of the form
$\begin{pmatrix}0&-\lambda^{\prime}_{i}\\\
\lambda^{\prime}_{i}&0\end{pmatrix}$ and $(0)$ (with the last type of block
only occurring once). We thus get that $(Q_{1}\oplus Q_{2}\oplus\dotsb\oplus
Q_{n})M^{\prime\prime}(Q_{1}\oplus Q_{2}\oplus\dotsb\oplus Q_{n})^{*}$ is of
the required form. ∎

##  10 Spectral theorems for $Cl_{p,q,0}(\mathbb{R})$ (for 1 involution) and
$Cl_{p,q,1}(\mathbb{R})$ (for 2 involutions)

The results here follow easily from those of the previous sections, and the
classification theorem for Clifford algebras over $\mathbb{R}$.

######  Definition 10.1.

By $(Cl_{p,q,0}(\mathbb{R}),*)$, we mean the $*$-algebra whose underlying
algebra is (of course) $Cl_{p,q,0}(\mathbb{R})$ and whose involution is
defined (uniquely) to satisfy $z^{*}=-z$ if $z^{2}=-1$ and $z^{*}=z$ if
$z^{2}=1$.

######  Definition 10.2.

By $(Cl_{p,q,1}(\mathbb{R}),*_{1})$, we mean the $*$-algebra whose underlying
algebra is (of course) $Cl_{p,q,1}(\mathbb{R})$ and whose involution is
defined the same as in $(Cl_{p,q,0}(\mathbb{R}),*)$, but with
$\delta^{*_{1}}=\delta$.

######  Definition 10.3.

By $(Cl_{p,q,1}(\mathbb{R}),*_{-1})$, we mean the $*$-algebra whose underlying
algebra is (of course) $Cl_{p,q,1}(\mathbb{R})$ and whose involution is
defined the same as in $(Cl_{p,q,0}(\mathbb{R}),*)$, but with
$\delta^{*_{-1}}=-\delta$.

The following has been proven elsewhere [4], and we recall it.

######  Lemma 10.1.

Every $(Cl_{p,q,0}(\mathbb{R}),*)$ is isomorphic to one of the following
$*$-algebras:

  1. 1.

$\mathbb{R}$ understood to have the trivial involution,

  2. 2.

$\mathbb{C}$ understood to be equipped with complex-conjugation as its
involution (which coincides with the transpose of its $\mathbb{R}$-matrix
representation),

  3. 3.

$\mathbb{H}$ understood to be equipped with its standard involution (which
coincides with the conjugate-transpose of its $\mathbb{R}$-matrix and
$\mathbb{C}$-matrix representations),

  4. 4.

$\mathbb{R}\oplus\mathbb{R}$ understood as a direct sum of $*$-algebras,

  5. 5.

$\mathbb{C}\oplus\mathbb{C}$ understood as a direct sum of $*$-algebras,

  6. 6.

$\mathbb{H}\oplus\mathbb{H}$ understood as a direct sum of $*$-algebras,

  7. 7.

$M_{2^{n}}(R)$ understood as having the conjugate-transpose as its involution,
with the “conjugate” (more correctly, involution) being inherited from the
$*$-algebra $R$,

for some values of $n$, and where $R$ is any of the $*$-algebras above. This
is a bit stronger than the usual statement of the classification theorem for
$Cl_{p,q,0}(\mathbb{R})$ in that it’s an isomorphism of $*$-algebras and not
just algebras.

We can now prove:

######  Proposition 10.1.

Every self-adjoint matrix $H$ over $(Cl_{p,q,0}(\mathbb{R}),*)$ admits a
$(Cl_{p,q,0}(\mathbb{R}),*)$-unitary matrix $U$ and a
$(Cl_{p,q,0}(\mathbb{R}),*)$-diagonal real matrix $D$ such that $UDU^{*}=H$.

###### Proof.

Using lemma 10.1, we prove our theorem for each of the seven cases within that
lemma in turn:

  1. 1.

First, $H$ has a real eigenvector $v$ with real eigenvalue. This follows
either from the Fundamental Theorem of Algebra or from Lagrange multipliers
(we won’t show the details here because they’re well covered elsewhere).
Finally, restrict $H$ to the orthogonal complement of $v$. $H$ over this
orthogonal complement is still self-adjoint, so the construction can be
repeated.

  2. 2.

We use the unpack-and-unwind method. Let
$\operatorname{unpack}_{\mathbb{C}}(A+Bi):=\begin{pmatrix}A&-B\\\
B&A\end{pmatrix}$ for $A$ and $B$ any $\mathbb{R}$-matrices. Observe that
$\operatorname{unpack}_{\mathbb{C}}(H)$ is self-adjoint over $\mathbb{R}$, and
hence has a real eigenvector $v$ of real eigenvalue $\lambda$. Let
$\operatorname{unwind}_{\mathbb{C}}(w):=\begin{pmatrix}\Re(w)\\\
\Im(w)\end{pmatrix}$. We have that $v=\operatorname{unwind}(v^{\prime})$ for
some $\mathbb{C}$-vector $v^{\prime}$. We thus have that
$\operatorname{unwind}(Hv^{\prime})=\operatorname{unpack}(H)\operatorname{unwind}(v^{\prime})=\lambda\operatorname{unwind}(v^{\prime})=\operatorname{unwind}(\lambda
v^{\prime})$. Since $\operatorname{unwind}$ is injective, we may cancel to get
$Hv^{\prime}=\lambda v^{\prime}$. We restrict $H$ to the orthogonal complement
of $v^{\prime}$, and repeat.

  3. 3.

We use the unpack-and-unwind method. Let
$\operatorname{unpack}_{\mathbb{H}}(A+Bi+Cj+Dk)=\operatorname{unpack}_{\mathbb{C}}\begin{pmatrix}A-Bi&-C+Di\\\
C+Di&A+Bi\end{pmatrix}$. Observe that $\operatorname{unpack}_{\mathbb{H}}(H)$
is self-adjoint over $\mathbb{R}$, and hence has a real eigenvector $v$ of
real eigenvalue $\lambda$. Let
$\operatorname{unwind}_{\mathbb{H}}(ai+bi+cj+dk):=\begin{pmatrix}a\\\ b\\\
c\\\ d\end{pmatrix}$, where $a,b,c,d$ are $\mathbb{R}$-vectors. We have that
$v=\operatorname{unwind}(v^{\prime})$ for some $\mathbb{H}$-vector
$v^{\prime}$. We thus have that
$\operatorname{unwind}(Hv^{\prime})=\operatorname{unpack}(H)\operatorname{unwind}(v^{\prime})=\lambda\operatorname{unwind}(v^{\prime})=\operatorname{unwind}(\lambda
v^{\prime})$. We restruct $H$ to the orthogonal complement of $v^{\prime}$,
and repeat.

  4. 4.

If $M$ is a self-adjoint over $\mathbb{R}\oplus\mathbb{R}$ then $M=(L,K)$
where $L$ and $K$ are self-adjoint matrices over $\mathbb{R}$. Then this
reduces to case 1.

  5. 5.

If $M$ is a self-adjoint over $\mathbb{C}\oplus\mathbb{C}$ then $M=(L,K)$
where $L$ and $K$ are self-adjoint matrices over $\mathbb{C}$. Then this
reduces to case 2.

  6. 6.

If $M$ is a self-adjoint over $\mathbb{H}\oplus\mathbb{H}$ then $M=(L,K)$
where $L$ and $K$ are self-adjoint matrices over $\mathbb{H}$. Then this
reduces to case 3.

  7. 7.

By block matrices, this reduces to the first six cases.∎

######  Proposition 10.2.

Every self-adjoint matrix $H$ over $(Cl_{p,q,1}(\mathbb{R}),*_{1})$ admits a
$(Cl_{p,q,1}(\mathbb{R}),*_{1})$-matrix $U$ and a diagonal
$(Cl_{p,q,1}(\mathbb{R}),*_{1})$-matrix $D$ such that $UDU^{*}=H$.

###### Proof.

We add an element $\delta$ to each of the seven cases in lemma 10.1. This
element $\delta$ satisfies:

  1. 1.

$\delta$ is in the centre.

  2. 2.

$\delta i=-i\delta$

  3. 3.

$\delta$ is in the centre.

  4. 4.

$\delta(x,y)=(y,x)\delta$.

  5. 5.

$\delta$ is in the centre.

  6. 6.

$\delta$ is in the centre.

  7. 7.

The algebra is $M_{2^{n}}(R\cup\\{\delta\\})$.

In each case, the involution is extended so that $\delta^{*}=\delta$. We now
prove the theorem for each of the seven cases in turn:

  1. 1.

This has been proven before in my paper on the dual number spectral theorem.
[2]

  2. 2.

See theorem 8.1.

  3. 3.

We use the unpack-and-unwind method. Let
$\operatorname{unpack}_{\mathbb{H}\otimes\mathbb{D}}(A+Bi+Cj+Dk)=\operatorname{unpack}_{\mathbb{C}\otimes\mathbb{D}}\begin{pmatrix}A-Bi&-C+Di\\\
C+Di&A+Bi\end{pmatrix}$ (where $A,B,C,D$ are $\mathbb{D}$-matrices). Observe
that $\operatorname{unpack}(H)$ is self-adjoint over $\mathbb{D}$, and hence
has a $\mathbb{D}$-eigenvector $v$ of $\mathbb{D}$-eigenvalue $\lambda$ (a
fact which follows from the dual-number spectral theorem [2]). Let
$\operatorname{unwind}_{\mathbb{H}\otimes\mathbb{D}}(ai+bi+cj+dk):=\begin{pmatrix}a\\\
b\\\ c\\\ d\end{pmatrix}$, where $a,b,c,d$ are $\mathbb{D}$-vectors. We have
that $v=\operatorname{unwind}(v^{\prime})$ for some
$\mathbb{H}\otimes\mathbb{D}$-vector $v^{\prime}$. We thus have that
$\operatorname{unwind}(Hv^{\prime})=\operatorname{unpack}(H)\operatorname{unwind}(v^{\prime})=\lambda\operatorname{unwind}(v^{\prime})=\operatorname{unwind}(\lambda
v^{\prime})$. We restruct $H$ to the orthogonal complement of $v^{\prime}$,
and repeat.

  4. 4.

See theorem 7.1.

  5. 5.

If $M$ is a self-adjoint over
$(\mathbb{C}\otimes\mathbb{D})\oplus(\mathbb{C}\otimes\mathbb{D})$ then
$M=(L,K)$ where $L$ and $K$ are self-adjoint matrices over
$\mathbb{C}\otimes\mathbb{D}$. Then this reduces to case 2.

  6. 6.

If $M$ is a self-adjoint over
$(\mathbb{H}\otimes\mathbb{D})\oplus(\mathbb{H}\otimes\mathbb{D})$ then
$M=(L,K)$ where $L$ and $K$ are self-adjoint matrices over
$\mathbb{H}\otimes\mathbb{D}$. Then this reduces to case 3.

  7. 7.

By block matrices, this reduces to the first six cases.∎

######  Lemma 10.2.

Every self-adjoint $(Cl_{1,1,1}(\mathbb{R}),*_{-1})$-matrix $H$ is unitarily
similar to a direct sum of matrices of the form
$\begin{pmatrix}\lambda_{i}&-\lambda^{\prime}_{i}\delta\\\
\lambda^{\prime}_{i}\delta&\lambda_{i}\end{pmatrix}$ and $(\lambda_{i})$
(where $\lambda_{i}\in\mathbb{R}$ and $\lambda^{\prime}_{i}>0$).

###### Proof.

Let $M$ be a self-adjoint $(Cl_{1,1,1}(\mathbb{R}),*_{-1})$-matrix. We find a
quaternion unitary matrix $S$ such that $S\operatorname{st}(A)S^{*}$ is
diagonal. We let $M^{\prime}=SMS^{*}$, which we write as a block matrix

$M^{\prime}=\begin{pmatrix}\lambda_{1}I+B_{11}\delta&B_{12}\delta&\dotsb&B_{1n}\delta\\\
-B_{12}^{T}\delta&\lambda_{2}I+B_{22}\delta&\ddots&\vdots\\\
\vdots&\ddots&\ddots&B_{n-1,n}\delta\\\
-B_{1n}^{T}\delta&\dotsb&-B_{n-1,n}^{T}\delta&\lambda_{n}I+B_{nn}\delta\end{pmatrix}.$

where each $B_{ii}$ is complex skew-Hermitian. We let

$P=\begin{pmatrix}I&\frac{B_{12}\delta}{\lambda_{1}-\lambda_{2}}&\dotsb&\frac{B_{1n}\delta}{\lambda_{1}-\lambda_{n}}\\\
\frac{B_{12}^{*}\delta}{\lambda_{1}-\lambda_{2}}&I&\ddots&\vdots\\\
\vdots&\ddots&\ddots&\frac{B_{n-1,n}\delta}{\lambda_{n-1}-\lambda_{n}}\\\
\frac{B_{1n}^{*}\delta}{\lambda_{1}-\lambda_{n}}&\dotsb&\frac{B_{n-1,n}^{*}\delta}{\lambda_{n-1}-\lambda_{n}}&I\end{pmatrix},$

and let $M^{\prime\prime}=PM^{\prime}P^{*}$. We end up with $M^{\prime\prime}$
being equal to a direct sum of matrices:
$M^{\prime\prime}=(\lambda_{1}I+B_{11}\delta)\oplus(\lambda_{2}I+B_{22}\delta)\oplus\dotsb\oplus(\lambda_{n}I+B_{nn}\delta)$.
We finally use proposition 6.4 to find matrices $Q_{i}$ such that
$Q_{i}B_{ii}Q_{i}^{*}$ is equal to a direct sum of matrices of the form
$\begin{pmatrix}0&-\lambda^{\prime}_{i}\\\
\lambda^{\prime}_{i}&0\end{pmatrix}$ and $(0)$. We thus get that $(Q_{1}\oplus
Q_{2}\oplus\dotsb\oplus Q_{n})M^{\prime\prime}(Q_{1}\oplus
Q_{2}\oplus\dotsb\oplus Q_{n})^{*}$ is of the required form. ∎

######  Proposition 10.3.

Every self-adjoint matrix $H$ over $(Cl_{p,q,1}(\mathbb{R}),*_{-1})$ admits a
unitary $(Cl_{p,q,1}(\mathbb{R}),*_{-1})$-matrix $U$ such that:

  * •

when $p=q=0$ or $q=1$ and $p=0$, there is a block-diagonal $D$ such
$UDU^{*}=H$ where $D$’s blocks are of the form
$\begin{pmatrix}\lambda_{i}&-\lambda^{\prime}_{i}\delta\\\
\lambda^{\prime}_{i}\delta&\lambda_{i}\end{pmatrix}$, $(\lambda_{i})$,
$\begin{pmatrix}\lambda^{\prime}_{i}\delta\end{pmatrix}$ or $(0)$, where
$\lambda_{i},\lambda^{\prime}_{i}$ are real and strictly positive.

  * •

when $p=1$ and $q=0$, there is a block-diagonal $D$ such $UDU^{*}=H$ where
$D$’s blocks are of the form
$\begin{pmatrix}\lambda_{i}&-\lambda^{\prime}_{i}\delta\\\
\lambda^{\prime}_{i}\delta&\lambda_{i}\end{pmatrix}$, $(\lambda_{i})$,
$\begin{pmatrix}\lambda^{\prime}_{i}\delta\end{pmatrix}$ or $(0)$, where the
$\lambda_{i}$ are in $\mathbb{R}\oplus\mathbb{R}$ and the
$\lambda^{\prime}_{i}$ are real and strictly positive.

  * •

when $p+q>1$, $H=UDU^{*}$ where $D$ is
$(Cl_{p,q,1}(\mathbb{R}),*_{-1})$-diagonal.

###### Proof.

We add an element $\delta$ to each of the seven cases in lemma 10.1. This
element $\delta$ satisfies:

  1. 1.

$\delta$ is in the centre.

  2. 2.

$\delta i=-i\delta$

  3. 3.

$\delta$ is in the centre.

  4. 4.

$\delta(x,y)=(y,x)\delta$.

  5. 5.

$\delta$ is in the centre.

  6. 6.

$\delta$ is in the centre.

  7. 7.

The algebra is $M_{2^{n}}(R\cup\\{\delta\\})$.

This exhausts all Clifford algebras of the form $Cl_{p,q,1}(\mathbb{R})$.
Using this, we prove our theorem for each of the seven cases in turn:

  1. 1.

This has been proven before in my paper on the dual number spectral theorem.
[2]

  2. 2.

See theorem 9.1.

  3. 3.

See lemma 10.2.

  4. 4.

See corollary 7.2.

  5. 5.

If $M$ is a self-adjoint over
$(\mathbb{C}\otimes\mathbb{D})\oplus(\mathbb{C}\otimes\mathbb{D})$ then
$M=(L,K)$ where $L$ and $K$ are self-adjoint matrices over
$\mathbb{C}\otimes\mathbb{D}$. Then this reduces to case 2.

  6. 6.

If $M$ is a self-adjoint over
$(\mathbb{H}\otimes\mathbb{D})\oplus(\mathbb{H}\otimes\mathbb{D})$ then
$M=(L,K)$ where $L$ and $K$ are self-adjoint matrices over
$\mathbb{H}\otimes\mathbb{D}$. Then this reduces to case 3.

  7. 7.

By block matrices, this reduces to the first six cases.∎

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