# Bounding the number of edges of matchstick graphs

Jérémy Lavollée  Department of Mathematics, The London School of Economics and
Political Science, Houghton Street, London WC2A 2AE  and  Konrad J. Swanepoel

###### Abstract.

We show that a matchstick graph with $n$ vertices has no more than
$3n-c\sqrt{n-1/4}$ edges, where
$c=\frac{1}{2}(\sqrt{12}+\sqrt{2\pi\sqrt{3}})$. The main tools in the proof
are the Euler formula, the isoperimetric inequality, and an upper bound for
the number of edges in terms of $n$ and the number of non-triangular faces. We
also find a sharp upper bound for the number of triangular faces in a
matchstick graph.

###### Key words and phrases:

matchstick graph, penny graph, plane unit-distance graph

###### 2020 Mathematics Subject Classification:

Primary 52C10. Secondary 05C10

##  1\. Introduction

Matchstick graphs, first introduced by Harborth in 1981 [6, 7], are graphs
drawn in the plane with each edge a straight-line segment of unit length, such
that no two edges have a point in common, unless the common point is an
endpoint of both edges (Figure 1).

Figure 1. A (disconnected) matchstick graph

Harborth posed various problems about matchstick graphs. The one that drew
most attention in the literature is that of finding $k$-regular matchstick
graphs with the smallest number of vertices. For example, it is known that
there are no $5$-regular matchstick graphs [2, 8], and the currently smallest
known $4$-regular matchstick graph, described in [7] and known as the Harborth
graph, has $52$ vertices and $104$ edges. Another of Harborth’s problems in
[7] is to find the maximum number of edges in a matchstick graph on $n$
vertices, for which he conjectured the following.

######  Conjecture 1 (Harborth [7]).

For each $n\geqslant 1$, the maximum number of edges in a matchstick graph on
$n$ vertices is $3n-\lceil\\!\sqrt{12n-3}\,\rceil$.

In [5] Harborth proved that this is indeed the maximum number of edges if we
furthermore assume that any two vertices are separated by a distance of at
least $1$ (the so-called penny graphs). This maximum is attained by an
appropriately chosen set of $n$ points on the triangular lattice (Figure 2).

Figure 2. A penny graph attaining equality in Conjecture 1

In [3, p. 225] this problem is again mentioned, where it is stated that it
“seems very likely that the maximum number of edges in a crossing-free unit-
distance graph is again $\lfloor 3n-\sqrt{12n-3}\rfloor$…” It is not hard to
show from the Euler formula and the isoperimetric inequality that the number
of edges is at most
$3n-\sqrt{\rule[-1.00006pt]{0.0pt}{10.00002pt}\smash{2\pi\sqrt{\smash[b]{3}}\cdot
n}}+O(1)$. Our main result is the following improvement.

######  Theorem 1.

In a matchstick graph with $n$ vertices, the number $e$ of edges satisfies

$\textstyle e\leqslant 3n-c\sqrt{n-1/4},$

where
$c=\frac{1}{2}\Bigl{(}\\!\sqrt{12}+\sqrt{\rule[-0.50003pt]{0.0pt}{8.99994pt}\smash{2\pi\sqrt{\smash[b]{3}}}}\Bigr{)}\approx
3.3815\dots$.

As a calculation shows, this bound turns out to be strong enough that it
settles Conjecture 1 for all $n\leqslant 14$, as well as for

$\displaystyle n$
$\displaystyle=16,18,19,20,21,23,24,26,27,29,30,32,33,36,37,39,40,$
$\displaystyle 43,44,47,48,51,52,55,56,60,61,65,69,70,74,75,79,80,85,$
$\displaystyle 90,91,96,102,108,114,120,127.$

The proof of Theorem 1 uses the isoperimetric inequality (Lemma 5 below), as
well as the following result which bounds the number of edges in terms of the
number of bounded non-triangular faces of the matchstick graph. Its proof is
based on Harborth’s induction proof [5] of Conjecture 1 for penny graphs.

######  Theorem 2.

In a matchstick graph with $n$ vertices, $e$ edges, and $g$ bounded non-
triangular faces, we have

$e\leqslant 3n-\sqrt{12n-3}+g.$

We do not assume that the graph is $2$-connected or even connected in Theorem
2. This is not just for the sake of generality, as this general statement is
needed when the induction hypothesis is applied in the proof.

Even though Theorem 2 does not seem to be sharp if there are bounded non-
triangular faces ($g>0$), when combined with the Euler formula, it gives the
following sharp upper bound for the number of triangular faces of a matchstick
graph.

######  Corollary 3.

In a matchstick graph with $n$ vertices, the number of bounded triangular
faces is at most $2n+1-\sqrt{12n-3}$.

Note that for each $n\geqslant 1$ there is a matchstick graph on the
triangular lattice with $n$ vertices and $2n+1-\lceil\\!\sqrt{12n-3}\,\rceil$
triangular faces (Figure 2).

The above results are proved in Section 3. In the next section, we establish
our terminology and introduce the fundamental tools we’ll need: the Euler
formula, a double-counting identity, and the isoperimetric inequality.

##  2\. Plane graphs and matchstick graphs

A _plane graph_ $G=(V,E)$ is defined to be a drawing of a graph in the plane
such that each vertex $v\in V$ is a different point in the plane, and each
edge $uv\in E$ is represented by a simple arc joining $u$ and $v$, in such a
way that two arcs only intersect in a common endpoint. The _faces_ of a plane
graph are the connected components of the complement of the plane graph in the
plane. One of the faces is unbounded. Throughout this paper, we will denote
the number of vertices by $n$, the number of edges by $e$, and the number of
bounded faces by $f$.

By the Euler formula, whenever $G$ is connected, we have $n-e+f=1$. If $G$ is
furthermore $2$-connected, then each face is bounded by a cycle. Denote the
number of vertices of the unbounded face by $b$, and the number of bounded
faces with exactly $i$ boundary vertices by $f_{i}$. We have the following
well-known relation.

######  Lemma 4.

For any $2$-connected plane graph with $n$ vertices, $e$ edges, $b$ boundary
vertices, and $f_{i}$ bounded faces with $i$ vertices, $i\geqslant 3$, we have
$e=3n-3-b-\sum_{i\geqslant 4}(i-3)f_{i}$.

###### Proof.

If we add up the number of vertices of each face, including the unbounded
face, we count each edge twice, thus obtaining $2e=b+\sum_{i\geqslant
3}if_{i}$. By Euler’s formula, $3e=3n+3f-3$. Subtracting these two identities,
we obtain the result. ∎

A _matchstick graph_ is a plane graph in which each edge is represented by a
straight-line segment of unit length. A matchstick graph is called a _penny
graph_ if the distance between any two vertices is at least $1$, and there is
an edge between all pairs of vertices at distance $1$.

In the proof of Theorem 1 we will need the following consequence of the
isoperimetric inequality that asserts that among all simple closed curves in
the plane of a fixed length, the circle is the unique curve that encloses the
largest area [1].

######  Lemma 5.

Let $G$ be a $2$-connected matchstick graph with $b$ vertices on the outer
boundary and $f_{3}$ bounded triangular faces. Then $b^{2}>\pi\sqrt{3}f_{3}$.

###### Proof.

The polygon bounding the unbounded face has $b$ edges and encloses $f_{3}$
equilateral triangles of unit side length. Each of these triangles has area
$\sqrt{3}/4$. Thus the polygon has area $A\geqslant\frac{\sqrt{3}}{4}f_{3}$
and perimeter $b$. By the isoperimetric inequality, any region of area $A$
bounded by a simple closed curve of length $b$ satisfies $b^{2}\geqslant 4\pi
A$, with equality only if the curve is a circle. It follows that $b^{2}>4\pi
A\geqslant\pi\sqrt{3}f_{3}$. ∎

##  3\. Proofs

We will repeatedly use the following inequality involving sums of square
roots.

######  Lemma 6.

Let $\alpha,\beta,\gamma,\delta$ be non-negative real numbers with
$\beta\leqslant\alpha\leqslant\gamma$ and $\alpha+\delta=\beta+\gamma$. Then

$\sqrt{\beta}+\sqrt{\gamma}\leqslant\sqrt{\alpha}+\sqrt{\delta},$

with equality iff $\alpha\in\\{\beta,\gamma\\}$.

###### Proof.

From

$0\leqslant(\alpha-\beta)(\gamma-\alpha)=\alpha\delta-\beta\gamma$

we obtain

$(\sqrt{\beta}+\sqrt{\gamma})^{2}=\beta+\gamma+2\sqrt{\beta\gamma}\leqslant\alpha+\delta+2\sqrt{\alpha\delta}=(\sqrt{\alpha}+\sqrt{\delta})^{2}.\qed$

######  Lemma 7.

If $n,n_{1},n_{2}$ are integers such that $n_{1},n_{2}\geqslant 3$,
$n_{1}+n_{2}=n+2$, then

$\lfloor 3n-\sqrt{12n-3}\rfloor+1\geqslant\lfloor
3n_{1}-\sqrt{12n_{1}-3}\rfloor+\lfloor 3n_{2}-\sqrt{12n_{2}-3}\rfloor.$

###### Proof.

The required inequality is equivalent to

$\sqrt{12n-3}+5\leqslant\lceil\\!\sqrt{12n_{1}-3}\,\rceil+\lceil\\!\sqrt{12n_{2}-3}\,\rceil.$
(1)

Without loss of generality, $n_{2}\geqslant n_{1}$. If $n_{1}\geqslant 6$,
then we set $\alpha=12n_{1}-3$, $\beta=69$, $\gamma=12n-51$,
$\delta=12n_{2}-3$ in Lemma 6 and obtain

$\sqrt{69}+\sqrt{12n-51}\leqslant\sqrt{12n_{1}-3}+\sqrt{12n_{2}-3}.$ (2)

Apply Lemma 6 again with $\alpha=12n-51$, $\beta=33$, $\gamma=12n-3$,
$\delta=81$ (since $n\geqslant 7$), we get

$\sqrt{12n-3}+\sqrt{33}\leqslant\sqrt{12n-51}+9.$ (3)

It follows from (2) and (3) that

$\sqrt{12n-3}+\sqrt{69}+\sqrt{33}-9\leqslant\sqrt{12n_{1}-3}+\sqrt{12n_{2}-3}.$

Finally, a calculation shows that $\sqrt{69}+\sqrt{33}-9>5$, and (1) follows.

For the remaining cases $n_{1}=3,4,5$, we need to round $\sqrt{12n_{1}-3}$ up
to obtain (1). For instance, if $n_{1}=3$, then we need to show
$\sqrt{12n-3}+5\leqslant 6+\sqrt{12n-15}$, which follows from Lemma 6 by
setting $\alpha=12n-15$, $\beta=33$, $\gamma=12n-3$, $\delta=45$ (since
$n\geqslant 4$), and noting that $\sqrt{45}-\sqrt{33}<1$. Similarly, when
$n_{1}=4$ then $n\geqslant 5$, and we obtain $\sqrt{12n-3}+5\leqslant
7+\sqrt{12n-27}$ by setting $\alpha=12n-27$, $\beta=25$, $\gamma=12n-3$,
$\delta=49$, and when $n_{1}=5$ then $n\geqslant 6$, and
$\sqrt{12n-3}+5\leqslant 8+\sqrt{12n-39}$ follows by setting $\alpha=12n-39$,
$\beta=25$, $\gamma=12n-3$, $\delta=61$. ∎

###### Proof of Theorem 2.

We use induction on the number of vertices $n\geqslant 1$. The theorem clearly
holds when $n=1$ or $n=2$. We now assume that $n\geqslant 3$ and that the
theorem holds for all smaller values of $n$ as induction hypothesis.

If the matchstick graph $G$ is not connected, let $G^{\prime}$ be a connected
component of $G$. If $G^{\prime}$ is in a bounded face of $G-G^{\prime}$, we
can move $G^{\prime}$ to the unbounded face of $G-G^{\prime}$. Note that this
does not change the number of non-triangular faces, unless $G^{\prime}$ was
originally inside a triangular face of $G-G^{\prime}$. However, then
$G^{\prime}$ cannot have any edges, and we are done by applying induction to
$G-G^{\prime}$. Thus we may assume that neither $G^{\prime}$ nor
$G-G^{\prime}$ lies in a bounded face of the other graph. Then it is easy to
move $G^{\prime}$ so that one of its vertices is at distance $1$ from a vertex
of $G-G^{\prime}$, while keeping $G^{\prime}$ and $G-G^{\prime}$ disjoint.
This creates a new edge that joins two connected components of $G$. This
process can be repeated until $G$ is connected, without decreasing the number
of edges.

We now assume without loss of generality that $G$ is connected. If $G$ is not
$2$-connected, then there is a vertex $v$ such that $G-v$ is disconnected
(Figure 3(a)). We can then decompose $G$ into two induced subgraphs $G_{1}$
and $G_{2}$ having only $v$ in common. If $G_{i}$ has $n_{i}$ vertices,
$e_{i}$ edges, and $g_{i}$ non-triangular faces ($i=1,2$), then
$n_{1},n_{2}\geqslant 2$, $n_{1}+n_{2}=n+1$, and $e_{1}+e_{2}=e$. It is clear
that $g_{1}+g_{2}=g$ if $G_{1}$ lies in the unbounded face of $G_{2}$ and
$G_{2}$ in the unbounded face of $G_{1}$. Suppose that $G_{1}$ (say) lies in a
bounded face of $G_{2}$. Then this face cannot be a triangle, as then $G_{1}$
would not have any edges, contradicting the connectedness of $G$. It follows
that $g_{1}+g_{2}=g$ in this case too. By induction,

$\displaystyle e=e_{1}+e_{2}$ $\displaystyle\leqslant
3n_{1}-\sqrt{12n_{1}-3}+g_{1}+3n_{2}-\sqrt{12n_{2}-3}+g_{2}$
$\displaystyle=3n+3-\sqrt{12n_{1}-3}-\sqrt{12n_{2}-3}+g$
$\displaystyle\leqslant 3n-\sqrt{12n-3}+g,$

by Lemma 6 with $\alpha=12n_{1}-3$, $\beta=9$, $\gamma=12n-3$,
$\delta=12n_{2}-3$.

For the remainder of the proof we assume without loss of generality that $G$
is $2$-connected. In particular, the boundary of the unbounded face is a
cycle.

$v$

(a) Cut vertex $v$

$u$$v$

(b) Chord $uv$

$u$$v$

(c) Face with non-neighbouring vertices $u$ and $v$ on the boundary

Figure 3. Three cases in the proof of Theorem 2

Suppose that the boundary cycle has a chord (Figure 3(b)). Let $G_{1}$ and
$G_{2}$ be two induced subgraphs covering $G$ such that the chord is their
only common edge and the endpoints of the chord their only two common
vertices. Using the same notation as before, with $G_{i}$ having $n_{i}$
vertices, $e_{i}$ edges, and $g_{i}$ non-triangular faces ($i=1,2$), we now
have $n_{1},n_{2}\geqslant 3$, $n_{1}+n_{2}=n+2$, $e_{1}+e_{2}=e+1$, and
$g_{1}+g_{2}=g$. Again we use induction on these two subgraphs to obtain

$\displaystyle e=e_{1}+e_{2}-1$ $\displaystyle\leqslant\lfloor
3n_{1}-\sqrt{12n_{1}-3}\rfloor+g_{1}+\lfloor
3n_{2}-\sqrt{12n_{2}-3}\rfloor+g_{2}-1$ $\displaystyle\leqslant
3n-\sqrt{12n-3}+g\quad\text{by Lemma~{}\ref{sqrt2}.}$

From now on we assume without loss of generality that the boundary cycle does
not have a chord.

We next show that if a bounded non-triangular face shares more than one vertex
with the unbounded face, then we can assume without loss of generality that
only two of its vertices are on the boundary cycle, they are adjacent, and at
most one of the two interior angles of the face at these two vertices is
smaller than $60^{\circ}$.

First suppose that a non-triangular bounded face has two non-adjacent vertices
on the boundary (Figure 3(c)). We decompose $G$ into two induced subgraphs
$G_{1}$ and $G_{2}$ such they only have these two vertices in common, with no
common edge. Again using the notation with $G_{i}$ having $n_{i}$ vertices,
$e_{i}$ edges, and $g_{i}$ non-triangular faces ($i=1,2$), we have
$n_{1},n_{2}\geqslant 3$, $n_{1}+n_{2}=n+2$, $e_{1}+e_{2}=e$, and
$g_{1}+g_{2}=g-1$. By induction,

$\displaystyle e=e_{1}+e_{2}$ $\displaystyle\leqslant\lfloor
3n_{1}-\sqrt{12n_{1}-3}\rfloor+g_{1}+\lfloor
3n_{2}-\sqrt{12n_{2}-3}\rfloor+g_{2}$ $\displaystyle\leqslant
3n-\sqrt{12n-3}+g\quad\text{by Lemma~{}\ref{sqrt2}.}$

From now on we assume without loss of generality that whenever a non-
triangular bounded face has more than one vertex on the boundary, it has only
two vertices on the boundary and they are adjacent. If both angles at these
boundary vertices are $<60^{\circ}$, then two of the edges of the face will be
forced to intersect. It follows that at most one angle of a non-triangular
face is less than $60^{\circ}$.

Let $g_{b}$ denote the number of non-triangular bounded faces that share a
vertex with the unbounded face. At each vertex of of the boundary cycle of
degree $i$ there are $i-1$ angles interior to bounded faces. If we denote the
number of boundary vertices of degree $i$ by $b_{i}$, then the total number of
boundary vertices is $b=\sum_{i\geqslant 2}b_{i}$, the number of angles is
$\sum_{i\geqslant 2}(i-1)b_{i}$, and the sum of these angles equals
$180^{\circ}(b-2)$. By the previous paragraph we have that at most $g_{b}$ of
these angles are smaller than $60^{\circ}$. Therefore,
$180^{\circ}(b-2)\geqslant 60^{\circ}(\sum_{i\geqslant 2}(i-1)b_{i}-g_{b})$,
hence

$\sum_{i\geqslant 2}(i-1)b_{i}-g_{b}\leqslant 3b-6.$ (4)

Since the boundary cycle has no chord, when we remove the $b$ vertices on the
boundary together with their incident edges, we remove exactly
$\sum_{i\geqslant 2}(i-1)b_{i}$ edges. We also remove $g_{b}$ non-triangular
faces and $b$ vertices. Without loss of generality, $n-b>0$, otherwise $G$ is
just the boundary cycle, hence $e=n\leqslant 3n-\sqrt{12n-3}+g$ since
$n\geqslant 3$. We use induction on the remaining graph of $n-b$ vertices to
obtain

$e-\sum_{i\geqslant 2}(i-1)b_{i}\leqslant 3(n-b)-\sqrt{12(n-b)-3}+g-g_{b},$

hence

$\displaystyle e$ $\displaystyle\leqslant
3(n-b)-\sqrt{12(n-b)-3}+g-g_{b}+\sum_{i\geqslant 2}(i-1)b_{i}$
$\displaystyle\leqslant 3(n-b)-\sqrt{12(n-b)-3}+g+3b-6\quad\text{by
\eqref{two}}$ $\displaystyle=3n-\sqrt{12(n-b)-3}+g-6,$

and in order to conclude that $e\leqslant 3n-\sqrt{12n-3}+g$, we need to show
that

$\sqrt{12n-3}\leqslant\sqrt{12(n-b)-3}+6.$ (5)

If $b\leqslant\sqrt{12n-3}-3$, then

$\sqrt{12(n-b)-3}\geqslant\sqrt{12\bigl{(}n-\sqrt{12n-3}+3\bigr{)}-3}=\sqrt{\bigl{(}\sqrt{12n-3}-6\bigr{)}^{2}},$

which gives (5), since $n>b\geqslant 3$, hence $n\geqslant 4$ and
$\sqrt{12n-3}-6>0$. Otherwise, $b>\sqrt{12n-3}-3$, and by Lemma 4, $e\leqslant
3n-b-3<3n-\sqrt{12n-3}$. ∎

######  Remark 8.

In the last step of the above proof, there is some slack, as we actually have
$e\leqslant 3n-b-3-g$ from Lemma 4. By taking this into account, it is
possible to prove the slightly stronger inequality $e\leqslant
3n-\sqrt{12(n+2g)-3}+g$. However, there are then more boundary cases to deal
with, and as this is not much of an improvement, we settled for the weaker
inequality in Theorem 2.

######  Remark 9.

When we removed the outer boundary cycle in the above proof, we needed this
cycle not to have a chord in order to count the number of edges that are
removed. This point is overlooked in Harborth’s original proof [5] on which
this proof is based.

###### Proof of Corollary 3.

By the Euler formula, $n-e+f_{3}+g=1$, and by Theorem 2, $e-g\leqslant
3n-\sqrt{12n-3}$. It follows that $f_{3}\leqslant 2n+1-\sqrt{12n-3}$. ∎

###### Proof of Theorem 1.

As in the proof of Theorem 2 we use induction on $n$. The theorem is easy to
verify for $n=1,2$, and as in the proof of Theorem 2, we can assume that $G$
is connected.

To show that we can furthermore assume that $G$ is $2$-connected, we also
proceed as in the proof of Theorem 2. If $G$ is not $2$-connected, we can
decompose $G$ into two induced subgraphs $G_{1}$ and $G_{2}$ having only a
single vertex in common. Let $G_{i}$ have $n_{i}$ vertices and $e_{i}$ edges
$(i=1,2)$. Then $n_{1},n_{2}\geqslant 2$, $n_{1}+n_{2}=n+1$ and
$e_{1}+e_{2}=e$. By induction and using the shorthand
$c=\frac{1}{2}\Bigl{(}\\!\sqrt{12}+\sqrt{\rule[-0.50003pt]{0.0pt}{8.99994pt}\smash{2\pi\sqrt{\smash[b]{3}}}}\Bigr{)}$
we have

$\displaystyle e=e_{1}+e_{2}$ $\displaystyle\leqslant
3n_{1}-c\sqrt{n_{1}-1/4}+3n_{2}-c\sqrt{n_{2}-1/4}$
$\displaystyle=3n+3-c\sqrt{n_{1}-1/4}-c\sqrt{n_{2}-1/4}.$

To conclude that $e\leqslant 3n-c\sqrt{n-1/4}$ we need to show that

$\frac{3}{c}+\sqrt{n-1/4}\leqslant\sqrt{n_{1}-1/4}+\sqrt{n_{2}-1/4}.$

Since $n_{i}\geqslant 2$, we can apply Lemma 6 with $\alpha=n_{1}-1/4$,
$\beta=3/2$, $\gamma=n-1$, $\delta=n_{2}-1/4$ to get

$\sqrt{3/2}+\sqrt{n-1}\leqslant\sqrt{n_{1}-1/4}+\sqrt{n_{2}-1/4}.$

Since $n\geqslant 3$, we can again apply Lemma 6 with $\alpha=n-1$,
$\beta=5/4$, $\gamma=n-1/4$, $\delta=2$, to obtain

$\sqrt{5}/2+\sqrt{n-1/4}\leqslant\sqrt{n-1}+\sqrt{2}.$

Combining these two equations together gives

$\sqrt{n-1/4}+\sqrt{5}/2-\sqrt{2}+\sqrt{3/2}\leqslant\sqrt{n_{1}-1/4}+\sqrt{n_{2}-1/4}$

which shows the required inequality since
$\sqrt{5}/2-\sqrt{2}+\sqrt{3/2}>3/c$.

We now assume that $G$ is 2-connected. Thus the unbounded face is bounded by a
cycle with $b$ edges. As before, denote the number of bounded faces with $i$
vertices by $f_{i}$ ($i\geqslant 3$), and the number of non-triangular faces
by $g$. By Lemma 4, noting that $g=\sum_{i\geqslant
4}f_{i}\leqslant\sum_{i\geqslant 4}(i-3)f_{i}$, we have

$e\leqslant 3n-3-b-g.$ (6)

Suppose for the sake of contradiction that

$e>3n-c\sqrt{n-1/4}.$ (7)

Then (6) and (7) give the following upper bound for $g$:

$\displaystyle g$ $\displaystyle<c\sqrt{n-1/4}-3-b.$ (8)

We obtain the following lower bound for $f_{3}$ from (7), (8), and the Euler
formula:

$\displaystyle f_{3}$ $\displaystyle=e-n-g+1$
$\displaystyle>3n-c\sqrt{n-1/4}-n-c\sqrt{n-1/4}+3+b+1$
$\displaystyle=2n-2c\sqrt{n-1/4}+4+b.$

Substitute this into the inequality $b^{2}>\pi\sqrt{3}f_{3}$ from Lemma 5 to
obtain

$\displaystyle b^{2}-\pi\sqrt{3}b$
$\displaystyle>2\pi\sqrt{3}\left(n-c\sqrt{n-1/4}+2\right).$

By completing the square,

$\displaystyle\left(b-\frac{\pi\sqrt{3}}{2}\right)^{2}$
$\displaystyle>\frac{3\pi^{2}}{4}+2\pi\sqrt{3}\left(n-c\sqrt{n-1/4}+2\right),$

we get the following lower bound for $b$:

$\displaystyle b$
$\displaystyle>\frac{\pi\sqrt{3}}{2}+\sqrt{\frac{3\pi^{2}}{4}+2\pi\sqrt{3}\left(n-c\sqrt{n-1/4}+2\right)}.$

We would like to deduce from this that
$b\geqslant\sqrt{2\pi\sqrt{3}\left(n-1/4\right)}-3$. It is sufficient to show
the following:

$\sqrt{\frac{3\pi^{2}}{4}+2\pi\sqrt{3}\left(n-c\sqrt{n-1/4}+2\right)}\geqslant\sqrt{2\pi\sqrt{3}\left(n-1/4\right)}-\frac{\pi\sqrt{3}}{2}-3.$
(9)

Since the left-hand side is non-negative, we can assume without loss of
generality that

$\sqrt{2\pi\sqrt{3}\left(n-1/4\right)}\geqslant\frac{\pi\sqrt{3}}{2}+3.$ (10)

Then we can square both sides of (9) and rearrange to obtain the equivalent

$\left(\pi\sqrt{3}+6-c\sqrt{\vphantom{\bigl{\\{}\bigr{\\}}}\smash[t]{2\pi\sqrt{3}}}\right)\sqrt{2\pi\sqrt{3}\left(n-1/4\right)}\geqslant
9-\frac{3\pi\sqrt{3}}{2}.$

This follows from (10), upon checking that
$\pi\sqrt{3}+6-c\sqrt{\vphantom{\bigl{\\{}\bigr{\\}}}\smash[t]{2\pi\sqrt{3}}}>0$
and

$\left(\\!\pi\sqrt{3}+6-c\sqrt{\vphantom{\bigl{\\{}\bigr{\\}}}\smash[t]{2\pi\sqrt{3}}}\,\right)\left(\frac{\pi\sqrt{3}}{2}+3\right)\geqslant
9-\frac{3\pi\sqrt{3}}{2}.$

So we have shown that
$b\geqslant\sqrt{\vphantom{\bigl{\\{}\bigr{\\}}}\smash[t]{2\pi\sqrt{3}\left(n-1/4\right)}}-3$,
which, together with (6) gives

$\displaystyle e$ $\displaystyle\leqslant
3n-\sqrt{\vphantom{\bigl{\\{}\bigr{\\}}}\smash[t]{2\pi\sqrt{3}\left(n-1/4\right)}}-g.$

By Theorem 2 we also have $e\leqslant 3n-\sqrt{12n-3}+g$. Adding these two
bounds, we obtain $e\leqslant 3n-c\sqrt{n-1/4}$, which contradicts the
assumption (7). Thus the assumption (7) is false and the theorem follows. ∎

######  Remark 10.

Eppstein [4] uses the isoperimetric inequality to find an upper bound of the
form $2n-c\sqrt{n}$ for the number of edges in a triangle-free penny graph on
$n$ vertices. To show an upper bound of this form for triangle-free matchstick
graphs will need a new idea, as there is no obvious way to bound the area of
the bounded faces from below.

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# Schauder Bases and Operator Theory II: strongly irreducible Schauder
Operators

Yang Cao  Yang Cao, Department of Mathematics , Jilin university, 130012,
Changchun, P.R.China<EMAIL_ADDRESS>,  Youqing Ji  Youqing Ji, Department
of Mathematics , Jilin university, 130012, Changchun, P.R.China
<EMAIL_ADDRESS>and  Geng Tian  Geng Tian, Department of Mathematics , Jilin
university, 130012, Changchun, P.R.China<EMAIL_ADDRESS>
(Date: Feb. 14, 2012)

###### Abstract.

In this paper, we will show that for an operator $T$ which is injective and
has dense range, there exists an invertible operator $X$ (in fact we can find
$U+K$, where $U$ is an unitary operator and $K$ is a compact operator with
norm less than a given positive real number) such that $XT$ is strongly
irreducible. As its application, strongly irreducible operators always exist
in the orbit of Schauder matrices.

###### Key words and phrases:

.

###### 2000 Mathematics Subject Classification:

Primary 47A55, 47A53, 47A16; Secondary 54H20, 37B99.

##  1\. Introduction and preliminaries

From the viewpoint of matrices, Schauder bases and operator theory have
natural relations. For example, the column vector sequence of the matrix of an
invertible operator comprise a Riesz basis under any orthonormal basis (ONB).
In our series paper, we focus on the operator which has a matrix
representation consisting of a Schauder basis under appropriate ONB (We shall
call them Schauder operators and Schauder matrices respectively).

In the matrix theory of finite dimensional space, the famous Jordan canonical
form theorem sufficiently reveals the internal structure of matrices. Works of
Jiang C. L., Ji Y. Q. etc. show that the strongly irreducible operators can be
seen as the generalized Jordan block in the case of separable infinite
dimensional Hilbert space. It will be shown that multiplying the matrix of any
invertible operator on the left of a Schauder matrix gets an equivalent
Schauder matrix (Theorem 3.3). Since Jordan blocks are the objects with
relatively simple properties, a natural question can be raised as follows:

######  Question 1.1.

Given any Schauder matrix $M$, does there exist a matrix $X$ of some
invertible operator such that $XM$ is strongly irreducible?

In this paper, we shall give an affirmative answer to this question. Following
theorem is our main result in operator theory which will be used to solve the
question.

######  Theorem 1.2.

Let $T\in\mathcal{L}(\mathcal{H})$ satisfy $KerT=(0)$ and
$\overline{RanT}=\mathcal{H}$. Then for any $\epsilon>0$, there exist an
unitary operator $U$ and a compact operator $K$ with $||K||<\epsilon$ such
that $(U+K)T$ is strongly irreducible. Moreover, since $\epsilon$ can be
chosen small enough, $U+K$ could be invertible.

######  Remark 1.3.

This theorem does not only motivated by basis theory observations. There exist
some other observations. We summarize them as follows:

1\. Physical backgrounds. In the Mathematics and Physics Interdisciplinary
Seminar at Jilin University, Professor Hai-Jun Wang of Physics Department told
us that multiplying an operator $X$ on the left of a Hamiltonian operator $T$
can be viewed as an evolution (or an evolving step) of a physics system which
is encoded by the Hamiltonian operator $T$. Moreover, the strong
irreducibility of an operator $T$ ensures that there is no subsystem
developing independently to the rest of the system. He also suggested us to
consider the Schauder basis as a sequence of "superposition states". From this
viewpoint, the special case of theorem 1.2 in which $T$ is a self-adjoint
operator may be more interesting. It could be useful in characterizing
evolution of observables.

2\. Compare to the classical results in the theory of strongly irreducible
operators. D.A.Herrero very concerned about compact perturbations of strongly
irreducible operators and asked that for any operator $T$ with connected
spectrum and any $\epsilon>0$, does there exist a compact operator $K$ with
$||K||<\epsilon$ such that $T+K$ is strongly irreducible? This question is the
essential strengthen of the question asked by G.H.Gong. Among many years,
C.L.Jiang, Y.Q.Ji, Z.Y.Wang, S.H.Sun and S.Power obtain the affirmative answer
step by step.

######  Theorem 1.4.

[14, 9, 10, 11, 13, 6, 8] Let $T\in\mathcal{L}(\mathcal{H})$, $\sigma(T)$
connected. Then for any $\epsilon>0$, there exists a compact operator
$K\in\mathcal{L}(\mathcal{H})$ with $||K||<\epsilon$ such that $T+K$ is
strongly irreducible.

Theorem 1.2 is a parallel consideration along the line of above theorem.
Theorem 1.2 focus on the action of multiplying an operator on the left while
theorem 1.4 focus on the action of adding an operator.

3\. Basis theory observations. Operator which is injective and has dense range
has a natural basis theory understanding, that is the main topic of our next
paper. We introduce the main result here:

######  Theorem 1.5.

Let $T\in\mathcal{L}(\mathcal{H})$, then $T$ is a Schauder operator if and
only if $T$ is injective and has dense range.

With this theorem in mind, question 1.1 naturally stimulate us to consider it
towards theorem 1.2.

We organize this paper as follows. The second section contains the proof of
our main theorem 1.2. In the last section we apply the main theorem into basis
theory. To generalize the class of bases which can be studied by operator
theory, we also introduce the blowing up matrix in that section.

###  1.1. Notations

Let $\mathcal{H}_{1},\mathcal{H}_{2},\mathcal{H}$ be complex separable Hilbert
spaces. Denote by $\mathcal{L}(\mathcal{H}_{1},\mathcal{H}_{2})$ the set of
all bounded linear operators mapping $\mathcal{H}_{1}$ into $\mathcal{H}_{2}$.
Denote by $\mathcal{K}(\mathcal{H}_{1},\mathcal{H}_{2})$ the subset of
$\mathcal{L}(\mathcal{H}_{1},\mathcal{H}_{2})$ of all compact operators. We
simply write $\mathcal{L}(\mathcal{H})$ and $\mathcal{K}(\mathcal{H})$ instead
of $\mathcal{L}(\mathcal{H},\mathcal{H})$ and
$\mathcal{K}(\mathcal{H},\mathcal{H})$ respectively. For
$T\in\mathcal{L}(\mathcal{H}_{1},\mathcal{H}_{2})$, denote the kernel of $T$
and the range of $T$ by Ker$T$ and Ran$T$ respectively. Let
$T\in\mathcal{L}(\mathcal{H})$, denote by $\sigma(T)$, $\sigma_{p}(T)$,
$\sigma_{e}(T)$, $\sigma_{lre}(T)$ and $\sigma_{W}(T)$ the spectrum, the point
spectrum, the essential spectrum, the Wolf spectrum and the Weyl spectrum of
$T$ respectively. Denote by $\sigma_{0}(T)$ the set of isolated points of
$\sigma(T)\backslash\sigma_{e}(T)$. For $\lambda\in\rho_{s-F}(T)$ $(\ext@arrow
0359\arrowfill@\Relbar\Relbar\Relbar{}{\mbox{def}}\mathbb{C}\backslash\sigma_{lre}(T))$,
${\rm ind}(\lambda-T)={\rm dimKer}(\lambda-T)-{\rm dimKer}(\lambda-T)^{*}$.
Denote
$\rho_{s-F}^{(n)}(T)=\\{\lambda\in\rho_{s-F}(T);~{}\text{ind}(\lambda-T)=n\\}$,
where $-\infty\leq n\leq\infty$. $T$ is said to be quasi-triangular if there
is a sequence $\\{P_{n}\\}_{n=1}^{\infty}$ of finite rank projections
increasing to the unit operator $I$ with respect to the strong operator
topology such that $\lim\limits_{n\rightarrow\infty}||(I-P_{n})TP_{n}||=0$. It
is well-known that $T$ is quasi-triangular if and only if ind$(T-\lambda)\geq
0$ for all $\lambda\in\rho_{s-F}(T)$. $T$ is said to be strongly irreducible
if there are no nontrivial idempotents commuting with $T$. A Cowen-Douglas
operator is an operator $T$ satisfying the following conditions:

(1) There is a nonempty connected open subset $\Omega$ of
$\rho_{s-F}^{(n)}(T)$ for a natural number n;

(2) $T-\lambda$ is surjective for each $\lambda\in\Omega$;

(3) $\bigvee_{\lambda\in\Omega}{\rm ker}(T-\lambda)=\mathcal{H}$.

If the conditions above are satisfied, we shall write
$T\in\mathcal{B}_{n}(\Omega)$.

##  2\. Main Results

### 2.1.

First, let us introduce some known results. Denote by $r(T)$ the spectral
radius of $T$. It is well-known that

$r(T)=\lim\limits_{n\rightarrow\infty}||T^{n}||^{\frac{1}{n}}.$

A bilateral weighted shift $T\in\mathcal{L}(\mathcal{H})$ is an operator that
maps each vector in some orthonormal basis $\\{e_{n}\\}_{n\in\mathbb{Z}}$ into
a scalar multiple of the next vector, $Te_{n}=\omega_{n}e_{n+1}$, for all
$n\in\mathbb{Z}$.

######  Lemma 2.1.

[18] 1) If $T\in\mathcal{L}(\mathcal{H})$ is an invertible bilateral weighted
shift, then the spectrum of $T$ is the annulus
$\\{\lambda\in\mathbb{C};[r(T^{-1})]^{-1}\leq|\lambda|\leq r(T)\\}$.

2) If $T\in\mathcal{L}(\mathcal{H})$ is a bilateral weighted shift that is not
invertible, then the spectrum of $T$ is the disc
$\\{\lambda\in\mathbb{C};|\lambda|\leq r(T)\\}$.

######  Lemma 2.2.

[12] Let $T$ be a bilateral weighted shift operator with weight sequence
$\\{\omega_{n}\\}_{n\in\mathbb{Z}}$. Then $T$ is a Cowen-Douglas operator if
and only if there exists a $\lambda_{0}\in\rho_{F}(T)$ such that
ind$(T-\lambda_{0})=1$.

######  Remark 2.3.

In the above lemma, if there exists a $\lambda_{0}\in\rho_{F}(T)$ such that
ind$(T-\lambda_{0})=1$, then $T$ must belong to $\mathcal{B}_{1}(\Omega)$ for
some open connected subset of $\mathbb{C}$. One can see [12] for details.

######  Lemma 2.4.

[9] Any operator $T\in\mathcal{B}_{1}(\Omega)$ is strongly irreducible.

######  Lemma 2.5.

Let $T\in\mathcal{L}(\mathcal{H})$ be a bilateral weighted shift with weight
sequence $\\{\omega_{n}\\}_{n\in\mathbb{Z}}$, $\omega_{n}>0$ for
$n\in\mathbb{Z}$ and max$\\{\omega_{n};n\geq 0\\}<$ min$\\{\omega_{n};n<0\\}$.
Then $T\in\mathcal{B}_{1}(\Omega)$ for some open connected subset $\Omega$ of
$\mathbb{C}$ and hence strongly irreducible.

###### Proof.

Let $\mathcal{H}_{1}=\bigvee_{n\leq 0}\\{e_{n}\\}$,
$\mathcal{H}_{2}=\bigvee_{n\geq 1}\\{e_{n}\\}$, then

$T=\begin{bmatrix}\begin{bmatrix}\ddots\\\ \ddots&0\\\
&\omega_{-1}&0\end{bmatrix}&0\\\ \begin{bmatrix}\cdots&0&\omega_{0}\\\ &0&0\\\
{\mathinner{\mkern 2.0mu\raise 1.0pt\hbox{.}\mkern 2.0mu\raise
4.0pt\hbox{.}\mkern 2.0mu\raise 7.0pt\hbox{.}\mkern 1.0mu}}&&\vdots\\\
\end{bmatrix}&\begin{bmatrix}0\\\ \omega_{1}&0\\\
&\ddots&\ddots\end{bmatrix}\end{bmatrix}\begin{matrix}\vdots\\\ e_{-1}\\\
e_{0}\\\ e_{1}\\\ e_{2}\\\ \vdots\end{matrix}\ext@arrow
0359\arrowfill@\Relbar\Relbar\Relbar{}{\mbox{def}}\begin{bmatrix}A\\\
C&B\end{bmatrix}\begin{matrix}\mathcal{H}_{1}\\\ \mathcal{H}_{2}\end{matrix}.$

Choose any $\lambda$ such that max$\\{\omega_{n};n\geq 0\\}<\lambda<$
min$\\{\omega_{n};n<0\\}$. We will show that $T-\lambda$ is a Fredholm
operator with index 1.

On the one hand, ${\rm inf}_{x\neq 0}\frac{||A^{*}x||}{||x||}\geq$
min$\\{\omega_{n};n<0\\}>\lambda$,
$||(A^{*}-\lambda)(x)||\geq||A^{*}x||-\lambda||x||\geq({\rm
min}\\{\omega_{n};n<0\\}-\lambda)||x||$ for any $x\in\mathcal{H}_{1}$, hence
$(A-\lambda)^{*}$ is bounded from below and $A-\lambda$ is right invertible.
Let $X_{11}\in\mathcal{L}(\mathcal{H}_{1})$ satisfy $(A-\lambda)X_{11}=I$.

On the other hand, $r(B)=\lim\limits_{k\rightarrow\infty}({\rm sup}_{n\geq
0}\\{\omega_{n+1}\omega_{n+2}\cdots\omega_{n+k}\\})^{\frac{1}{k}}<\lambda$,
hence $B-\lambda$ is invertible.

Let

$X=\begin{bmatrix}X_{11}&0\\\
-(B-\lambda)^{-1}CX_{11}&(B-\lambda)^{-1}\end{bmatrix}\begin{matrix}\mathcal{H}_{1}\\\
\mathcal{H}_{2}\end{matrix},$

then $(T-\lambda)X=I$. It is easy to compute that Ker$(T-\lambda)=\\{\alpha
x;\alpha\in\mathbb{C}\\}$, where

$\displaystyle
x=\sum_{n\in\mathbb{Z}}x_{n}e_{n},~{}x_{0}=1,~{}x_{n}=\dfrac{\omega_{0}\omega_{1}\cdots\omega_{n-1}}{\lambda^{n}},~{}x_{-n}=\dfrac{\lambda^{n}}{\omega_{-1}\omega_{-2}\cdots\omega_{-n}},~{}\forall~{}n>0.$

Hence dimKer$(T-\lambda)=1$ and $T-\lambda$ is a Fredholm operator with index
1. From lemma 2.2, remark 2.3, lemma 2.4, we know that
$T\in\mathcal{B}_{1}(\Omega)$ for some open connected subset $\Omega$ of
$\mathbb{C}$ and strongly irreducible. ∎

######  Lemma 2.6.

[4] Let $A\in\mathcal{L}(\mathcal{H}_{1}),B\in\mathcal{L}(\mathcal{H}_{2})$,
denote by $\mathcal{T}_{A,B}$ the Rosenblum operator, then

(1) the following three statements equivalent:

i) Ran$\mathcal{T}_{A,B}=\mathcal{L}(\mathcal{H}_{2},\mathcal{H}_{1})$;

ii) $\sigma_{r}(A)\cap\sigma_{l}(B)=\emptyset$, where $\sigma_{r}(A)$ and
$\sigma_{l}(B)$ are right spectrum and left spectrum respectively;

iii) $\mathcal{K}(\mathcal{H}_{2},\mathcal{H}_{1})\subseteq$
Ran$\mathcal{T}_{A,B}$.

(2) If $\sigma_{l}(A)\cap\sigma_{r}(B)=\emptyset$, then
Ker$\mathcal{T}_{A,B}=\\{0\\}$.

######  Lemma 2.7.

[15] Let $0<l_{j}\leq\infty,j=1,2,$
$\mathcal{H}=\oplus_{j=1}^{2}(\oplus_{0\leq i<l_{j}}\mathcal{H}_{i}^{j})$,
then

$T=\begin{bmatrix}A&Q\\\ &B\end{bmatrix}\begin{matrix}\oplus_{0\leq
i<l_{1}}\mathcal{H}_{i}^{1}\\\ \oplus_{0\leq
i<l_{2}}\mathcal{H}_{i}^{2}\end{matrix}\in\mathcal{L}(\mathcal{H})$

is a strongly irreducible operator when the following conditions satisfy:

i) $A=\oplus_{0\leq i<l_{1}}A_{i},B=\oplus_{0\leq
i<l_{2}}B_{i},Q=(Q_{ij})_{ij}$, where
$A_{i}\in\mathcal{L}(\mathcal{H}_{i}^{1}),B_{i}\in\mathcal{L}(\mathcal{H}_{i}^{2})$,
$Q_{ij}\in\mathcal{L}(\mathcal{H}_{j}^{2},\mathcal{H}_{i}^{1})$;

ii) $A_{i},B_{j}$ are all strongly irreducible operators;

iii) Ker$\mathcal{T}_{B_{j},A_{i}}=\\{0\\}$ for all $i,j$,
Ker$\mathcal{T}_{A_{i},A_{j}}=\\{0\\}$, Ker$\mathcal{T}_{B_{i},B_{j}}=\\{0\\}$
for all $i,j,i\neq j$;

iv) If $\sigma_{r}(A_{i})\cap\sigma_{l}(B_{j})\neq\emptyset$, then
$Q_{ij}\overline{\in}Ran\mathcal{T}_{A_{i},B_{j}}$;

v) For any $0\leq m,n<l_{1}$, there exist $k_{0}<\infty$,
$\\{i_{k}\\}_{k=1}^{k_{0}+1}$, $\\{j_{k}\\}_{k=1}^{k_{0}}$ such that
$\sigma_{r}(A_{i_{k}})\cap\sigma_{l}(B_{j_{k}})\neq\emptyset$,
$\sigma_{r}(A_{i_{k+1}})\cap\sigma_{l}(B_{j_{k}})\neq\emptyset$, $i_{1}=m$,
$i_{k_{0}+1}=n$;

vi) For any $0\leq m,n<l_{2}$, there exist $k_{1}<\infty$,
$\\{i_{k}\\}_{k=1}^{k_{1}}$, $\\{j_{k}\\}_{k=1}^{k_{1}+1}$ such that
$\sigma_{r}(A_{i_{k}})\cap\sigma_{l}(B_{j_{k}})\neq\emptyset$,
$\sigma_{r}(A_{i_{k}})\cap\sigma_{l}(B_{j_{k+1}})\neq\emptyset$, $j_{1}=m$,
$j_{k_{1}+1}=n$.

######  Lemma 2.8.

[7, 6] Let $T\in\mathcal{L}(\mathcal{H})$ be quasi-triangular,
$\sigma(T),\sigma_{W}(T)$ are all connected and
$\lambda_{0}\in\partial\sigma_{W}(T)$. Then for any given $\epsilon>0$, there
exists a compact operator $K\in\mathcal{L}(\mathcal{H})$, $||K||<\epsilon$
such that

i) $T+K$ is strongly irreducible;

ii) If $\lambda_{0}\overline{\in}\sigma_{p}(B)$, then
Ker$\mathcal{T}_{B,T+K}=\\{0\\}$.

###  2.2. Proof of Main Theorem

###### Proof.

From the polar decomposition theorem, we have $T=V|T|.$ Since $KerT=(0)$ and
$\overline{RanT}=\mathcal{H}$, $V$ is an unitarily operator. If the theorem is
correct for positive operator $|T|$, i.e. there exist unitarily operator $U$
and compact operator $K$, $||K||<\epsilon$ such that $(U+K)|T|$ is strongly
irreducible, then $(UV^{*}+KV^{*})T=(U+K)|T|$ is strongly irreducible and the
norm of compact operator $KV^{*}$ is less than $\epsilon$. Hence it suffices
to prove the theorem when T is a positive operator. We will break it into four
cases.

Case 1. Let $T$ be an invertible operator. Since $\sigma_{e}(T)\neq\emptyset$,
we denote
$\lambda_{min}=min\\{\lambda;\lambda\in\sigma_{e}(T)\\},\lambda_{min}=min\\{\lambda;\lambda\in\sigma_{e}(T)\\}$.
From the Weyl-Von Neumann theorem, for $\epsilon>0$, there exists a compact
operator $K_{1}$, $||K_{1}||<\frac{\epsilon}{2||T^{-1}||}$ such that,

1) $T+K_{1}$ is a diagonal operator with entries
$\\{\lambda_{1},\lambda_{2},\ldots\\}$ under an ONB
$\\{e_{n}\\}_{n=1}^{\infty}$;

2) there exist two subsequences $\\{\alpha_{n}\\}_{n=1}^{\infty}$ and
$\\{\beta_{n}\\}_{n=1}^{\infty}$ of $\\{\lambda_{n}\\}_{n=1}^{\infty}$ which
satisfy

$\displaystyle\lim\limits_{n\rightarrow\infty}\alpha_{n}=\lambda_{min},~{}\alpha_{n}<\alpha_{n+1},$
$\displaystyle\lim\limits_{n\rightarrow\infty}\beta_{n}=\lambda_{max},~{}\beta_{n}>\beta_{n+1};$

3)
$\\{\lambda_{1},\lambda_{2},\ldots\\}\backslash\\{\\{\alpha_{1},\alpha_{2},\ldots\\}\bigcup\\{\beta_{1},\beta_{2},\ldots\\}\\}\subseteq[\lambda_{min},\lambda_{max}]$.

Denote the elements of set
$\\{\lambda_{1},\lambda_{2},\ldots\\}\backslash\\{\\{\alpha_{1},\alpha_{2},\ldots\\}\bigcup\\{\beta_{1},\beta_{2},\ldots\\}\\}$
by $\\{\gamma_{n}\\}_{n=1}^{\infty}$. Then we have

$T+K_{1}=\begin{bmatrix}\begin{bmatrix}\ddots&&&&&\\\ &\beta_{2}&\\\
&&\beta_{1}&\\\ &&&\alpha_{1}&\\\ &&&&\alpha_{2}&\\\ &&&&&\ddots\\\
\end{bmatrix}&\\\ &\begin{bmatrix}\gamma_{1}\\\ &\gamma_{2}\\\ &&\ddots\\\
\end{bmatrix}\ext@arrow 0359\arrowfill@\Relbar\Relbar\Relbar{}{\mbox{def}}D\\\
\end{bmatrix}\begin{matrix}\vdots\\\ f_{4}\\\ f_{2}\\\ f_{1}\\\ f_{5}\\\
\vdots\\\ f_{3}\\\ f_{7}\\\ \vdots\\\ \end{matrix},$

where $\\{f_{n}\\}_{n=1}^{\infty}$ is a rearrangement of ONB
$\\{e_{n}\\}_{n=1}^{\infty}$.

Let

$U=\begin{bmatrix}\ddots\\\ \ddots&0&\\\ &1&0&\\\ &&1&0&\\\ &&&1&0&\\\
&&&&\ddots&\ddots&\\\ \end{bmatrix}\begin{matrix}\vdots\\\ f_{4}\\\ f_{2}\\\
f_{1}\\\ f_{5}\\\ \vdots\\\ \end{matrix}~{}{\rm
and}~{}U_{1}=\begin{bmatrix}U&\\\ &I\\\
\end{bmatrix}\begin{matrix}\mathcal{H}_{1}\\\ \mathcal{H}_{2}\end{matrix},$

where $\mathcal{H}_{1}=\bigvee\\{f_{2n+2},f_{4n+1},n=0,1,2,\ldots\\}$,
$\mathcal{H}_{2}=\bigvee\\{f_{4n+3},n=0,1,2,\ldots\\}$, then we have

$U_{1}(T+K_{1})=\begin{bmatrix}\begin{bmatrix}\ddots\\\ \ddots&0\\\
&\beta_{2}&0&\\\ &&\beta_{1}&0\\\ &&&\alpha_{1}&0\\\ &&&&\ddots&\ddots\\\
\end{bmatrix}\ext@arrow
0359\arrowfill@\Relbar\Relbar\Relbar{}{\mbox{def}}B&\\\ &D\\\
\end{bmatrix}\begin{matrix}\vdots\\\ f_{4}\\\ f_{2}\\\ f_{1}\\\ f_{5}\\\
\vdots\\\ \mathcal{H}_{2}\end{matrix}.$

It is easy to compute that

$\displaystyle
r(B)=\lim\limits_{n\rightarrow\infty}(\beta_{1}\beta_{2}\cdots\beta_{n})^{\frac{1}{n}}=\lim\limits_{n\rightarrow\infty}\beta_{n}=\lambda_{max},$
$\displaystyle
r(B^{-1})^{-1}=\lim\limits_{n\rightarrow\infty}((\alpha_{1}^{-1}\alpha_{2}^{-1}\cdots\alpha_{n}^{-1})^{\frac{1}{n}})^{-1}=\lim\limits_{n\rightarrow\infty}\alpha_{n}=\lambda_{min}.$

From lemma 2.1, the spectrum of $B$ is the annulus
$\\{\lambda\in\mathbb{C};\lambda_{min}\leq|\lambda|\leq\lambda_{max}\\}$.
Moreover, the diagonal operator $D$ is contained in this annulus. Hence the
spectrum $\sigma(U_{1}(T+K_{1}))$ is connected. It follows from the theorem
1.4 that there exists a compact operator $K_{2}$,
$||K_{2}||<\frac{\epsilon}{2||T^{-1}||}$ such that $U_{1}(T+K_{1})+K_{2}$ is
strongly irreducible. Notice that
$\\{U_{1}+(U_{1}K_{1}T^{-1}+K_{2}T^{-1})\\}T=U_{1}(T+K_{1})+K_{2}$ and
$U_{1}K_{1}T^{-1}+K_{2}T^{-1}$ is a compact operator with
$||U_{1}K_{1}T^{-1}+K_{2}T^{-1}||<\epsilon$, we complete the proof of this
case.

Case 2. Let $T$ be an operator with $\sigma_{e}(T)=\\{0\\}$. Since $KerT=(0)$
and $\overline{RanT}=\mathcal{H}$, $T$ must have the matrix representation as
follows,

$T=\begin{bmatrix}\lambda_{0}&\\\ &\lambda_{1}\\\ &&\lambda_{-1}\\\
&&&\lambda_{2}\\\ &&&&\lambda_{-2}\\\ &&&&&\ddots\\\
\end{bmatrix}\begin{matrix}e_{0}\\\ e_{1}\\\ e_{2}\\\ e_{3}\\\ e_{4}\\\
\vdots\end{matrix},$

where $\lambda_{n}\neq 0$,
$\lim\limits_{n\rightarrow+\infty}\lambda_{n}=\lim\limits_{n\rightarrow-\infty}\lambda_{n}=0.$
As case 1, we can find an unitarily operator $U$ such that

$UT=\begin{bmatrix}\ddots\\\ \ddots&0\\\ &\lambda_{-1}&0\\\ &&\lambda_{0}&0\\\
&&&\lambda_{1}&0\\\ &&&&\ddots&\ddots\\\ \end{bmatrix}\begin{matrix}\vdots\\\
e_{2}\\\ e_{0}\\\ e_{1}\\\ e_{3}\\\ \vdots\end{matrix}.$

We will show that $UT$ is strongly irreducible by directly computation. Let

$P=\begin{bmatrix}\ddots&\vdots&\vdots&\vdots&\vdots&{\mathinner{\mkern
2.0mu\raise 1.0pt\hbox{.}\mkern 2.0mu\raise 4.0pt\hbox{.}\mkern 2.0mu\raise
7.0pt\hbox{.}\mkern 1.0mu}}\\\
\cdots&a_{-1,-1}&a_{-1,0}&a_{-1,1}&a_{-1,2}&\cdots\\\
\cdots&a_{0,-1}&a_{0,0}&a_{0,1}&a_{0,2}&\cdots\\\
\cdots&a_{1,-1}&a_{1,0}&a_{1,1}&a_{1,2}&\cdots\\\
\cdots&a_{2,-1}&a_{2,0}&a_{2,1}&a_{2,2}&\cdots\\\ {\mathinner{\mkern
2.0mu\raise 1.0pt\hbox{.}\mkern 2.0mu\raise 4.0pt\hbox{.}\mkern 2.0mu\raise
7.0pt\hbox{.}\mkern 1.0mu}}\par&\vdots&\vdots&\vdots&\vdots&\ddots\\\
\end{bmatrix}\begin{matrix}\vdots\\\ e_{2}\\\ e_{0}\\\ e_{1}\\\ e_{3}\\\
\vdots\end{matrix}$

be an operator commuting with $UT$. Then we have

(2.1) $\displaystyle\lambda_{i}a_{i,j}=\lambda_{j}a_{i+1,j+1},~{}\forall
i,j\in\mathbb{Z}.$

First, $a_{i,j}=0$ when $i\neq j$. If $k>0$, then

(2.2) $\displaystyle
a_{i,i+k}=\dfrac{\lambda_{k-1}\lambda_{k-2}\cdots\lambda_{0}}{\lambda_{i+k-1}\lambda_{i+k-2}\cdots\lambda_{i}}a_{0,k},~{}\forall~{}i\geq
k,$ (2.3) $\displaystyle
a_{i,i+k}=\dfrac{\lambda_{i+k}\lambda_{i+k+1}\cdots\lambda_{k-1}}{\lambda_{i}\lambda_{i+1}\cdots\lambda_{-1}}a_{0,k},~{}\forall~{}i\leq-1.$

If $k<0$, then

(2.4) $\displaystyle
a_{i,i+k}=\dfrac{\lambda_{k+1}\lambda_{k+2}\cdots\lambda_{0}}{\lambda_{i+k-1}\lambda_{i+k}\cdots\lambda_{i-2}}a_{0,k},~{}\forall~{}i\leq
k+2,$ (2.5) $\displaystyle
a_{i,i+k}=\dfrac{\lambda_{i-1}\lambda_{i-2}\cdots\lambda_{0}}{\lambda_{i+k-1}\lambda_{i+k-2}\cdots\lambda_{k}}a_{0,k},~{}\forall~{}i\geq
1.$

From $(2.2),(2.4)$, we know that $a_{0,k}=0$ for all $k\in\mathbb{Z}$. Hence
$a_{i,j}=0$ when $i\neq j$.

Second, $a_{i,i}=a_{j,j}$ for all $i,j\in\mathbb{Z}$. It is easily seen from
$(2.1)$. In summary, $P=I$ or $P=0$.

Case 3. Let $T$ be an operator which satisfy that $0$ is an isolate point of
$\sigma_{e}(T)$ and $\sigma_{e}(T)\backslash\\{0\\}\neq\emptyset$.

From the spectral decomposition theorem of self-adjoint operators, we have

$T=\begin{bmatrix}T_{1}&\\\ &T_{2}\\\
\end{bmatrix}\begin{matrix}\mathcal{H}_{1}\\\ \mathcal{H}_{2}\end{matrix},$

where $\sigma_{e}(T_{1})=\\{0\\}$ and $0\overline{\in}\sigma(T_{2})$,
$\mathcal{H}_{1}\bot\mathcal{H}_{2}$ and
$\mathcal{H}_{1}\oplus\mathcal{H}_{2}=\mathcal{H}$.

Notice that dim$\mathcal{H}_{2}=\infty$.

As case 2, since $KerT=(0)$ and $\overline{RanT}=\mathcal{H}$, $T_{1}$ must be
a compact operator and

$T_{1}=\begin{bmatrix}\lambda_{0}\\\ &\lambda_{1}\\\ &&\lambda_{2}\\\
&&&\ddots\end{bmatrix}\begin{matrix}e_{0}\\\ e_{1}\\\ e_{2}\\\
\vdots\end{matrix},$

where $\\{e_{n}\\}_{n=0}^{\infty}$ is an ONB of $\mathcal{H}_{1}$,
$\lambda_{n}\neq 0$, $\lim\limits_{n\rightarrow\infty}\lambda_{n}=0.$

Since $T_{2}$ is an invertible self-adjoint operator, there exist a compact
operator $K$ with $||K||<\frac{\epsilon}{||T_{2}^{-1}||}$, such that $T_{2}+K$
is also invertible, and

$T_{2}+K=\begin{bmatrix}\eta_{1}\\\ &\eta_{2}\\\ &&\eta_{3}\\\
&&&\ddots\end{bmatrix}\begin{matrix}f_{1}\\\ f_{2}\\\ f_{3}\\\
\vdots\end{matrix},$

where $\\{f_{n}\\}_{n=1}^{\infty}$ is an ONB of $\mathcal{H}_{2}$.

Let

$K_{1}=\begin{bmatrix}0&\\\ &K\\\
\end{bmatrix}\begin{matrix}\mathcal{H}_{1}\\\ \mathcal{H}_{2}\end{matrix}.$

Choose $U$ as case 1 such that

$U(T+K_{1})=\begin{bmatrix}\ddots\\\ \ddots&0\\\ &\eta_{1}&0\\\
&&\lambda_{0}&0\\\ &&&\lambda_{1}&0&\\\ &&&&\ddots&\ddots\\\
\end{bmatrix}\begin{matrix}\vdots\\\ f_{1}\\\ e_{0}\\\ e_{1}\\\ e_{2}\\\
\vdots\\\ \end{matrix}.$

From lemma 2.5, we know that $U(T+K_{1})$ is strongly irreducible.

Let $K_{2}=UK_{1}T^{-1},$ then $||K_{2}||<\epsilon$ and
$(U+K_{2})T=U(T+K_{1})$ is strongly irreducible. So we complete the proof of
this case.

Case 4. This case is a little harder. Let $T$ be an operator which satisfy
that $0$ is not an isolate point of $\sigma_{e}(T)$, i.e. there exists a
sequence $\lambda_{n}$ such that
$\\{\lambda_{n}\\}_{n=1}^{\infty}\subseteq\sigma_{e}(T)$,
$\lambda_{n}>\lambda_{n+1}$ and
$\lim\limits_{n\rightarrow\infty}\lambda_{n}=0.$

From the spectral decomposition theorem of self-adjoint operators, we have

$T=\begin{bmatrix}T_{1}\\\ &T_{2}\\\ &&T_{3}\\\ &&&\ddots\\\
\end{bmatrix}\begin{matrix}\mathcal{H}_{1}\\\ \mathcal{H}_{2}\\\
\mathcal{H}_{3}\\\ \vdots\end{matrix},$

where $\\{\mathcal{H}_{j}\\}_{j\geq 1}$ is a pairwise orthogonal family of
subspaces and $\bigoplus_{j\geq 1}\mathcal{H}_{j}=\mathcal{H}$,
$\sigma(T_{j})\subseteq[\lambda_{j+1},\lambda_{j}]$,
$\lambda_{j+1},\lambda_{j}\in\sigma_{e}(T_{j})$ for all $j\geq 1$.

First step, we will find unitarily operators and compact operators $U_{j}$,
$K_{j}$, $||K_{j}||<\frac{\epsilon}{2j}$ such that each of
$(U_{j}+K_{j})T_{j}$ is strongly irreducible.

1) If there exists a point $\lambda_{j+1}<\alpha_{j}<\lambda_{j}$ such that
$\alpha_{j}-T_{j}$ is invertible, then we deal with it as follows. Let
$[\alpha_{j}-\delta_{j},\alpha_{j}+\delta_{j}]$ be the small interval
contained in $(\lambda_{j+1},\lambda_{j})$ such that
$[\alpha_{j}-\delta_{j},\alpha_{j}+\delta_{j}]\subseteq\rho(T_{j})$. From the
Weyl-Von Neumann theorem, there exists a compact operator $K_{j}$,
$||K_{j}||<\frac{\epsilon}{2j||T_{j}^{-1}||}$ such that

$T_{j}+K_{j}=\begin{bmatrix}\ddots\\\ &&\beta_{2}^{j}\\\ &&&\beta_{0}^{j}\\\
&&&&\beta_{1}^{j}\\\ &&&&&&\ddots\\\ \end{bmatrix}\begin{matrix}\vdots\\\
e_{2}^{j}\\\ e_{0}^{j}\\\ e_{1}^{j}\\\ \vdots\end{matrix},$

where
$\\{\beta_{2n}^{j}\\}\subseteq(\alpha_{j}+\delta_{j},\lambda_{j}],\\{\beta_{2n+1}^{j}\\}\subseteq[\lambda_{j+1},\alpha_{j}-\delta_{j})$
and $\bigvee_{i\geq 0}\\{e_{i}^{j}\\}=\mathcal{H}_{j}$.

Let $U_{j}\in\mathcal{L}(\mathcal{H}_{j})$ as case 1 such that

$U_{j}(T_{j}+K_{j})=\begin{bmatrix}\ddots&\\\ \ddots&0&\\\
&\beta_{2}^{j}&0&\\\ &&\beta_{0}^{j}&0\\\ &&&\beta_{1}^{j}&0\\\
&&&&\ddots&\ddots\\\ \end{bmatrix}\begin{matrix}\vdots\\\ e_{2}^{j}\\\
e_{0}^{j}\\\ e_{1}^{j}\\\ e_{3}^{j}\\\ \vdots\end{matrix}\ext@arrow
<EMAIL_ADDRESS>
Then from lemma 2.5, $B_{j}$ is a bilateral shift operator which is strongly
irreducible. Denote $\widetilde{K_{j}}=U_{j}K_{j}T_{j}^{-1}$, then
$||\widetilde{K_{j}}||<\frac{\epsilon}{2j}$ and
$(U_{j}+\widetilde{K_{j}})T_{j}=B_{j}$ is strongly irreducible.

2) If $[\lambda_{j+1},\lambda_{j}]\subseteq\sigma(T_{j})$, then we deal with
it as follows. Choose an interval
$[\eta_{j}^{2},\eta_{j}^{1}]\subseteq(\lambda_{j+1},\lambda_{j})$ arbitrarily.
Then

$T_{j}=\begin{bmatrix}T_{j1}&\\\ &T_{j2}\\\ &&T_{j3}\\\
\end{bmatrix}\begin{matrix}\mathcal{H}_{j}^{1}\\\ \mathcal{H}_{j}^{2}\\\
\mathcal{H}_{j}^{3}\end{matrix},$

where $\\{\mathcal{H}_{j}^{i}\\}_{i=1}^{3}$ is pairwise orthogonal family of
subspaces of $\mathcal{H}_{j}$ and
$\bigoplus_{i=1}^{3}\mathcal{H}_{j}^{i}=\mathcal{H}_{j}$,
$\sigma(T_{j1})=[\eta_{j}^{1},\lambda_{j}]$,
$\sigma(T_{j2})=[\lambda_{j+1},\eta_{j}^{2}]$,
$\sigma(T_{j3})=[\eta_{j}^{2},\eta_{j}^{1}]$. From the Weyl-Von Neumann
theorem, there exist compact operators
$C_{j1}\in\mathcal{L}(\mathcal{H}_{j}^{1})$,
$C_{j2}\in\mathcal{L}(\mathcal{H}_{j}^{2})$,
$||C_{j1}||<\frac{\epsilon}{4j||T_{j}^{-1}||},||C_{j2}||<\frac{\epsilon}{4j||T_{j}^{-1}||}$
such that, $T_{j1}+C_{j1}$ is a diagonal operator with entries
$\\{\gamma_{n}^{j}\\}_{n\leq 0}\subseteq[\eta_{j}^{1},\lambda_{j}]$,
$T_{j2}+C_{j2}$ is a diagonal operator with entries
$\\{\gamma_{n+1}^{j}\\}_{n\geq 0}\subseteq[\lambda_{j+1},\eta_{j}^{2}]$,
Card$\\{n;\gamma_{n+1}^{j}=\eta_{j}^{2}\\}=\infty$ and

(2.6)
$\displaystyle\dfrac{\gamma_{0}^{j}\gamma_{1}^{j}\cdots\gamma_{n-1}^{j}}{(\eta_{j}^{2})^{n}}\geq\dfrac{1}{\sqrt{n}},~{}\forall
n\geq 1.$

As case 1, denote
$V_{j}\in\mathcal{L}(\mathcal{H}_{j}^{1}\oplus\mathcal{H}_{j}^{2})$ be the
unitarily operator such that

$V_{j}\begin{bmatrix}T_{j1}+C_{j1}&\\\ &T_{j2}+C_{j2}\\\
\end{bmatrix}=\begin{bmatrix}\ddots&\\\ \ddots&0\\\ &\gamma_{-1}^{j}&0\\\
&&\gamma_{0}^{j}&0\\\ &&&\gamma_{1}^{j}&0\\\ &&&&\ddots&\ddots\\\
\end{bmatrix}\begin{matrix}\vdots\\\ e_{-1}^{j}\\\ e_{0}^{j}\\\ e_{1}^{j}\\\
e_{2}^{j}\\\ \vdots\end{matrix}\ext@arrow
0359\arrowfill@\Relbar\Relbar\Relbar{}{\mbox{def}}D_{j},$

where
$\bigvee_{i\in\mathbb{Z}}\\{e_{i}^{j}\\}=\mathcal{H}_{j}^{1}\oplus\mathcal{H}_{j}^{2}$.

From lemma 2.5, $D_{j}$ is strongly irreducible and moreover
$\eta_{j+1}\overline{\in}\sigma_{p}(D_{j})$. If $(D_{j}-\eta_{j}^{2})(x)=0$,
$x=\Sigma_{i\in\mathbb{Z}}x_{i}e_{i}^{j}\in\mathcal{H}_{j}^{1}\oplus\mathcal{H}_{j}^{2}$,
then for $n\geq 0$

(2.7) $\displaystyle
x_{n+1}=\dfrac{\gamma_{0}^{j}\gamma_{1}^{j}\cdots\gamma_{n}^{j}}{(\eta_{j}^{2})^{n+1}}x_{0},~{}x_{-n}=\dfrac{(\eta_{j}^{2})^{n}}{\gamma_{-1}^{j}\gamma_{-2}^{j}\cdots\gamma_{-n}^{j}}x_{0}.$

From $(2.6)$ and $(2.7)$, $x_{n}=0,~{}\forall n\in\mathbb{Z}$.

Since $T_{j3}$ is a quasi-triangular operator,
$\sigma(T_{j3})=\sigma_{W}(T_{j3})=[\eta_{j}^{2},\eta_{j}^{1}]$ and
$\eta_{j}^{2}\in\partial\sigma_{W}(T_{j3})$, from lemma 2.8, there exists a
compact operator $C_{j3}$, $||C_{j3}||<\frac{\epsilon}{8j||T_{j}^{-1}||}$ such
that $T_{j3}+C_{j3}$ is strongly irreducible and
Ker$(\mathcal{T}_{D_{j},T_{j3}+C_{j3}})=\\{0\\}$. Moreover, since
$\sigma_{r}(T_{j3}+C_{j3})\cap\sigma_{l}(D_{j})\neq\emptyset$, from lemma 2.6,
there exists a compact operator
$E_{j}:\mathcal{H}_{j}^{1}\oplus\mathcal{H}_{j}^{2}\longrightarrow\mathcal{H}_{j}^{3}$,
$||E_{j}||<\frac{\epsilon}{8j||T_{j}^{-1}||}$ such that
$E_{j}\overline{\in}Ran\mathcal{T}_{T_{j3}+C_{j3},D_{j}}$.

Hence from lemma 2.7, we obtain

$S_{j}\ext@arrow
0359\arrowfill@\Relbar\Relbar\Relbar{}{\mbox{def}}\begin{bmatrix}D_{j}&\\\
E_{j}&T_{j3}+C_{j3}\\\
\end{bmatrix}\begin{matrix}\mathcal{H}_{j}^{1}\oplus\mathcal{H}_{j}^{2}\\\
\mathcal{H}_{j}^{3}\end{matrix}$

is strongly irreducible.

Let

$W_{j}=\begin{bmatrix}V_{j}&\\\ &I\\\
\end{bmatrix}\begin{matrix}\mathcal{H}_{j}^{1}\oplus\mathcal{H}_{j}^{2}\\\
\mathcal{H}_{j}^{3}\end{matrix},$

and

$C_{j}=W_{j}\begin{bmatrix}C_{j1}\\\ &C_{j2}\\\ &&0\\\
\end{bmatrix}T_{j}^{-1}+\begin{bmatrix}0&0\\\ E_{j}&C_{j3}\\\
\end{bmatrix}T_{j}^{-1},$

then $C_{j}$ is a compact operator with $||C_{j}||<\frac{\epsilon}{2j}$ and
$(W_{j}+C_{j})T_{j}=S_{j}$ is strongly irreducible.

In summary, we obtain $(U_{j}+\widetilde{K_{j}})T_{j}=B_{j}$ or
$(W_{j}+C_{j})T_{j}=S_{j}$ for any chosen $T_{j}$. For the sake of brevity, we
denote $U_{j}$ or $W_{j}$ by $\overline{U}_{j}$, denote $\widetilde{K_{j}}$ or
$C_{j}$ by $\overline{K}_{j}$, denote $B_{j}$ or $S_{j}$ by
$\overline{B}_{j}$.

Let $U=\bigoplus_{j\geq 1}\overline{U}_{j}$, $K=\bigoplus_{j\geq
1}\overline{K}_{j}$, then $U$ is an unitarily operator, $K$ is a compact
operator with $||K||<\frac{\epsilon}{2}$ and

$(U+K)T=\begin{bmatrix}\overline{B}_{1}\\\ &\overline{B}_{2}\\\
&&\overline{B}_{3}\\\ &&&\ddots\\\
\end{bmatrix}\begin{matrix}\mathcal{H}_{1}\\\ \mathcal{H}_{2}\\\
\mathcal{H}_{3}\\\ \vdots\end{matrix}.$

Second step, we will use lemma 2.7 to complete the proof. Since
$\sigma_{r}(\overline{B}_{2n+1})\cap\sigma_{l}(\overline{B}_{2n})\neq\emptyset$,
$\sigma_{r}(\overline{B}_{2n+1})\cap\sigma_{l}(\overline{B}_{2n+2})\neq\emptyset$,
there exist compacts operator $E_{n}$, $F_{n}$,
$||E_{n}||<\frac{\epsilon}{4n||T_{2n}^{-1}||}$,
$||F_{n}||<\frac{\epsilon}{4n||T_{2n}^{-1}||}$ such that $E_{n}\overline{\in}$
Ran$\mathcal{T}_{\overline{B}_{2n-1},\overline{B}_{2n}}$ and
$F_{n}\overline{\in}$
Ran$\mathcal{T}_{\overline{B}_{2n+1},\overline{B}_{2n}}$.

Let

$\displaystyle\overline{K}$ $\displaystyle=$ $\displaystyle
K+\begin{bmatrix}0&\begin{bmatrix}E_{1}T_{2}^{-1}\\\
F_{1}T_{2}^{-1}&E_{2}T_{4}^{-1}\\\ &F_{2}T_{4}^{-1}&E_{3}T_{6}^{-1}\\\
&&\ddots&\ddots\\\ \end{bmatrix}\\\
&0\end{bmatrix}\begin{matrix}\mathcal{H}_{1}\\\ \mathcal{H}_{3}\\\
\mathcal{H}_{5}\\\ \vdots\\\ \bigoplus_{n\geq 1}\mathcal{H}_{2n}\\\
\end{matrix}$ $\displaystyle=$
$\displaystyle\begin{bmatrix}\begin{bmatrix}\overline{K}_{1}\\\
&\overline{K}_{3}\\\ &&\overline{K}_{5}\\\ &&&\ddots\\\
\end{bmatrix}&\begin{bmatrix}E_{1}T_{2}^{-1}\\\
F_{1}T_{2}^{-1}&E_{2}T_{4}^{-1}\\\ &F_{2}T_{4}^{-1}&E_{3}T_{6}^{-1}\\\
&&\ddots&\ddots\\\ \end{bmatrix}\\\ &\begin{bmatrix}\overline{K}_{2}\\\
&\overline{K}_{4}\\\ &&\overline{K}_{6}\\\ &&&\ddots\\\
\end{bmatrix}\end{bmatrix}\begin{matrix}\mathcal{H}_{1}\\\ \mathcal{H}_{3}\\\
\mathcal{H}_{5}\\\ \vdots\\\ \mathcal{H}_{2}\\\ \mathcal{H}_{4}\\\
\mathcal{H}_{6}\\\ \vdots\end{matrix},$

then $\overline{K}$ is a compact operator with $||\overline{K}||<\epsilon$ and

$\displaystyle(U+\overline{K})T$ $\displaystyle=$
$\displaystyle\begin{bmatrix}\begin{bmatrix}\begin{smallmatrix}\overline{U}_{1}+\overline{K}_{1}\\\
&\overline{U}_{3}+\overline{K}_{3}\\\ &&\overline{U}_{5}+\overline{K}_{5}\\\
&&&\ddots\\\
\end{smallmatrix}\end{bmatrix}&\begin{bmatrix}\begin{smallmatrix}E_{1}T_{2}^{-1}\\\
F_{1}T_{2}^{-1}&E_{2}T_{4}^{-1}\\\ &F_{2}T_{4}^{-1}&E_{3}T_{6}^{-1}\\\
&&\ddots&\ddots\\\ \end{smallmatrix}\end{bmatrix}\\\
&\begin{bmatrix}\begin{smallmatrix}\overline{U}_{2}+\overline{K}_{2}\\\
&\overline{U}_{4}+\overline{K}_{4}\\\ &&\overline{U}_{6}+\overline{K}_{6}\\\
&&&\ddots\\\
\end{smallmatrix}\end{bmatrix}\end{bmatrix}\begin{matrix}\begin{smallmatrix}\mathcal{H}_{1}\\\
\mathcal{H}_{3}\\\ \mathcal{H}_{5}\\\ \vdots\\\ \mathcal{H}_{2}\\\
\mathcal{H}_{4}\\\ \mathcal{H}_{6}\\\ \vdots\end{smallmatrix}\end{matrix}$
$\displaystyle\begin{array}[]{c}\times\end{array}$
$\displaystyle\begin{bmatrix}\begin{bmatrix}T_{1}\\\ &T_{3}\\\ &&T_{5}\\\
&&&\ddots\\\ \end{bmatrix}&0\\\ 0&\begin{bmatrix}T_{2}\\\ &T_{4}\\\ &&T_{6}\\\
&&&\ddots\\\ \end{bmatrix}\end{bmatrix}\begin{matrix}\mathcal{H}_{1}\\\
\mathcal{H}_{3}\\\ \mathcal{H}_{5}\\\ \vdots\\\ \mathcal{H}_{2}\\\
\mathcal{H}_{4}\\\ \mathcal{H}_{6}\\\ \vdots\end{matrix}$ $\displaystyle=$
$\displaystyle\begin{bmatrix}\begin{bmatrix}\overline{B}_{1}\\\
&\overline{B}_{3}\\\ &&\overline{B}_{5}\\\ &&&\ddots\\\
\end{bmatrix}&\begin{bmatrix}E_{1}\\\ F_{1}&E_{2}\\\ &F_{2}&E_{3}\\\
&&\ddots&\ddots\\\ \end{bmatrix}\\\ &\begin{bmatrix}\overline{B}_{2}\\\
&\overline{B}_{4}\\\ &&\overline{B}_{6}\\\ &&&\ddots\\\
\end{bmatrix}\end{bmatrix}\begin{matrix}\mathcal{H}_{1}\\\ \mathcal{H}_{3}\\\
\mathcal{H}_{5}\\\ \vdots\\\ \mathcal{H}_{2}\\\ \mathcal{H}_{4}\\\
\mathcal{H}_{6}\\\ \vdots\end{matrix}.$

In order to obtain the result, we only need to show the conditions $i)-vi)$ of
lemma 2.7.

From the construction above, conditions $i),ii),iv),v),vi)$ are satisfied.
Since $\sigma(\overline{B}_{i})\cap\sigma(\overline{B}_{j})=\emptyset$, when
$|i-j|\geq 2$, from lemma 2.6,
Ker$\mathcal{T}_{\overline{B}_{i},\overline{B}_{j}}=\\{0\\}$ for $i\neq j$,
$|i-j|\geq 2$.

For $j$ even, $|i-j|=1$, we should consider
Ker$\mathcal{T}_{\overline{B}_{j},\overline{B}_{i}}=\\{0\\}$.

Since each $\overline{B}_{j}$ maybe $B_{j}$ or $S_{j}$, we should consider
four cases for any Ker$\mathcal{T}_{\ast,\ast}$. But the proof are all the
same, we just consider one case that
Ker$\mathcal{T}_{\overline{B}_{j},\overline{B}_{j+1}}=\\{0\\}$ where

$\displaystyle\overline{B}_{j}=S_{j}=\begin{bmatrix}D_{j}&\\\
E_{j}&T_{j3}+C_{j3}\\\
\end{bmatrix}\begin{matrix}\mathcal{H}_{j}^{1}\oplus\mathcal{H}_{j}^{2}\\\
\mathcal{H}_{j}^{3}\end{matrix},$
$\displaystyle\overline{B}_{j+1}=S_{j+1}=\begin{bmatrix}D_{j+1}&\\\
E_{j+1}&T_{j+1,3}+C_{j+1,3}\\\
\end{bmatrix}\begin{matrix}\mathcal{H}_{j+1}^{1}\oplus\mathcal{H}_{j+1}^{2}\\\
\mathcal{H}_{j+1}^{3}\end{matrix}.$

Let

$X=\begin{bmatrix}X_{11}&X_{12}\\\
X_{21}&X_{22}\end{bmatrix}\begin{matrix}\mathcal{H}_{j+1}^{1}\oplus\mathcal{H}_{j+1}^{2}\\\
\oplus\\\
\mathcal{H}_{j+1}^{3}\end{matrix}\mapsto\begin{matrix}\mathcal{H}_{j}^{1}\oplus\mathcal{H}_{j}^{2}\\\
\oplus\\\ \mathcal{H}_{j}^{3}\end{matrix}$

belongs to Ker$\mathcal{T}_{\overline{B}_{j},\overline{B}_{j+1}}$. Then

(2.9) $\displaystyle D_{j}X_{11}=X_{11}D_{j+1}+X_{12}E_{j+1},$ (2.10)
$\displaystyle D_{j}X_{12}=X_{12}(T_{j+1,3}+C_{j+1,3}),$ (2.11) $\displaystyle
E_{j}X_{11}+(T_{j3}+C_{j3})X_{21}=X_{21}D_{j+1}+X_{22}E_{j+1},$ (2.12)
$\displaystyle E_{j}X_{12}+(T_{j3}+C_{j3})X_{22}=X_{22}(T_{j+1,3}+C_{j+1,3}).$

Since $\sigma(D_{j})\cap\sigma(T_{j+1,3}+C_{j+1,3})=\emptyset$, from lemma 2.6
and $(2.9)$, $X_{12}=0$. Since
$\sigma(T_{j3}+C_{j3})\cap\sigma(T_{j+1,3}+C_{j+1,3})=\emptyset$, from lemma
2.6, $X_{12}=0$ and $(2.11)$, $X_{22}=0$.

If $X_{11}=0$, then since
$\sigma(T_{j3}+C_{j3})\cap\sigma(D_{j+1})=\emptyset$, $X_{21}=0$, and hence
$X=0$.

Let

$X_{11}=\begin{bmatrix}\ddots&\vdots&\vdots&\vdots&\vdots&{\mathinner{\mkern
2.0mu\raise 1.0pt\hbox{.}\mkern 2.0mu\raise 4.0pt\hbox{.}\mkern 2.0mu\raise
7.0pt\hbox{.}\mkern 1.0mu}}\\\
\cdots&a_{-1,-1}&a_{-1,0}&a_{-1,1}&a_{-1,2}&\cdots\\\
\cdots&a_{0,-1}&a_{0,0}&a_{0,1}&a_{0,2}&\cdots\\\
\cdots&a_{1,-1}&a_{1,0}&a_{1,1}&a_{1,2}&\cdots\\\
\cdots&a_{2,-1}&a_{2,0}&a_{2,1}&a_{2,2}&\cdots\\\ {\mathinner{\mkern
2.0mu\raise 1.0pt\hbox{.}\mkern 2.0mu\raise 4.0pt\hbox{.}\mkern 2.0mu\raise
7.0pt\hbox{.}\mkern 1.0mu}}\par&\vdots&\vdots&\vdots&\vdots&\ddots\\\
\end{bmatrix}\begin{matrix}\vdots\\\ e_{-1}^{j+1}\\\ e_{0}^{j+1}\\\
e_{1}^{j+1}\\\ e_{2}^{j+1}\\\ \vdots\end{matrix}\mapsto\begin{matrix}\vdots\\\
e_{-1}^{j}\\\ e_{0}^{j}\\\ e_{1}^{j}\\\ e_{2}^{j}\\\ \vdots\end{matrix},$

then from $(2.8)$, we have

(2.13)
$\displaystyle\gamma_{m}^{j}a_{m,n}=\gamma_{n}^{j+1}a_{m+1,n+1},~{}\forall
m,n\in\mathbb{Z}.$

Hence

(2.14) $\displaystyle
a_{m+1,m+k+1}=\dfrac{\gamma_{0}^{j}\gamma_{1}^{j}\cdots\gamma_{m}^{j}}{\gamma_{k}^{j+1}\gamma_{k+1}^{j+1}\cdots\gamma_{m+k}^{j+1}}a_{0,k},~{}\forall
m\geq 0,~{}k\in\mathbb{Z},$ (2.15) $\displaystyle
a_{m,m+k}=\dfrac{\gamma_{m+k}^{j+1}\gamma_{m+k+1}^{j+1}\cdots\gamma_{k-1}^{j+1}}{\gamma_{m}^{j}\gamma_{m+1}^{j}\cdots\gamma_{-1}^{j}}a_{0,k},~{}\forall
m<0,~{}k\in\mathbb{Z}.$

Since

$\dfrac{\gamma_{n}^{j}}{\gamma_{m}^{j+1}}\geq\dfrac{\lambda_{j+1}}{\eta_{j+1}^{2}}>1,~{}\forall
m,n\geq 1,$
$\lim\limits_{m\rightarrow+\infty}\dfrac{\gamma_{0}^{j}\gamma_{1}^{j}\cdots\gamma_{m}^{j}}{\gamma_{k}^{j+1}\gamma_{k+1}^{j+1}\cdots\gamma_{m+k}^{j+1}}=\infty.$

Hence from $(2.13)$ and $(2.14)$, $a_{m,n}=0,~{}\forall m,n\in\mathbb{Z}$. ∎

######  Remark 2.9.

If $T$ is a positive operator which satisfies
$(0,\epsilon)\cap\sigma_{e}(T)=\emptyset$ for some $\epsilon>0$,
Ker$T=\\{0\\}$ and $\sigma_{e}(T)$ is not connected, then there exist an
unitary operator $U$ and a compact operator $K$ with norm less than a given
positive number such that $(U+K)T$ is a strongly irreducible Cowen-douglas
operator. If $T$ is a compact operator which is injective and has dense range,
then for any unitary operator $U$ and any compact operator $K$, $(U+K)T$ can
not be a Cowen-Douglas operator. It is an easy corollary of Case 1,2,3 of the
above proof.

##  3\. Application in Basis Theory

### 3.1.

In this subsection, we discuss existence of operators satisfying both certain
basis theory property and strongly irreducibility.

Recall that a sequence $\psi=\\{f_{n}\\}_{n=1}^{\infty}$ is called a Schauder
basis of the Hilbert space $\mathcal{H}$ if and only if for every vector
$x\in\mathcal{H}$ there exists a unique sequence
$\\{\alpha_{n}\\}_{n=1}^{\infty}$ of complex numbers such that the partial sum
sequence $x_{k}=\sum_{n=1}^{k}\alpha_{n}f_{n}$ converges to $x$ in norm.

######  Definition 3.1.

An $\omega\times\omega$ matrix $F$ is said to be a Schauder matrix if and only
if the sequence $\psi=\\{f_{n}\\}_{n=1}^{\infty}$ of its column vectors in
$\mathcal{H}$ comprises a Schauder basis.

Note that from the definition, each column vector $f_{k}$ is automatically a
$l^{2}-$sequence since it represents a vector in $\mathcal{H}$. Given an ONB
$\varphi=\\{e_{n}\\}_{n=1}^{\infty}$ and a basis sequence
$\psi=\\{f_{n}=\sum_{k=1}^{\infty}f_{kn}e_{k}\\}_{n=1}^{\infty}$ of
$\mathcal{H}$, then the matrix $F_{\psi}=(f_{kn})$ is a Schauder matrix by
above definition. It shall be called the Schauder matrix corresponding to the
basis $\psi$.

######  Definition 3.2.

A matrix $F$ is called a unconditional, conditional, Riesz, normalized or
quasinormal matrix respectively if and only if the sequence of its column
vectors comprise a unconditional, conditional, Riesz, normalized or
quasinormal basis. Two Schauder matrices $F_{\psi},F_{\varphi}$ are called
equivalent if and only if the corresponding bases $\psi$ and $\varphi$ are
equivalent.

######  Theorem 3.3.

Assume that $F$ is a Schauder matrix and $G^{*}$ is its inverse matrix. We
have
1\. For each invertible matrix $X$, $XF$ is also a Schauder matrix. Moreover,
$XF$ is unconditional(conditional) if and only if $F$ is
unconditional(conditional);
2\. For each diagonal matrix $D=diag(\alpha_{1},\alpha_{2},\cdots)$ in which
each diagonal element $\alpha_{k}$ is nonzero, $FD$ is also a Schauder matrix.
Moreover, $FD$ is unconditional(conditional) if and only if $F$ is
unconditional(conditional);
3\. For a unconditional matrix $F$, $FU$ is also a unconditional matrix for
$U\in\pi_{\infty}$;
4\. Two Schauder matrix $F$ and $F^{{}^{\prime}}$ are equivalent if and only
if there is a invertible matrix $X$ such that $XF=F^{{}^{\prime}}$.

For a Schauder matrix $M_{\psi}$ and a matrix $X$ of some invertible operator
$T$, $M_{\psi^{{}^{\prime}}}=XM_{\psi}$ is the Schauder matrix of the Schauder
basis $\psi^{{}^{\prime}}=\\{Tf_{n}\\}_{n=1}^{\infty}$. Given an ONB
$\varphi$, the orbit set

$O_{gl}(M_{\psi})=\\{XM_{\psi};\mbox{matrix $X$ represents an invertible
operator $T\in Gl(\mathcal{H})$}\\}$

consists of all equivalent bases of $\psi$. Our main theorem 1.2 and above
theorem 3.3 tell us that we always can pick a strongly irreducible operator as
the representation element in the orbit $O_{gl}(M_{\psi})$.

######  Theorem 3.4.

Suppose that $\psi=\\{f_{n}\\}_{n=1}^{\infty}$ is a basis sequence and its
corresponding Schauder matrix $M_{\psi}$ under some given ONB represents a
bounded operator $T_{\psi}$. Then there is a matrix $X$ of some invertible
operator such that $XM_{\psi}$ represents an strongly irreducible bounded
compact operator in $L(\mathcal{H})$.

######  Corollary 3.5.

For a Schauder matrix $M_{\psi}$ representing a bounded operator, there always
be matrices of strongly irreducible operators in its orbit $O_{gl}(M_{\psi})$.

By the pole decomposition theorem of operators, there is also a unitary matrix
$U$ such that $UM_{\psi}$ represents a self-adjoint operator. Hence the orbit
$O_{gl}(M_{\psi})$ has both the operators having a large number of strongly
reducible subspaces and the operators having no nontrivial strongly reducible
subspace.

###  3.2. The blowing up matrix

To observe the relations between basis theory and operator theory, is it
enough to just consider bounded operators? Let
$\psi=\\{f_{n}\\}_{n=1}^{\infty}$ be a Schauder basis and $M_{\psi}$ is the
corresponding Schauder matrix under some ONB
$\varphi=\\{e_{n}\\}_{n=1}^{\infty}$. In general, $M_{\psi}$ does not
represents a bounded operator even for a quasinormal basis $\psi$. Following
example show this phenomenon.

######  Example 3.6.

(see [19], Example14.5, p429.) Let $\\{\alpha_{n}\\}_{n=1}^{\infty}$ be a
sequence of positive numbers such that
$\sum_{n=1}^{\infty}n\alpha_{n}^{2}<\infty$ and
$\sum_{n=1}^{\infty}\alpha_{n}=\infty$. Then the sequence
$\\{f_{n}\\}_{n=1}^{\infty}$ defined as

$f_{2n-1}=e_{2n-1}+\sum_{i=n}^{\infty}\alpha_{i-n+1}e_{2i},~{}~{}f_{2n}=e_{2n},~{}~{}n=1,2,\cdots$

is a conditional basis of $\mathcal{H}$. Denote by $F$ the corresponding
Schauder matrix and $T$ the operator it represented. Now we shall show that
$T$ is a unbounded operator indeed. To do this, we rewrite the matrix of $T$
under new ONB. Here we use the fact that $T$ is bounded if and only if
$U^{*}TU$ is bounded for each unitary operator $U$. Denote by
$\mathcal{H}_{1}=span\\{e_{2n-1};n=1,2,\cdots\\}$ and
$\mathcal{H}_{2}=span\\{e_{2n};n=1,2,\cdots\\}$. Now we can rewrite $F$ into
the form

$\begin{array}[]{rll}T=&\left(\begin{array}[]{cc}I&0\\\
T_{1}&I\end{array}\right)&\begin{array}[]{c}\mathcal{H}_{1}\\\
\mathcal{H}_{2}\end{array}\end{array}$

in which the operator $T_{1}$ has a matrix as

$\left(\begin{array}[]{cccc}\alpha_{1}&0&0&0\\\ \alpha_{2}&\alpha_{1}&0&0\\\
\alpha_{3}&\alpha_{2}&\alpha_{1}&0\\\
\vdots&\ddots&\ddots&\ddots\end{array}\right).$

Denote by $S^{(2)}$ the shift operator defined as $S^{(2)}e_{2n}=e_{2n+2}$. It
is trivial to check that we have $S^{(2)}T_{1}=T_{1}S^{(2)}$. Hence we have
$T_{2}$ is in the commutant $\mathcal{A}^{{}^{\prime}}(S^{(2)})$ if $T_{2}$ is
a bounded operator. But it is impossible since the holomorphic function
defined by series $\sum_{n=1}^{\infty}\alpha_{n}z^{n}$ is not in the class
$H^{\infty}$ by the fact $\sum_{n=1}^{\infty}\alpha_{n}=\infty$(cf, [18],
theorem 3, p62).

However, we always can relate a basis to a bounded operator as follows. Assume
that $M=(f_{kn})$ is an $\omega\times\omega$ matrix such that each column
vector $f_{n}=\\{f_{kn}\\}_{k=1}^{\infty}$ is an $l^{2}-sequence$. Then under
a fixed ONB sequence $\\{e_{n}\\}_{n=1}^{\infty}$,
$\mathbf{f}_{n}=\sum_{n=1}^{\infty}f_{kn}e_{k}$ is a vector in $\mathcal{H}$.
Then for a nonzero complex number sequence
$\alpha=\\{\alpha_{k}\\}_{k=1}^{\infty}$, it is clear that
$M_{\alpha}=(\alpha_{1}f_{1},\alpha_{2}f_{2},\alpha_{3}f_{3},\cdots)$ is also
a matrix with $l^{2}-$sequences as its column vectors. Moreover, if $M$
represents a bounded operator and $\alpha$ be a bounded sequence, then
$M_{\alpha}$ also be a matrix of a bounded operator since $M^{{}^{\prime}}=MD$
in which $D$ is the bounded diagonal operator with $\alpha_{k}$ as its
diagonal elements. Inspired by the properties of bases(cf, proposition 4.1.5
and 4.2.12 in the book [16]), we have the following definition.

######  Definition 3.7.

$M_{\alpha}$ is called the blowing up matrix of $M$ with sequence $\alpha$.

######  Theorem 3.8.

Suppose that $\\{f_{n}\\}_{n=1}^{\infty}$ is a basis sequence and
$\\{\alpha_{n}\\}_{n=1}^{\infty}$ is a sequence of complex numbers such that
the sum $\sum_{n=1}^{\infty}||\alpha_{n}f_{n}||$ be finite. Moreover, let
$\psi=\\{\alpha_{n}f_{n}\\}_{n=1}^{\infty}$. Then the matrix $M_{\psi}$
represents a bounded compact operator under any ONB.

###### Proof.

Fixed an ONB $\varphi=\\{e_{n}\\}_{n=1}^{\infty}$. Then each vector $f_{n}$
has a unique $l^{2}-$coodinate $f_{n}=\\{f_{kn}e_{k}\\}_{k=1}^{\infty}$. Let
$g_{n}=\\{g_{kn}\\}_{k=1}^{\infty}=\alpha_{n}f_{n}=\\{\alpha
f_{kn}e_{k}\\}_{k=1}^{\infty}$. Then the matrix $M_{\psi}=(g_{kn})$ under the
ONB $\varphi$. For any vector $x=\sum_{k=1}^{\infty}\xi_{k}e_{k}$, the series
$\sum_{k=1}^{\infty}\xi_{k}g_{k}$ converges since we have

$||\sum_{k=m}^{\infty}\xi_{k}g_{k}||\leq\sum_{k=m}^{\infty}||\xi_{k}g_{k}||\leq\sup_{k}\\{|\xi_{k}|\\}(\sum_{k=m}^{\infty}||g_{k}||)\rightarrow
0$

as $m\rightarrow 0$. Hence $M_{\psi}$ represents an operator $T_{\psi}$ well-
defined everywhere on $\mathcal{H}$. Now by the closed graph theorem, we just
need to show that $M_{\psi}$ also represents a closed operator to finish the
proof. Suppose that $x_{n}=\\{\xi_{k}^{(n)}\\}_{k=1}^{\infty}\rightarrow
x_{0}=\\{\xi_{k}^{(0)}\\}_{k=1}^{\infty}$ in the norm. Now for any
$\epsilon>0$, there is some integer $n_{0}$ such that
$|\xi^{(n)}_{k}-\xi^{(0)}_{k}|<1$ holds for any $n>n_{0}$. Let $k_{1}$ be the
integer satisfying $\sum_{k=k_{1}+1}^{\infty}||g_{k}||<\frac{\epsilon}{2}$.
Then there is an integer $n_{1}$ such that we have
$\sum_{k=1}^{k_{1}}|\xi_{k}^{(n)}-\xi_{k}^{(0)}|||g_{k}||<\frac{\epsilon}{2}$
for all $n>n_{1}$. Let $N=\max\\{n_{0},n_{1}\\}$. Hence for $n>N$ we have

$\begin{array}[]{rl}||T_{\psi}(x_{n}-x_{0})||&=||\sum_{k=1}^{\infty}(\xi_{k}^{(n)}-\xi_{k}^{(0)})g_{k}||\\\
&\leq||\sum_{k=1}^{k_{1}}(\xi_{k}^{(n)}-\xi_{k}^{(0)})g_{k}||+||\sum_{k=k_{1}+1}^{\infty}(\xi_{k}^{(n)}-\xi_{k}^{(0)})g_{k}||\\\
&\leq\sum_{k=1}^{k_{1}}|\xi_{k}^{(n)}-\xi_{k}^{(0)}|||g_{k}||+\sum_{k=k_{1}+1}^{\infty}||g_{k}||<\frac{\epsilon}{2}\\\
&\leq\frac{\epsilon}{2}+\frac{\epsilon}{2}.\end{array}$

Now it is left to show that $M_{\psi}$ is also a compact operator. Let
$K_{n}=\sum_{l=1}^{n}g_{l}\otimes e_{l}$ in which the operator $g_{l}\otimes
e_{l}$ is defined as $(g_{l}\otimes e_{l})x=(x,e_{l})g_{l}$. Then we have

$\begin{array}[]{rl}||(T_{\psi}-K_{n})||&=\sup_{||x||=1}||(T_{\psi}-K_{n})x||\\\
&=\sup_{||x||=1}||(\sum_{l=n+1}^{\infty}g_{l}\otimes e_{l})x||\\\
&\leq\sup_{||x||=1}\sum_{l=n+1}^{\infty}||g_{l}\otimes e_{l})x||\\\
&\leq\sum_{l=n+1}^{\infty}||g_{l}||\rightarrow 0\\\ \end{array}$

as $n\rightarrow\infty$. Hence $T_{\psi}$ is a norm limit of a sequence of
compact operators, that is, $T_{\psi}$ is a compact operator. ∎

Above theorem enlarge the class of bases that can be studied by bounded
operators.

######  Theorem 3.9.

Suppose that $\\{f_{n}\\}_{n=1}^{\infty}$ is a basis sequence and
$\\{\alpha_{n}\\}_{n=1}^{\infty}$ is a sequence of complex numbers such that
the sum $\sum_{n=1}^{\infty}||\alpha_{n}f_{n}||$ be finite. Moreover, let
$\psi=\\{\alpha_{n}f_{n}\\}_{n=1}^{\infty}$. Then there is a matrix $X$ of
some invertible operator such that $XM_{\psi}$ represents an strongly
irreducible bounded compact operator in $L(\mathcal{H})$.

### 3.3.

Now we turn to study the existence of strongly irreducible Schauder operator.
Recall that a Schauder operator $T$ is an operator mapping some ONB into a
Schauder basis. In his paper [17], Olevskii call an operator to be generating
if and only if it maps some ONB into a quasinormal conditional basis. Hence
our definition of Schauder operator is a generalization of Olevskii’s one. A
Schauder operator is said to be conditional if it maps some ONB sequence into
a conditional basis sequence. Our theorem 1.2 ensure the existence of
operators satisfying special properties. For convenience and self-sufficiency,
we list some results on an operator theory description of Schauder basis
appearing in paper [1] without proof.

######  Theorem 3.10.

Following conditions are equivalent:
1\. $T$ is a Schauder operator;
2\. $T$ maps some ONB $\\{e_{n}\\}_{n=1}^{\infty}$ into a basis;
3\. $T$ has a polar decomposition $T=UA$ in which $A$ is a Schauder operator;
4\. Assume that $T$ has a matrix representation $F$ under a fixed ONB
$\\{e_{n}\\}_{n=1}^{\infty}$. There is some unitary matrix $U$ such that $FU$
is a Schauder matrix.

######  Corollary 3.11.

([1]) Compact operator $K=diag\\{1,\frac{1}{2},\frac{1}{3},\cdots\\}$ is a
conditional operator.

######  Corollary 3.12.

There is an operator $T\in L(\mathcal{H})$ satisfying following properties:
1\. $T$ is a strongly irreducible compact operator;
2\. There exists some ONB such that $T$ has a matrix which is a conditional
Schauder matrix.

###### Proof.

Fixed an ONB, suppose that $K=diag\\{1,\frac{1}{2},\frac{1}{3},\cdots\\}$ be
the diagonal operator appearing in corollary 3.11. Then there is some unitary
matrix $U$ such that $KU$ is a conditional Schauder matrix. Then apply theorem
1.2 and theorem 3.10 to the matrix $KU$. ∎

######  Corollary 3.13.

There is an operator $T\in L(\mathcal{H})$ satisfying following properties:
1\. $T$ is a strongly irreducible compact operator;
2\. There exists some ONB such that $T$ has a matrix which is a unconditional
Schauder matrix.

###### Proof.

Fixed an ONB, suppose that $K=diag\\{1,\frac{1}{2},\frac{1}{3},\cdots\\}$ be
the diagonal operator appearing in corollary 3.11. Then apply theorem 1.2 and
theorem 3.10 to the matrix $K$. ∎

By the following main theorem 1 of [17], we can get the same results about the
non-compact case.

######  Theorem 3.14.

([17], p479) A bounded operator $T$ is generating if and only if the following
conditions hold:

a) the operators $T$ and $T^{*}$ do not admit the eigenvalue $\lambda=0$;

b) there exists a number $q,0<q<1$ such that to each segment $[q^{n+1},q^{n}]$
in the spectral decomposition of the positive operator
$(T^{*}T)^{\frac{1}{2}}$, there corresponds an infinite dimensional invariant
subspace.

######  Corollary 3.15.

There is an operator $T\in L(\mathcal{H})$ satisfying following properties:
1\. $T$ is a strongly irreducible non-compact operator;
2\. There exists some ONB such that $T$ has a matrix which is a conditional
Schauder matrix.

###### Proof.

Let $T$ be a self-adjopint operator satisfying the conditional b). Then there
is some unitary matrix $U$ such that $TU$ is a conditional Schauder matrix. It
is clear that $TU$ can not be compact. Then apply theorem 1.2 and theorem 3.10
to the operator $T$. ∎

######  Corollary 3.16.

There is an operator $T\in L(\mathcal{H})$ satisfying following properties:
1\. $T$ is a strongly irreducible non-compact operator;
2\. There exists some ONB such that $T$ has a matrix which is a unconditional
Schauder matrix.

###### Proof.

In fact by proposition 2.20 in paper [1] and theorem 1.2, all invertible and
strongly irreducible operators satisfy the requirements of corollary. For
example, given an irreducible operator $T$ and a complex number
$\lambda\notin\sigma(T)$, then $\lambda-T$ is such one operator. ∎

The first requirement of above lemmas are in the operator theory category
while the others are in basis theory category. The first one ask that the
matrix be good in the sense of Jordan block; the second one ask that the
matrix be nice from the basis viewpoint, that is, the column vectors comprise
a basis.

###  3.4. Small disturbance on the basis const

Now we consider the “small” condition appearing in the theorem 1.2, it ensure
us to get a new basis with a small disturbance on both of the basis const and
the unconditional const.

Recall that a sequence $\psi=\\{f_{n}\\}_{n=1}^{\infty}$ is called a Schauder
basis of the Hilbert space $\mathcal{H}$ if and only if for every vector
$x\in\mathcal{H}$ there exists a unique sequence
$\\{\alpha_{n}\\}_{n=1}^{\infty}$ of complex numbers such that the partial sum
sequence $x_{k}=\sum_{n=1}^{k}\alpha_{n}f_{n}$ converges to $x$ in norm.

Denote by $P_{k}$ the the diagonal matrix with the first $k-$th entries on
diagonal line equal to 1 and $0$ for others. Then as an operator $P_{k}$
represents the orthogonal projection from $\mathcal{H}$ to the subspace
$\mathcal{H}^{(k)}=span\\{e_{1},e_{2},\cdots,e_{k}\\}$.

######  Lemma 3.17.

Assume that $\\{e_{k}\\}_{k=1}^{\infty}$ is a fixed ONB of $\mathcal{H}$.
Suppose that an $\omega\times\omega$ matrix $F=(f_{ij})$ satisfies the
following properties:

1\. Each column of the matrix $F$ is a $l^{2}-$sequence;

2\. $F$ has a unique left inverse matrix $G^{*}=(g_{kl})$ such that each row
of $G^{*}$ is also a $l^{2}-$sequence;

3\. Operators $Q_{k}$ defined by the matrix $Q_{k}=FP_{k}G^{*}$ are well-
defined projections on $\mathcal{H}$ and converges to the unit operator $I$ in
the strong operator topology.

Then the sequence
$\\{f_{k}\\}_{k=1}^{\infty},f_{k}=\sum_{j=1}^{\infty}f_{ij}e_{i}$ must be a
Schauder basis.

The projection $FP_{n}G^{*}$ is just the $n-$th “natural projection” so called
in [16](p354). It is also the $n-$th partial sum operator so called in
[19](definition 4.4, p25). Now we can translate theorem 4.1.15 and corollary
4.1.17 in [16] into the following

######  Proposition 3.18.

If $F$ is a Schauder matrix, then $M=\sup_{n}\\{||FP_{n}G^{*}||\\}$ is a
finite const.

The const $M$ is called the basis const for the basis
$\\{f_{n}\\}_{n=1}^{\infty}$.

Assume that $\psi=\\{f_{n}\\}_{n=1}^{\infty}$ is a basis. For a subset
$\Delta$ of $\mathbb{N}$, denote by $P_{\Delta}$ the diagonal matrix defined
as $P_{\Delta}(nn)=1$ for $n\in\Delta$ and $P_{\Delta}(nn)=0$ for
$n\notin\Delta$. The projection $Q_{\Delta}=F_{\psi}P_{\Delta}G_{\psi}^{*}$
defined in above lemmas is called a natural projection(see, definition 4.2.24,
[16], p378). In fact for a vector $x=\sum_{n=1}^{\infty}x_{n}f_{n}$, it is
trivial to check $Q_{\Delta}x=\sum_{n\in\Delta}x_{n}f_{n}$. Then we have a
same result for the  unconditional basis const(cf, definition4.2.28, [16],
p379):

######  Proposition 3.19.

If $F_{\psi}$ is a Schauder matrix, then the unconditional basis const of the
basis $\psi$ is
$M_{ub}=\sup_{\Delta\subseteq\mathbb{N}}\\{||F_{\psi}P_{\Delta}G_{\psi}^{*}||\\}$.

In virtue of the proposition 4.2.29 and theorem 4.2.32 in the book [16], we
have

######  Proposition 3.20.

For a Schauder basis $\psi$, it is a unconditional basis if and only if
$\sup_{\Delta\subseteq\mathbb{N}}\\{||F_{\psi}P_{\Delta}G_{\psi}^{*}||\\}<\infty$.

Now we can show that we can change a Schauder matrix to an equivalent Schauder
matrix which represents a strongly irreducible operator with small disturbance
on basis const.

######  Theorem 3.21.

Suppose that $\psi=\\{f_{n}\\}_{n=1}^{\infty}$ be a basis and its
corresponding Schauder matrix $F_{\psi}$ under a given ONB
$\varphi=\\{e_{n}\\}_{n=1}^{\infty}$ represents a bounded operator $T_{\psi}$
in $L(\mathcal{H})$. Then for any positive number $\epsilon>0$, there is a
matrix $X$ of some invertible operator $T\in L(\mathcal{H})$ such that
following properties hold:
1\. $XF_{\psi}$ is a strongly irreducible operator;
2\. The column vector sequence $\psi^{{}^{\prime}}$ of $XF_{\psi}$ comprise a
basis sequence equivalent to $\psi$. Moreover, if we denote by
$M^{{}^{\prime}}$ the basis const of $\psi^{{}^{\prime}}$, then we can ask
that the condition $|M-M^{{}^{\prime}}|<\epsilon$ holds; If $\psi$ is also a
unconditional basis, we can also ask that the additional condition
$|M_{ub}^{{}^{\prime}}-M_{ub}|<\epsilon$ holds(Here $M_{ub}^{{}^{\prime}}$
denote the unconditional basis const of the basis $\psi^{{}^{\prime}}$).

###### Proof.

By our main theorem 1.2, we only need to show that the second property holds
when $X=U+K$ been chosen carefully. Clearly for a given $\delta>0$, we can
choose a $X=U+K$ such that both
$1-\frac{\delta}{2}<||U+K||<1+\frac{\delta}{2}$ and
$1-\frac{\delta}{2}<||(U+K)^{-1}||<1+\frac{\delta}{2}$ hold. Then we have

$\begin{array}[]{c}||XF_{\psi}P_{k}G^{*}_{\psi}X^{-1}||\leq||X||\cdot||F_{\psi}P_{k}G^{*}_{\psi}||\cdot||X^{-1}||,\\\
||F_{\psi}P_{k}G^{*}_{\psi}||\leq||X^{-1}||\cdot||XF_{\psi}P_{k}G^{*}_{\psi}X^{-1}||\cdot||X||.\end{array}$

Hence following inequality holds for any $k\in\mathbb{N}$:

$(1+\delta)^{-2}\cdot||F_{\psi}P_{k}G^{*}_{\psi}||\leq||XF_{\psi}P_{k}G^{*}_{\psi}X^{-1}||\leq(1+\delta)^{2}\cdot||F_{\psi}P_{k}G^{*}_{\psi}||.$

Now by proposition 3.18, we have

$M=\sup_{n}\\{||FP_{n}G^{*}||\\}\hbox{ and
}M^{{}^{\prime}}=\sup_{n}\\{||XFP_{n}G^{*}X^{-1}||\\}.$

Therefore by choosing $\delta<\frac{\epsilon}{4M}$ we can prove the first part
of property 2; the second part of property 2 can be proved in just the same
way by lemma 3.20. ∎

Acknowledgements A large part of this article was developed during the seminar
on operator theory held at Jilin University in China.

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