# Nonparametric Tests in Linear Model with Autoregressive Errors

Olcay Arslan,Yeşim Güney
Jana Jurečková, Yetkin Tuaç
University of Ankara and Charles University, Prague The research of J.
Jurečková was supported by the Grant GAČR 18-01137S

###### Abstract

In the linear regression model with possibly autoregressive errors, we propose
a family of nonparametric tests for regression under a nuisance
autoregression. The tests avoid the estimation of nuisance parameters, in
contrast to the tests proposed in the literature. A simulation study, as well
as an application of tests to real data, illustrate their good performance.
Keywords: Autoregression Linear regression Rank test Regression rank scores
Autoregression rank scores

##  1 Introduction

The standard assumption of the independent and identically distributed errors
in the linear regression model is often violated. Some authors admit the
autoregressive structure of model errors. McKnight et al. (2000) applied a
double bootstrap method to analyze linear models with autoregressive errors.
The authors mostly estimated the regression parameters under autoregressive
errors with a known innovation distribution. Alpuim and El-Shaarawi (2008)
used the ordinary least squares (OLS) estimator under the $p$-order
autoregressive (AR(p)) error term and the normal innovations. Tuaç et al.
(2018) considered linear regression model with AR(p) errors with Student’s
t-distribution and used conditional maximum likelihood estimation of model
parameters. In (2020), Tuaç et al. proposed an autoregressive regression
procedure based on the skew-normal and skew-$t$ distributions. Güney et al.
(2020a) considered the conditional maximum Lq-likelihood (CMLq) estimation
method for the autoregressive error terms regression models under normality
assumption.

In the real life applications, the data sets may contain outliers and their
distribution can be heavy-tailed. Then we should take recourse to
nonparametric models without a specific distribution assumptions. The most
powerful tools for estimation and other inference in this area are the
regression and autoregression quantiles. However, the most convenient for
testing are the ranks of observations or the ranks of residuals and their
extensions, as the regression and autoregression rank scores. The quantile
regression was introduced by Koenker and Bassett (1978) and by their
followers. It is an approach to model the conditional quantile function of
response variable depending on covariates. The regression rank scores were
introduced by Gutenbrunner and Jurečková (1992) and a generalization of this
concept to the autoregressive model is due to Koul and Saleh (1995). The class
of regression rank scores tests was developed by Gutenbrunner and Jurečková
(1992) and Gutenbrunner et al. (1992). The optimal autoregression rank scores
tests in the AR model were constructed by Hallin and Jurečková (1999). El
Bantli and Hallin (2001) constructed the Kolmogorov-Smirnov type test in AR
model based on the autoregression rank scores, following the KS test in the
linear model test by Jurečková (1991). The averaged autoregression quantiles
and their asymptotics were studied by Güney et al. (2020b). In the present
paper, we assume that our data follow a linear regression model whose model
errors can be possibly autoregressive. The corresponding probability
distributions are generally unknown, only satisfy some assumptions. In this
setup, we shall verify the hypothesis of no linear regression under possible
nuisance autoregression of model errors.

##  2 Statement of the model

We consider the linear regression model of order $s,$ whose model errors
follow a stationary autoregressive process of order $p:$

$\displaystyle
y_{t}=\beta_{0}+\mathbf{x}_{t}^{\top}\mbox{\boldmath$\beta$}^{\ast}+\varepsilon_{t}=\beta_{0}+x_{t1}\beta_{1}+...+x_{tp}\beta_{p}+\varepsilon_{t},$
(2.1)
$\displaystyle\varepsilon_{t}=\varphi_{0}+\varphi_{1}\varepsilon_{t-1}+u_{t}+\ldots+\varphi_{p}\varepsilon_{t-p},\;t=1,2,...,n,$
(2.2)
$\displaystyle\mbox{\boldmath$\beta$}^{\ast}=(\beta_{1},\ldots,\beta_{s})^{\top}.$

Here $y_{t}$ is the response variable,
$\mathbf{x}_{t}=(x_{t0},\ldots,x_{ts})^{\top}$ are the regressors and
$\beta_{j},\;j=0,\ldots,s$ are unknown regression parameters. We assume that
$x_{t0}=1$ for all $t,$ hence that $\beta_{0}$ is an intercept.

Moreover, $\varphi_{0},\varphi_{1},\ldots,\varphi_{p}$ are unknown
autoregression parameters, where the intercept $\varphi_{0}$ is added for
mathematical convenience and can be 0. The innovations $u_{t}$ are assumed
being independently and identically distributed (i.i.d.) with a continuous
distribution function $F$ and density $f,$ generally unknown but satisfying

$E\left(u_{t}\right)=0,\quad Var\left(u_{t}\right)=\sigma^{2}<\infty.$ (2.3)

The stationarity condition requires that all roots of the equation
$z^{2}-\varphi_{1}z^{1}=0-\varphi_{2}z^{p-2}-...-\varphi_{p}z^{t-p}=0$ are
inside the unit circle (Brockwell and Davis, 1987). Note that the error model
given in equation (2.2) is a strictly stationary process, and so the
$\varepsilon_{t}$ share a common marginal distribution and thus share the same
quantiles. The intercept term $\varphi_{0}$ in model (2.1) is included for the
identifiability of the autoregression quantiles, and can be equal to 0. The
distribution function $F$ of $u_{t}$ is unknown, but we assume that it is
increasing on the set $\\{u:0<F(u)<1\\}$. Because of the identifiability, we
assume that the starting observations $(y_{-p+1},\ldots,y_{0})$ are known.

For the convenience, we also write (2.2) in the form

$u_{t}=\Phi\left(B\right)\varepsilon_{t}$ (2.4)

where $B$ is called the backshift operator. Then the linear regression model
with AR(p) error term given in equation (2.1) can be expressed as

$\displaystyle\Phi\left(B\right)y_{t}=y_{t}-\varphi_{0}-\varphi_{1}y_{t-1}+...\varphi_{p}y_{t-p},$
(2.5)
$\displaystyle\mathbf{\Phi}\left(B\right)\mathbf{x}_{t}=\mathbf{x}_{t}-\varphi_{0}-\varphi_{1}\mathbf{x}_{t-1}+\ldots\varphi_{p}\mathbf{x}_{t-p},$

hence

$\Phi\left(B\right)y_{t}=\left(\mathbf{\Phi}\left(B\right)\mathbf{x}_{t}\right)^{T}\mbox{\boldmath$\beta$}+u_{t}.$
(2.6)

In the model (2.1), we construct the tests of the hypothesis:

$\mathbf{H}_{0}:\;\mbox{\boldmath$\beta$}^{\ast}=\mathbf{0},\;\mbox{ with
}\;\beta_{0},~{}(\varphi_{0},\varphi_{1},\ldots,\varphi_{p})^{\top}\neq\mathbf{0}\;\mbox{
unspecified.}$

Our tests are nonparametric; the test of $\mathbf{H}_{0}$ is based on the
autoregression rank scores and on the linear autoregression rank statistics
for the model (2.1), (2.2) without regression.

##  3 Rank tests for $\mathbf{H}_{0}$

We shall be testing the absence of regression

$\mathbf{H}_{0}:\;\mbox{\boldmath$\beta$}^{\ast}=\mathbf{0},\qquad\beta_{0},\;\varphi_{0},\;\varphi_{1},\ldots,\varphi_{p}\;\mbox{unspecified}.$

The usual alternative is the local (Pitman) regression

$\mathbf{K}_{n}:\;\mbox{\boldmath$\beta$}^{\ast}=\mbox{\boldmath$\beta$}_{n}^{\ast}=n^{-1/2}\mbox{\boldmath$\beta$}_{x}^{\ast},\;\mbox{
with }\;\mbox{\boldmath$\beta$}_{x}^{\ast}\in I\kern-3.69995ptR^{p}\;\mbox{
fixed.}$ (3.1)

Under $\mathbf{H}_{0},$ the observations follow the model

$y_{t}=\beta_{0}+\varepsilon_{t},\;t=1,\ldots,n.$

The hypothesis $\mathbf{H}_{0}$ in fact means the hypothetical autoregressive
model

$\mathbf{H}_{0}:\;y_{t}=\beta_{0}+\varphi_{0}+\varphi_{1}y_{t-1}+\varphi_{1}y_{t-1}+\ldots+\varphi_{p}y_{t-p}+u_{t},\;t=1,\ldots,n$
(3.2)

which we like to test against the alternative $\mathbf{K}_{n}.$ Let

${\mathbf{y}}_{t}^{*}=(y_{t},y_{t-1},\ldots,y_{t-p})^{\top},\;{\mathbf{y}}_{t}=(1,\,y_{t},y_{t-1},\ldots,y_{t-p})^{\top},\;t=0,\ldots,n-1$
(3.3)

and consider the random matrices of the respective orders $n\times p$ and
$n\times(p+1)$

$\displaystyle\mathbf{Y}_{n}^{\ast}=\left[\begin{array}[]{c}\mathbf{y}_{1}^{\ast\top}\\\
\cdots\\\ \mathbf{y}_{n}^{\ast\top}\\\
\end{array}\right],\qquad\mathbf{Y}_{n}=\left[\begin{array}[]{c}\mathbf{y}_{1}^{\top}\\\
\cdots\\\ \mathbf{y}_{n}^{\top}\\\ \end{array}\right].$ (3.10)

For convenience, denote also

$\displaystyle\mathbf{x}_{t}^{\ast}=(x_{t1},\ldots,x_{tp})^{\top},\quad\mathbf{x}_{t}=(1,x_{t1},\ldots,x_{ts})^{\top}=(1,\mathbf{x}_{t}^{\ast\top})^{\top}$
$\displaystyle\mathbf{X}_{n}^{\ast}=\left[\begin{array}[]{c}\mathbf{x}_{1}^{\ast\top}\\\
\cdots\\\ \mathbf{x}_{n}^{\ast\top}\\\
\end{array}\right],\qquad\mathbf{X}_{n}=\left[\begin{array}[]{c}\mathbf{x}_{1}^{\top}\\\
\cdots\\\ \mathbf{x}_{n}^{\top}\\\ \end{array}\right].$

The autoregression rank scores
$\widehat{\mathbf{a}}_{n}(\alpha)=\left(\hat{a}_{n1}(\alpha),\ldots,\hat{a}_{nn}(\alpha)\right)^{\top}$
under hypothesis $\mathbf{H}_{0}$ are defined as the solution vector of the
linear programming problem

$\left\\{\begin{array}[]{lll}\sum_{t=1}^{n}y_{t}\hat{a}_{nt}(\alpha):&=&\mbox{
max }\\\\[5.69054pt]
\sum_{t=1}^{n}\left(\hat{a}_{nt}(\alpha)-(1-\alpha)\right)&=&0\\\\[5.69054pt]
\mathbf{Y}_{n}^{\ast\top}\left(\widehat{\mathbf{a}}_{n}(\alpha)-(1-\alpha){\mathbf{1}}_{n}\right)&=&\mathbf{0}\\\\[5.69054pt]
\widehat{\mathbf{a}}_{n}(\alpha)\in[0,1]^{n},\quad 0\leq\alpha\leq 1.&&\\\
\end{array}\right.$ (3.12)

The autoregression rank scores are autoregression-invariant. More precisely,
(3.12) implies that $\widehat{{\mathbf{a}}}_{n}(\alpha)$ can be also formally
written as a solution of the linear program

$\left\\{\begin{array}[]{lll}\sum_{t=1}^{n}u_{t}\hat{a}_{nt}(\alpha):&=&\mbox{
max }\\\\[5.69054pt]
\sum_{t=1}^{n}\left(\hat{a}_{nt}(\alpha)-(1-\alpha)\right)&=&0\\\\[5.69054pt]
\mathbf{Y}_{n}^{\ast\top}\left(\widehat{\mathbf{a}}_{n}(\alpha)-(1-\alpha){\mathbf{1}}_{n}\right)&=&\mathbf{0}\\\\[5.69054pt]
\widehat{\mathbf{a}}_{n}(\alpha)\in[0,1]^{n},\quad 0\leq\alpha\leq 1.&&\\\
\end{array}\right.$

where $\mathbf{u}_{n}=(u_{1},\ldots,u_{n})^{\top}$ is the unobservable white
noise process.

We shall construct a family of new tests of the hypothesis $\mathbf{H}_{0}$
for the model (2.1), based on autoregression rank scores, and analyze the
asymptotic distribution of the test criterion under the null hypothesis as
well as under contiguous alternatives. Surprisingly, no preliminary estimation
of $\varphi$ is needed in order to compute autoregression rank score
statistics, in contrast with the aligned rank methods (Puri and Sen and
others).

The (unknown) density $f$ of $u_{t}$ is assumed to belong to the family
${\mathcal{F}}$ of exponentially tailed densities, satisfying (2.3) and the
following conditions on the tails:

  1. (F1)

$f$ is positive and absolutely continuous, with a.e. derivative $f^{\prime}$
and finite Fisher information
$\mathcal{I}(f)=\int\left(\frac{f^{\prime}(x)}{f(x)}\right)^{2}f(x){\rm
d}x<\infty$; moreover, there exists $K_{f}\geq 0$ such that $f$ has two
bounded derivatives $f^{\prime}$ and $f^{\prime\prime}$ for all $|x|>K_{f}$;

  2. (F2)

$f$ is monotonically decreasing to $0$ as $x\rightarrow\pm\infty$ and

$\lim_{x\rightarrow-\infty}\frac{-\log
F(x)}{b|x|^{r}}=\lim_{x\rightarrow\infty}\frac{-\log(1-F(x))}{b|x|^{r}}=1$

for some $b>0$ and $r\geq 1$.

Other properties of densities in $\mathcal{F}$ are summarized in [8].

Moreover, we impose the following conditions on the regression matrix
$\mathbf{X}_{n}^{\ast}:$

  1. (X1)

The matrix
$\mathbf{A}_{n}=n^{-1}\sum_{t=1}^{n}\mathbf{X}_{n}^{\ast\top}\mathbf{X}_{n}^{\ast}$
is positive definite of order $s$ for $n\geq n_{0}.$

  2. (X2)

$n^{-1}\sum_{t=1}^{n}\|\mathbf{x}_{nt}\|^{4}=O(1)$ as $n\rightarrow\infty.$

  3. (X3)

$\lim_{n\rightarrow\infty}\max_{1\leq t\leq
n}\left\\{n^{-1}\mathbf{x}_{nt}^{\top}\mathbf{A}_{n}^{-1}\mathbf{x}_{nt}\right\\}=\mathbf{0}.$

Define $\mathbf{D}_{n}=n^{-1}\mathbf{Y}_{n}^{\top}\mathbf{Y}_{n}$ and

$\mathbf{H}_{n}=\mathbf{Y}_{n}^{\top}(\mathbf{Y}_{n}^{\top}\mathbf{Y}_{n})^{-1}\mathbf{Y}_{n},\qquad\widehat{\mathbf{X}}_{n}^{\ast}=\mathbf{H}_{n}\mathbf{X}_{n}^{\ast}$
(3.13)

the projection matrix and the projection of $\mathbf{X}_{n}^{\ast}$ on the
space spanned by the columns of $\mathbf{Y}_{n}$, respectively. Moreover, let

$\mathbf{Q}_{n}=n^{-1}(\mathbf{X}_{n}^{\ast}-\widehat{\mathbf{X}}_{n}^{\ast})^{\top}(\mathbf{X}_{n}^{\ast}-\widehat{\mathbf{X}}_{n}^{\ast}).$
(3.14)

The (random) matrices $\mathbf{D}_{n}$ and $\mathbf{Q}_{n}$ are of the
respective orders $(p+1)\times(p+1)$ and $s\times s$. We shall assume that

$\lim_{n\rightarrow\infty}\mathbf{D}_{n}=\mathbf{D},\qquad\lim_{n\rightarrow\infty}I\kern-3.69995ptE(\mathbf{Q}_{n})=\mathbf{Q}$
(3.15)

where $\mathbf{D}$ and $\mathbf{Q}$ are positive definite matrices.

Choose a nondecreasing, square integrable score generating function
$J:(0,1)\rightarrow I\kern-3.69995ptR$, such that $J(1-u)=-J(u),\;0<u<1$ and
that $J^{\prime}(u)$ exists for $u\in(0,\,\alpha_{0})\cup(1-\alpha_{0},\,1)$
and, in this domain, satisfies the Chernoff-Savage condition

$|J^{\prime}(u)|\leq c(u(1-u))^{-1-\delta},\ 0<\delta<\frac{1}{4}.$ (3.16)

Define the scores generated by $J$ as
$\widehat{\mathbf{b}}_{n}=(\hat{b}_{n1},\ldots,\hat{b}_{nn})^{\top}$ with

$\hat{b}_{nt}=-\int_{0}^{1}J(u){\rm d}\hat{a}_{nt}(u),\quad t=1,\ldots,n.$
(3.17)

The proposed tests of $\mathbf{H}_{0}$ are based on the linear autoregression
rank statistics of the form

${\mathbf{S}}_{n}=n^{-\frac{1}{2}}(\mathbf{X}_{n}^{\ast}-\widehat{\mathbf{X}}_{n}^{\ast})^{\top}\widehat{\mathbf{b}}_{n},$
(3.18)

and for testing $\mathbf{H}_{0}$ against $\mathbf{K}_{n}$ we propose the
criterion

$\mathcal{T}_{n}=\mathbf{S}_{n}^{\top}\mathbf{Q}_{n}^{-1}\mathbf{S}_{n}/A^{2}(J)$
(3.19)

where

$A^{2}(J)=\int_{0}^{1}(J(t)-\bar{J})^{2}{\rm
d}t,\qquad\bar{J}=\int_{0}^{1}J(t){\rm d}t.$ (3.20)

The typical choices of $J$ are:

(i) Wilcoxon scores (optimal for $f$ logistic) : $J(u)=u-\frac{1}{2},\;0<u<1.$
The scores are
$\hat{b}_{n;t}=-\int_{0}^{1}(J(u)-\frac{1}{2})d\hat{a}_{t}(u)=\int_{0}^{1}J^{2}(u)du-\frac{1}{2}$
while $A^{2}(J)=\frac{1}{12}$ and $\gamma(J,F)=\int f^{2}(x)dx.$

(ii) Normal (van der Waerden) scores (asymptotically optimal for $f$ normal):
$J(t)=\Phi^{-1}(u),\;0<u<1,\Phi$ being the d.f. of standard normal
distribution. Here $A^{2}(J)=1$ and $\gamma(J,F)=\int f(F^{-1}(J(x)))dx$.

(iii) Median (sign) scores: $J(u)=\frac{1}{2}~{}sign(u-\frac{1}{2}),\;0<u<1.$

Notice that the test statistic $\mathcal{T}_{n}$ requires no estimation of
nuisance parameters, since the functional $A(J)$ depends only on the score
function and not on (the unknown) $F.$ We shall show that the asymptotic
distribution of $\mathcal{T}_{n}$ under $\mathbf{H}_{0}$ is central $\chi^{2}$
with $s$ degrees of freedom, hence it is asymptotically distribution free.
Under $\mathbf{K}_{n}$ it is noncentral $\chi^{2}$ with $s$ degrees of freedom
and noncentrality parameter dependent on $J$ and $F$ but not on the nuisance
parameters. In this way, it is asymptotically equivalent to the rank test of
$\mathbf{H}_{0}$ in the situation without nuisance autoregression.

##  4  Asymptotic behavior of the test of $\mathbf{H}_{0}$

Let us return to the model (2.1). Assume that the matrices
$\mathbf{X}_{n}^{\ast}$ and $\mathbf{Y}_{n}^{\ast}$ satisfy conditions
(3.13)–(3.15). We want to test the hypothesis

$\mathbf{H}_{0}:\;\mbox{\boldmath$\beta$}^{\ast}=\mathbf{0}\quad(\beta_{0},\;\mbox{\boldmath$\varphi$}\;\mbox{
unspecified })$

against the alternative

$\mathbf{K}_{n}:\;\mbox{\boldmath$\beta$}^{\ast}=n^{-1/2}\mbox{\boldmath$\beta$}_{x}^{\ast}\quad(\mbox{\boldmath$\beta$}_{x}^{\ast}\in
I\kern-3.69995ptR_{s}\;\mbox{ fixed }).$

Let
$\widehat{\mathbf{a}}_{n}(\alpha)=(\hat{a}_{n1}(\alpha),\ldots,\hat{a}_{nn}(\alpha))$
be the autoregression rank scores corresponding to the submodel under
$\mathbf{H}_{0},$ i.e.

$y_{t}=\beta_{0}+\varphi_{0}+\varphi_{1}y_{t-1}+\varphi_{1}y_{t-1}+\ldots+\varphi_{p}y_{t-p}+u_{t},\;t=1,\ldots,n.$

Let $J:(0,1)\mapsto I\kern-3.69995ptR$ be a nondecreasing and square
integrable score-generating function such that $J(1-u)=-J(u),\;0<u<1$,
satisfying (3.16). Define the scores
$\widehat{\mathbf{b}}_{n}=(\hat{b}_{n1},\ldots,\hat{b}_{nn})^{\top}$ by the
relation (3.17). Consider the test statistics
$\mathcal{T}_{n}=\mathbf{S}_{n}^{\top}\mathbf{Q}_{n}^{-1}\mathbf{S}_{n}/A^{2}(J)$
defined in (3.18)–(3.20). The test is based on the asymptotic distribution of
$\mathcal{T}_{n}$ under $\mathbf{H}_{O}$, described in the following theorem.

###### Theorem 4.1

Assume that the distribution $F$ of the innovations $u_{t}$ satisfies
(F1)–(F4) and the regression matrix $\mathbf{X}_{n}$ satisfies (X1)–(X4). Let
$\mathbf{T}_{n}$ be generated by the function $J$ satisfying (3.16),
nondecreasing and square integrable on $(0,1).$

(i) Then, under $\mathbf{H}_{0},$ the asymptotic distribution of
$\mathcal{T}_{n}$ is central $\chi^{2}$ with $s$ degrees of freedom.

(ii) Under $\mathbf{K}_{n},$ the asymptotic distribution of $\mathcal{T}_{n}$
is noncentral $\chi^{2}$ with $s$ degrees of freedom and the noncentrality
parameter

$\displaystyle\eta^{2}=\mbox{\boldmath$\beta$}_{x}^{\top}\mathbf{Q}\mbox{\boldmath$\beta$}_{x}\cdot\gamma^{2}(J,F)/A^{2}(J)$
where $\displaystyle\gamma(J,F)=-\int_{0}^{1}J(v){\rm d}f(F^{-1}(v)).$

Hence, the test rejects $\mathbf{H}_{0}$ on the significance level
$\tau\in(0,1)$ if $\mathcal{T}_{n}>\chi_{s}^{2}(1-\tau)$ where
$\chi_{s}^{2}(1-\tau)$ is the $100(1-\tau)\%$-quantile of the $\chi^{2}$
distribution with $s$ degrees of freedom. The asymptotic distribution under
$\mathbf{K}_{n}$ also shows that the Pitman efficiency of the test coincides
with that of the classical rank test in the situation without the
autoregressive errors.
Proof.

(i) It follows from [8], Theorem 3.3 and [7], Theorem 4.1, that under
$\mathbf{H}_{0},$ the linear autoregression rank scores statistic admits the
representation

$\mathbf{Q}_{n}^{-1/2}\mathbf{S}_{n}=n^{-1/2}\mathbf{Q}_{n}^{-1/2}(\mathbf{X}_{n}^{\ast}-\overline{\mathbf{X}}_{n}^{\ast})^{\top}\widetilde{\mathbf{b}}_{n}+o_{p}(1)$

as $n\rightarrow\infty,$ where
$\widetilde{\mathbf{b}}_{n}=(\tilde{b}_{n1},\ldots,\tilde{b}_{nn})^{\top}$ and
$\tilde{b}_{nt}=J(F(u_{nt})),\;t=1,\ldots,n.$ Then (i) follows from the
central limit theorem.

(ii) The same representation holds also under the sequence of alternatives
$\mathbf{K}_{n},$ which is contiguous with respect to the sequence of null
distributions with the densities $\prod_{t=1}^{n}f(u_{t}).$

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$\displaystyle-16\kappa^{12}\tilde{\beta}_{4}^{2}\left(72\tilde{\beta}_{4}-23\right)\left(8\tilde{\beta}_{4}+1\right)^{7}+512\kappa^{10}\tilde{\beta}_{4}^{3}\left(1216\tilde{\beta}_{4}^{2}-384\tilde{\beta}_{4}+15\right)\left(8\tilde{\beta}_{4}+1\right)^{6}$
$\displaystyle+128\kappa^{10}\tilde{\beta}_{4}^{2}\left(1216\tilde{\beta}_{4}^{2}-384\tilde{\beta}_{4}+15\right)\left(8\tilde{\beta}_{4}+1\right)^{6}+8\kappa^{10}\tilde{\beta}_{4}\left(1216\tilde{\beta}_{4}^{2}-384\tilde{\beta}_{4}+15\right)\left(8\tilde{\beta}_{4}+1\right)^{6}$
$\displaystyle+\left(8\tilde{\beta}_{4}\kappa+\kappa\right)^{8}\left(-196608\tilde{\beta}_{4}^{4}+39936\tilde{\beta}_{4}^{3}+4416\tilde{\beta}_{4}^{2}-384\tilde{\beta}_{4}+9\right)\Big{]}\Big{\\}}$
$\displaystyle\times\Big{\\{}\left(8\tilde{\beta}_{4}+1\right)^{14}\left(-4\left(\kappa^{2}-4\right)\tilde{\beta}_{4}+64\tilde{\beta}_{4}^{2}+1\right)\left[2\left(\kappa^{2}+2\right)\tilde{\beta}_{4}-32\tilde{\beta}_{4}^{2}+1\right]^{2}$
$\displaystyle+\kappa^{4}\tilde{\beta}_{3}^{4}\left(8\tilde{\beta}_{4}+1\right)^{10}\left(-16\tilde{\beta}_{4}+\kappa^{2}-2\right)^{2}\left[-4\left(\kappa^{2}-4\right)\tilde{\beta}_{4}+64\tilde{\beta}_{4}^{2}+1\right]$
$\displaystyle-2\tilde{\beta}_{3}^{2}\left(8\tilde{\beta}_{4}+1\right)^{10}\left(8\kappa\tilde{\beta}_{4}+\kappa\right)^{2}\left[-4\left(\kappa^{2}-4\right)\tilde{\beta}_{4}+64\tilde{\beta}_{4}^{2}+1\right]$
$\displaystyle\times\left(2\kappa^{4}\tilde{\beta}_{4}-64\kappa^{2}\tilde{\beta}_{4}^{2}+512\tilde{\beta}_{4}^{3}-24\tilde{\beta}_{4}+\kappa^{2}-2\right)\Big{\\}}^{-1}\,.$

Notice that, (134) will reduce to (118) by taking ${\tilde{\beta}}_{4}=0$.

The dimensionless form of $\mathcal{M}_{2}$ at the $r\to r_{h}$ limit for
Model 3 (cf. Sec.V.3) is given below

$\displaystyle\left.\mathcal{M}_{2}\right|_{r\to r_{h}}$ $\displaystyle=$
$\displaystyle\frac{1}{2}\Big{\\{}16\tilde{\beta}_{3}^{6}\tilde{\beta}_{4}^{3}\left(20\tilde{\beta}_{3}^{2}-31\tilde{\beta}_{4}\right)\kappa^{24}-8\tilde{\beta}_{3}^{4}\tilde{\beta}_{4}^{2}\left(4\left(76\tilde{\beta}_{4}-17\right)\tilde{\beta}_{3}^{4}+\tilde{\beta}_{4}\left(31-140\tilde{\beta}_{4}\right)\tilde{\beta}_{3}^{2}-368\tilde{\beta}_{4}^{3}\right)\kappa^{22}$
(135)
$\displaystyle-\tilde{\beta}_{3}^{2}\Big{[}64\left(1-4\tilde{\beta}_{4}\right)^{2}\tilde{\beta}_{3}^{8}-4\tilde{\beta}_{4}\left(1216\tilde{\beta}_{4}^{2}-752\tilde{\beta}_{4}+29\right)\tilde{\beta}_{3}^{6}+\tilde{\beta}_{4}^{2}\left(-2624\tilde{\beta}_{4}^{2}+1904\tilde{\beta}_{4}+31\right)\tilde{\beta}_{3}^{4}$
$\displaystyle+32\tilde{\beta}_{4}^{4}\left(248\tilde{\beta}_{4}-69\right)\tilde{\beta}_{3}^{2}+4864\tilde{\beta}_{4}^{6}\Big{]}\kappa^{20}$
$\displaystyle+2\tilde{\beta}_{3}^{2}\Big{[}64\left(32\tilde{\beta}_{4}^{2}-28\tilde{\beta}_{4}+5\right)\tilde{\beta}_{3}^{8}-4\tilde{\beta}_{4}\left(64\tilde{\beta}_{4}^{2}-304\tilde{\beta}_{4}+3\right)\tilde{\beta}_{3}^{6}$
$\displaystyle-7\tilde{\beta}_{4}^{2}\left(576\tilde{\beta}_{4}^{2}-720\tilde{\beta}_{4}+83\right)\tilde{\beta}_{3}^{4}-4\tilde{\beta}_{4}^{3}\left(832\tilde{\beta}_{4}^{2}+192\tilde{\beta}_{4}-69\right)\tilde{\beta}_{3}^{2}-128\tilde{\beta}_{4}^{5}\left(14\tilde{\beta}_{4}+19\right)\Big{]}\kappa^{18}$
$\displaystyle-2\Big{[}32\left(64\tilde{\beta}_{4}^{2}-128\tilde{\beta}_{4}+37\right)\tilde{\beta}_{3}^{10}+8\left(256\tilde{\beta}_{4}^{3}-560\tilde{\beta}_{4}^{2}+221\tilde{\beta}_{4}-9\right)\tilde{\beta}_{3}^{8}$
$\displaystyle+\tilde{\beta}_{4}\left(-2304\tilde{\beta}_{4}^{3}+4768\tilde{\beta}_{4}^{2}-1474\tilde{\beta}_{4}+77\right)\tilde{\beta}_{3}^{6}-\tilde{\beta}_{4}^{2}\left(9728\tilde{\beta}_{4}^{3}-9408\tilde{\beta}_{4}^{2}+912\tilde{\beta}_{4}+23\right)\tilde{\beta}_{3}^{4}$
$\displaystyle+16\tilde{\beta}_{4}^{4}\left(-576\tilde{\beta}_{4}^{2}+304\tilde{\beta}_{4}+57\right)\tilde{\beta}_{3}^{2}-2048\tilde{\beta}_{4}^{7}\Big{]}\kappa^{16}$
$\displaystyle-4\Big{[}192\left(8\tilde{\beta}_{4}-5\right)\tilde{\beta}_{3}^{10}+8\left(384\tilde{\beta}_{4}^{2}-335\tilde{\beta}_{4}+30\right)\tilde{\beta}_{3}^{8}+\tilde{\beta}_{4}\left(512\tilde{\beta}_{4}^{2}-786\tilde{\beta}_{4}+63\right)\tilde{\beta}_{3}^{6}$
$\displaystyle-8\tilde{\beta}_{4}^{2}\left(128\tilde{\beta}_{4}^{3}+720\tilde{\beta}_{4}^{2}-327\tilde{\beta}_{4}+22\right)\tilde{\beta}_{3}^{4}$
$\displaystyle+4\tilde{\beta}_{4}^{3}\left(-1024\tilde{\beta}_{4}^{3}-1536\tilde{\beta}_{4}^{2}+468\tilde{\beta}_{4}+19\right)\tilde{\beta}_{3}^{2}-512\tilde{\beta}_{4}^{6}\left(8\tilde{\beta}_{4}+3\right)\Big{]}\kappa^{14}$
$\displaystyle+\Big{[}-2304\tilde{\beta}_{3}^{10}-64\left(140\tilde{\beta}_{4}-33\right)\tilde{\beta}_{3}^{8}-4\left(2072\tilde{\beta}_{4}^{2}-714\tilde{\beta}_{4}+25\right)\tilde{\beta}_{3}^{6}$
$\displaystyle+\tilde{\beta}_{4}\left(6144\tilde{\beta}_{4}^{3}+7424\tilde{\beta}_{4}^{2}-1832\tilde{\beta}_{4}+69\right)\tilde{\beta}_{3}^{4}$
$\displaystyle+\tilde{\beta}_{4}^{2}\left(28672\tilde{\beta}_{4}^{3}+12032\tilde{\beta}_{4}^{2}-2528\tilde{\beta}_{4}-19\right)\tilde{\beta}_{3}^{2}+256\tilde{\beta}_{4}^{5}\left(128\tilde{\beta}_{4}+15\right)\Big{]}\kappa^{12}$
$\displaystyle+\Big{[}-1536\tilde{\beta}_{3}^{8}+\left(400-3472\tilde{\beta}_{4}\right)\tilde{\beta}_{3}^{6}+32\tilde{\beta}_{4}^{2}\left(112\tilde{\beta}_{4}-1\right)\tilde{\beta}_{3}^{4}$
$\displaystyle+2\tilde{\beta}_{4}^{2}\left(10240\tilde{\beta}_{4}^{2}+1216\tilde{\beta}_{4}-199\right)\tilde{\beta}_{3}^{2}+256\tilde{\beta}_{4}^{4}\left(108\tilde{\beta}_{4}+5\right)\Big{]}\kappa^{10}$
$\displaystyle+\left[-400\tilde{\beta}_{3}^{6}+2\left(512\tilde{\beta}_{4}^{2}-174\tilde{\beta}_{4}+9\right)\tilde{\beta}_{3}^{4}+24\tilde{\beta}_{4}\left(320\tilde{\beta}_{4}^{2}+3\tilde{\beta}_{4}-1\right)\tilde{\beta}_{3}^{2}+80\tilde{\beta}_{4}^{3}\left(160\tilde{\beta}_{4}+3\right)\right]\kappa^{8}$
$\displaystyle+8\left[\left(18\tilde{\beta}_{4}-5\right)\tilde{\beta}_{3}^{4}+5\tilde{\beta}_{4}\left(40\tilde{\beta}_{4}-1\right)\tilde{\beta}_{3}^{2}+\tilde{\beta}_{4}^{2}\left(440\tilde{\beta}_{4}+3\right)\right]\kappa^{6}$
$\displaystyle+\left[8\tilde{\beta}_{3}^{4}+4\left(44\tilde{\beta}_{4}-1\right)\tilde{\beta}_{3}^{2}+\tilde{\beta}_{4}\left(576\tilde{\beta}_{4}+1\right)\right]\kappa^{4}+\left(8\tilde{\beta}_{3}^{2}+52\tilde{\beta}_{4}\right)\kappa^{2}+2\Big{\\}}$
$\displaystyle\times\left(4\kappa^{2}\tilde{\beta}_{4}+1\right)^{-5}\left[\kappa^{4}\left(-\tilde{\beta}_{3}^{2}\right)+2\kappa^{2}\left(\tilde{\beta}_{3}^{2}+2\tilde{\beta}_{4}\right)+1\right]^{-2}.$

Notice that, (135) will reduce to (118) by taking ${\tilde{\beta}}_{4}=0$.

##  Appendix C Supplemental materials for the numerical analysis in Sec.V

Figure 1: The behavior of $\mathcal{M}_{2}$ in phase space $\\{\mu,{\tilde{\beta}}_{3}\\}$ at the $r\to r_{h}$ limit for Model 1 [cf., (118)]. Here the cyan shadowed region is where the second angular Laplacian stability condition gets satisfied, viz., $\mathcal{M}_{2}\geq 0$. According to the contour of this shadowed region, we have marked the upper bound of allowed $|{\tilde{\beta}}_{3}|$ by red dashed vertical lines (which are tangent to the boundary of the shadowed region and are located at $|{\tilde{\beta}}_{3}|\approx 0.441$) and the red solid pentagrams.  |
---|---
(a) | (b)
Figure 2: Panel (a): The parameter phase space spanned by $\\{{\tilde{\beta}}_{3},{\tilde{\beta}}_{4},\mu\\}$. The red solid part represents the allowed region defined by Eqs.(4.12)-(4.14), (4.20) as well as (4.21) of Heisenberg  _et al._ (2018) and by considering the constraints $\mu\in(0,1)$ plus ${\tilde{\beta}}_{3}\in(-0.441,0.441)$. The cyan semi-transparent part indicates the bounds of ${\tilde{\beta}}_{4}$, within which the $\\{{\tilde{\beta}}_{3},{\tilde{\beta}}_{4}\\}$ phase space is covered by the allowed region for arbitrary legal $\mu$, as required by the theory. Panel (b): The parameter phase space $\\{{\tilde{\beta}}_{3},{\tilde{\beta}}_{4}\\}$ by setting $\mu=0$. Here the cyan shadowed area represents the allowed region. According to the contour of the shadowed region, we have marked the upper and lower bounds of allowed ${\tilde{\beta}}_{4}$ by red dashed horizontal lines (which are tangent to the boundary of the shadowed region and are located at ${\tilde{\beta}}_{4}\approx 0.251$ and ${\tilde{\beta}}_{4}\approx-0.125$) and the red solid pentagrams. |
---|---
(a) | (b)
Figure 3: Panel (a): The numerical solutions for $f$ (red solid line) and $h$ (blue dot-dashed line) in Model 2, together with their GR limit (orange dotted line) as comparison (in GR we have $f=h$). Panel (b): The behavior of $\mathcal{M}_{1}$ and $\mathcal{M}_{2}$ in Model 2. Notice that, the results for $\mathcal{M}_{1}$ are marked by blue triangles while those for $\mathcal{M}_{2}$ are marked by red dots. These results are plotted as functions of $r_{h}/r$ (so that we have a larger scope) by choosing ${\mu}=0.5$, ${\tilde{\beta}}_{3}=0.441$ and ${\tilde{\beta}}_{4}=0.1$. |
---|---
(a) | (b)
Figure 4: Panel (a): The numerical solutions for $f$ (red solid line), $h$
(blue dot-dashed line) and the dimensionless background quantity
$r_{h}^{-1}dA_{0}/d\xi=-\xi^{-2}A_{0}^{\prime}$ [purple dashed line, cf.,
(133)] in Model 3. The GR limit for $f$ and $h$ (in GR we have $f=h$) is also
added as comparison (orange dotted line). Panel (b): The behavior of
$\mathcal{M}_{1}$ and $\mathcal{M}_{2}$ in Model 3. Notice that, the results
for $\mathcal{M}_{1}$ are marked by blue triangles while those for
$\mathcal{M}_{2}$ are marked by red dots. These results are plotted as
functions of $r_{h}/r$ (so that we have a larger scope) by choosing
${\mu}=0.5$, ${\tilde{\beta}}_{3}=0.441$ and ${\tilde{\beta}}_{4}=0.1$.

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# Octonionic Quadratic Equations

T.Kalpa Madhawa
Department of Mathematics
Southern Illinois University

###### Abstract

There are four division algebras over $\mathbb{R}$, namely real numbers,
complex numbers, quaternions, and octonions. Lack of commutativity and
associativity make it difficult to investigate algebraic and geometric
properties of octonions. It does not make sense to ask, for example, whether
the equation $x^{2}+1=0$ is solvable, without specifying the field in which we
want the solutions to be lie. The equation $x^{2}+1=0$ has no solutions in
$\mathbb{R}$, which is to say, there are no real numbers satisfying this
equation. On the other hand, there are complex numbers which do satisfy this
equation in the field $\mathbb{C}$ of all complex numbers. How about if we
extend the same idea to other two normed division algebras quaternions and
octonions. Liping Huang and Wasin So [6] derive explicit formulas for
computing the roots of quaternionic quadratic equations. We extend their work
to octonionic case and solve monic left octonionic quadratic equation of the
form $x^{2}+bx+c=0$, where $a,b$ are octonions in general.[ We called this
form of quadratic equation as left octonion quadratic equation because we can
consider $x^{2}+xb+c=0$ as a different case due to non-commutativity of
octonions]. Finally, we represent the left spectrum of $2\times 2$ octonionic
matrix as a set of solutions to a corresponding octonionic quadratic equation,
which is an application of deriving explicit formulas for computing the roots
of octonionic quadratic equations.

Keywords— octonionic quadratic equation, left spectrum.

##  1 Quaternions and Octonions

###  1.1 Algebra of quaternions

Quaternions arose historically from Sir William Rowan Hamilton’s attempts in
the midnineteenth century to generalize complex numbers in some way that would
applicable to three-dimensional space. [9] In complex numbers we have square
root of -1 called $i$, what happens if we include another, independent, square
root of -1? Call it $j$. Then the big question is, what is $ij$? William
Hamilton eventually proposed that $k=ij$ should be yet another square root of
-1, and that the multiplication table should be cyclic, that is

$\begin{split}ij=k=-ji,\\\ jk=i=-kj,\\\ ki=j=-ik,\end{split}$

Figure 1.1: The quaternionic multiplication table

We refer to $i$,$j$, and $k$ as imaginary quaternionic units. Notice that
these units anticommute. This multiplication table is shown schematically in
Figure $1.1$ multiplying two of these quaternionic units together in the
direction of the arrow yields the third; going against the arrow contributes
an additional minus sign. The quaternions are denoted by $\mathbb{H}$; the H
is for "Hamilton", they are spanned by the identity element 1 and three
imaginary units, that is , a quaternion $q$ can be represented as four real
numbers $(q_{1},q_{2},q_{3},q_{4})$, usually written

$q=q_{1}+q_{2}i+q_{3}j+q_{4}k$ (1.1)

which can be thought of as a point or vector in $\mathbb{R}^{4}$. since
$(1.1)$ can be written in the form

$q=(q_{1}+q_{2}i)+(q_{3}+q_{4}i)j$ (1.2)

we see that a quaternion can be viewed as a pair of complex numbers, we can
write $\mathbb{H}=\mathbb{C}\oplus\mathbb{C}j$ in direct analogy to the
construction of $\mathbb{C}$ from $\mathbb{R}$. The quaternionic conjugate
$\bar{q}$ of a quaternion $q$ is obtained via the(real) linear map which
reverses the sign of each imaginary unit, so that

$\bar{q}=q_{1}-q_{2}i-q_{3}j-q_{4}k$ (1.3)

if $q$ is given by $(1.1)$. Conjugation leads directly to the norm of
quaternion $|q|$, defined by

$|q|^{2}=q\bar{q}=q_{1}^{2}+q_{2}^{2}+q_{3}^{2}+q_{4}^{2}.$ (1.4)

again, the only quaternion with norm zero is zero, and every nonzero
quaternion has a unique inverse, namely

$q^{-1}=\bar{q}/|q|^{2}$ (1.5)

uaternionic conjugation satisfies the identity $\overline{pq}=\bar{q}\bar{p}$
from which it follows that the norm satisfies $|pq|=|p||q|$. Squaring both
sides and expanding the result in terms of components yields the $4-$ sqaures
rule,

$\begin{split}&(p_{1}q_{1}-p_{2}q_{2}-p_{3}q_{3}-p_{4}q_{4})^{2}+(p_{2}q_{1}+p_{1}q_{2}-p_{4}q_{3}+p_{3}q_{4})^{2}\\\
&+(p_{3}q_{1}+p_{4}q_{2}+p_{1}q_{3}-p_{2}q_{4})^{2}+(p_{4}q_{1}-p_{3}q_{2}+p_{2}q_{3}+p_{1}q_{4})^{2}\\\
&=(p_{1}^{2}+p_{2}^{2}+p_{3}^{2}+p_{4}^{4})(q_{1}^{2}+q_{2}^{2}+q_{3}^{2}+q_{4}^{2})\end{split}$
(1.6)

which is not quite as obvious as the $2-$ squars rule. This identity implies
that the quaternions form a normed division algebra, that is, not only are
there inverses, but there are no zero divisors-if a product is zero, one of
the factors must be zero. It is important to realize that $\pm i,\pm j$, and
$\pm k$ are not the only quaternionic square roots of $-1$,

any imaginary quaternion squares to a negative number, so it is only necessary
to choose its norm to be one in order to get a square root of $-1$. The
imaginary quaternions of norm one form a $2-$dimensional sphere($q_{1}=0$); in
the above notation, this is the set of points

$q_{2}^{2}+q_{3}^{2}+q_{4}^{2}=1$ (1.7)

any such unit imaginary quaternion $u$ can be used to construct a complex
subalgebra of $\mathbb{H}$, which we will also denote by $\mathbb{C}$, namely

$\mathbb{C}={a+bu}$

with $a,b\in\mathbb{R}$. Furthermore, we can use the identity to write

$e^{u\theta}=\cos\theta+u\sin\theta$

This means that any quaternion can be written in the form $q=re^{u\theta}$
where $r=|q|$ and $u$ denotes the direction of the imaginary part of $q.$

###  1.2 Algebra of octonions

In analogy to the previous construction of $\mathbb{C}$ and $\mathbb{H}$, an
octonion $x$ can be thought of as a pair of quaternions, so that

$\mathbb{O}=\mathbb{H}\oplus\mathbb{H}\,l$

we will denote $i$ times $l$ simply as $il$, and similarly with $j$ and $k$.
It is easy to see that $l$,$il,jl,$ and $kl$ all square to $-1$; there are now
seven independent imaginary units, and we could write

$x=x_{1}+x_{2}i+x_{3}j+x_{4}k+x_{5}l+x_{6}il+x_{7}jl+x_{8}kl\ \ where\
x_{m}\in\mathbb{R}\quad\ for\ i=1,2,...,8.$ (1.8)

which can be thought of as a point or vector in $\mathbb{R}^{8}$. The real
part of $x$ is just $x_{1}$; the imaginary part of $x$ is everything else.
Algebraically, we could define

${\rm Re}\,(x)=(x+\bar{x})/2$ (1.9)

$\hbox{Im}(x)=(x-\bar{x})/2$ (1.10)

The imaginary part is differs slightly from the standard usage of these terms
for complex numbers, where $\hbox{Im}(z)$ normally refers to a real numbers,
the coefficient of $i$. This convention is not possible here, since the
imaginary part has seven degrees of freedom, and can be thought of as a vector
in $\mathbb{R}^{7}$. The full multiplication table is summarized in Figure
$1.2$ by means of the $7-$point projective plane. Each point corresponds to an
imaginary unit.

Figure 1.2: The octonionic multiplication table

Each line corresponds to a quaternionic triple, much like $\\{i,j,k\\}$, with
the arrow giving the orientation. For Example,

$\begin{split}&kl=kl\\\ &lkl=k\\\ &klk=l\end{split}$ (1.11)

and each of these products anticommutes, that is, reversing the order
contributes a minus sign. We define the octonionic conjugate $\bar{x}$ of an
octonion $x$ as the (real) linear map which reverses the sign of each
imaginary unit. Thus,

$\bar{x}=x_{1}-x_{2}i-x_{3}j-x_{4}k-x_{5}l-x_{6}il-x_{7}jl-x_{8}kl$ (1.12)

if $x$ is given by $(1.8)$. Direct computation shows that

$\overline{xy}=\bar{y}\bar{x}$ (1.13)

The norm of an octonion $|x|$ is defined by

$|x|^{2}=x\bar{x}=x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}+x_{5}^{2}+x_{6}^{2}+x_{7}^{2}+x_{8}^{2}$
(1.14)

Again, the only octonion with norm zero is zero, and every nonzero octonion
has a unique inverse, namely

$x^{-1}=\bar{x}/|x|^{2}$ (1.15)

as with the other division algebras, the norm satisfies the identity

$|xy|=|x||y|$ (1.16)

writing out this expression in terms of components yields the $8-$squares
rule, which is no longer at all obvious. The octonions therefore also form a
normed division algebra.
A remarkable property of the octonions is that they are not associative, For
Example

$\begin{split}&(ij)(l)=+(k)(l)=+kl\\\ &(i)(jl)=(i)(jl)=-kl\end{split}$ (1.17)

##  2 Octonionic quadratic formulas.

We will start with two lemmas.

######  Lemma 2.1.

Let $B,E,$ and $D$ be real numbers such that

  1. 1.

$D\neq 0$, and

  2. 2.

$B<0$ implies $B^{2}<4E$

Then the cubic equation

$y^{3}+2By^{2}+(B^{2}-4E)y-D^{2}=0$

has exactly one positive solution.

###### Proof.

Let

$f(y)=y^{3}+2By^{2}+(B^{2}-4E)y-D^{2}=0$

Note that $f(0)=-D^{2}<0$ and $\lim_{y\to+\infty}f(y)=+\infty$. According to
the intermediate value Theorem: if $f$ is a continuous function over an
interval [a,b], then $f$ takes all values between $f(a)$ and $f(b)$. Since
above cubic polynomial is a continuous function, its graph must intersect the
x-axis at some finite points greater than zero. So the equation has at least
one positive root. Now let us prove that $f$ has only one positive root.
Suppose that $f$ has three real roots, $r_{1},r_{2}$ and $r_{3}$. Take
$r_{1}>0$ be the positive root we found above. We must show that
$r_{2},r_{3}<0$. Then we have the result.
we know,

$r_{1}.r_{2}.r_{3}=D^{2}>0$

this implies the product of $r_{2}$ and $r_{3}$ is positive. Therefore,
$r_{2}$ and $r_{3}$ should be in same sign( both positive or negative). Let’s
assuming $r_{2},r_{3}>0$,
if $B<0$ implies $r_{1}+r_{2}+r_{3}=-2B>0$ But

$r_{1}.r_{2}+r_{1}.r_{3}+r_{2}.r_{3}=B^{2}-4E$

and should be positive. This is a contradiction due to part $2$ of
$\textbf{Lemma 2.1}.$ thus, we have only one positive solution to $f(y)=0$. ∎

######  Lemma 2.2.

Let $B,E,$ and $D$ be real numbers such that

  1. 1.

$E\geq 0$, and

  2. 2.

$B<0$ implies $B^{2}<4E$

Then the real system

$N^{2}-(B+T^{2})N+E=0,$ (2.1) $T^{3}+(B-2N)T+D=0,$ (2.2)

has at most two solutions $(T,N)$ satisfying $T\in\mathbb{R}$ and $N\geq 0$ as
follows.

  1. 1.

$T=0,N=(B\pm\sqrt{B^{2}-4E)}/2$ provided that $D=0,B^{2}\geq 4E.$

  2. 2.

$T=\pm\sqrt{2\sqrt{E}-B},N=\sqrt{E}$ provided that $D=0,B^{2}<4E.$

  3. 3.

$T=\pm\sqrt{z},N=(T^{3}+BT+D)/2T$ provided that $D\neq 0$ and $z$ is the
unique positive root of the real polynomial $z^{3}+2Bz^{2}+(B^{2}-4E)z-D^{2}$.

###### Proof.

(a) if $D=0$ by (2.2) then we have $T=0$ and (2.1) gives
$N=(B\pm\sqrt{B^{2}-4E)}/2$
(b) If $D=0$ and $B^{2}<4E$, by (2.2)we have

$T(T^{2}-(2N-B))=0\ \ \textrm{Note that}\ \ 2N-B>0$

Therefore,

$T=\pm\sqrt{2N-B}$

By (2.1) we have,

$N^{2}-(B+2N-B)N+E=0$ $N^{2}=E\ \ \textrm{implies}\ \ N=\sqrt{E}$

Hence,

$T=\pm\sqrt{2\sqrt{E}-B}\ \ \textrm{and}\ \ N=\sqrt{E}$

(c) If $D\neq 0$ from (2.2) $N=(T^{3}+BT+D)/2T$ plug this $N$ value to (2.1)
we have

$\Big{(}\frac{T^{3}+BT+D}{2T}\Big{)}^{2}-(B+T^{2})\Big{(}\frac{T^{3}+BT+D}{2T}\Big{)}+E=0,$
$\Big{(}T^{3}+BT+D\Big{)}^{2}-2T(B+T^{2})\Big{(}T^{3}+BT+D\Big{)}+4T^{2}E=0,$
$\Big{(}T^{3}+BT+D\Big{)}^{2}-2T(B+T^{2})\Big{(}T^{3}+BT+D\Big{)}+4T^{2}E=0,$
$T^{6}+2BT^{4}+(B^{2}-4E)T^{2}-D^{2}=0,$

Let $T=\pm\sqrt{z}$ Then, by $\textbf{Lemma 2.1}.$ we have $z$ as unique
positive root satisfies cubic equation $z^{3}+2Bz^{2}+(B^{2}-4E)z-D^{2}$. ∎

######  Theorem 2.3.

The solutions of left octonionic quadratic equation $x^{2}+bx+c=0$ can be
obtained by formulas according to the following cases.
case 1. if $b,c\in\mathbb{R}$ and $b^{2}\geq 4c$, then

$x=\frac{-b\pm\sqrt{b^{2}-4c}}{2}$

case 2. if $b,c\in\mathbb{R}$ and $b^{2}<4c,$ then

$x=\frac{1}{2}(-b+\alpha\textbf{i}+\beta\textbf{j}+\gamma\textbf{k}+\delta\textbf{l}+p\textbf{il}+q\textbf{jl}+r\textbf{kl})$
$\ \ \textrm{where}\ \
\alpha^{2}+\beta^{2}+\gamma^{2}+\delta^{2}+p^{2}+q^{2}+r^{2}=4c-b^{2}\ \
\textrm{and}\ \ \alpha,\beta,\gamma,\delta,p,q,r\in\mathbb{R}$

case 3. if $b\in\mathbb{R}$ and $c\notin\mathbb{R}$, then

$\hskip
56.9055ptx={\frac{-b}{2}}\pm{\frac{\rho}{2}}\mp{\frac{c_{1}}{\rho}}\textbf{i}\mp{\frac{c_{2}}{\rho}}\textbf{j}\mp{\frac{c_{3}}{\rho}}\textbf{k}\mp{\frac{c_{4}}{\rho}}\textbf{l}\mp{\frac{c_{5}}{\rho}}\textbf{il}\mp{\frac{c_{6}}{\rho}}\textbf{jl}\mp{\frac{c_{7}}{\rho}}\textbf{kl}$
$\textrm{where}\hskip
2.84526ptc=c_{0}+c_{1}\textbf{i}+c_{2}\textbf{j}+c_{3}\textbf{k}+c_{4}\textbf{l}+c_{5}\textbf{il}+c_{6}\textbf{jl}+c_{7}\textbf{kl}\hskip
5.69054pt\textrm{and}\hskip 5.69054pt$
$\rho=\sqrt{{\bigg{(}b^{2}-4c_{0}+\sqrt{(b^{2}-4c_{0})^{2}+16(c_{1}^{2}+c_{2}^{2}+c_{3}^{2}+c_{4}^{2}+c_{5}^{2}+c_{6}^{2}+c_{7}^{2})}\bigg{)}}/2}$

case 4. if $b\notin\mathbb{R}$, then

$x=\frac{-{\rm Re}\,b}{2}-(b^{\prime}+T)^{-1}(c^{\prime}-N),$

where

$b^{\prime}=b-{\rm Re}\,\ b=\textrm{Im}\ b,\ c^{\prime}=c-\big{(}({\rm
Re}\,b)/2)(b-({\rm Re}\,b)/2\big{)},\ \textrm{and}\ (T,N)\ \textrm{is}\
\textrm{chosen}\ \textrm{as}\ \textrm{follows}.$

  1. 1.

$T=0,N=(B\pm\sqrt{B^{2}-4E})/2$ provided that $D=0$, $B^{2}\geq 4E.$

  2. 2.

$T=\pm\sqrt{2\sqrt{E}-B},N=\sqrt{E}$ provided that $D=0,B^{2}<4E.$

  3. 3.

$T=\pm\sqrt{z},N=(T^{3}+BT+D)/2T$ provided that $D\neq 0$ and $z$ is the
unique positive root of the real polynomial $z^{3}+2Bz^{2}+(B^{2}-4E)z-D^{2}$,
where
$B=b^{\prime}\bar{b^{\prime}}+c^{\prime}+\bar{c^{\prime}}=|b^{\prime}|^{2}+2{\rm
Re}\,c^{\prime},\ E=c^{\prime}\bar{c^{\prime}}=|c^{\prime}|^{2},\
D=\bar{b^{\prime}}c^{\prime}+\bar{c^{\prime}}b^{\prime}=2{\rm
Re}\,(\bar{b^{\prime}}c^{\prime})$, are real numbers.

###### Proof.

Case 1. $b,c\in\mathbb{R}$ and $b^{2}\geq 4c.$ Note that $x$ is a solution if
and only if $q^{-1}xq$ is also a solution for $q\neq 0,$ and hence, there are
at most two solutions, both are real

$\frac{-b\pm\sqrt{b^{2}-4c}}{2}$

Case 2. $b,c\in\mathbb{R}$ and $b^{2}<4c.$ Note that $x$ is a solution if and
only if $q^{-1}xq$ is also a solution for $q\neq 0,$ and there are at least
two complex solutions

$\frac{-b\pm\sqrt{4c-b^{2}}\textbf{i}}{2}$

Hence, the solution set is

$\Bigg{\\{}q^{-1}\Big{(}\frac{-b+\sqrt{4c-b^{2}}\textbf{i}}{2}\Big{)}q:q\neq
0\Bigg{\\}}$

Let $R^{2}=4c-b^{2}>0$ and $q\in\mathbb{O}$

$\Bigg{\\{}q^{-1}\Big{(}\frac{-b+R\textbf{i}}{2}\Big{)}q:q\neq 0\Bigg{\\}}$
$\Bigg{\\{}\frac{-b+q^{-1}R\textbf{i}q}{2}:q\neq 0\Bigg{\\}}$
$\Bigg{\\{}\frac{-b+Rq^{-1}\textbf{i}q}{2}:q\neq 0\Bigg{\\}}$
$Rq^{-1}iq=\alpha\textbf{i}+\beta\textbf{j}+\gamma\textbf{k}+\delta\textbf{l}+p\textbf{il}+q\textbf{jl}+r\textbf{kl}\in{\rm
Im}\,\mathbb{O}$

and

$\overline{Rq^{-1}iq}=-Rq^{-1}iq$

Therefore,

$|Rq^{-1}iq|=(Rq^{-1}iq)\overline{(Rq^{-1}iq)}=-(Rq^{-1}iq)(Rq^{-1}iq)=R^{2}=4c-b^{2}$

implies,

$\alpha^{2}+\beta^{2}+\gamma^{2}+\delta^{2}+p^{2}+q^{2}+r^{2}=4c-b^{2}$

Hence solution set is,

$=\Bigg{\\{}\frac{1}{2}(-b+\alpha\textbf{i}+\beta\textbf{j}+\gamma\textbf{k}+\delta\textbf{l}+p\textbf{il}+q\textbf{jl}+r\textbf{kl}):\alpha^{2}+\beta^{2}+\gamma^{2}+\delta^{2}+p^{2}+q^{2}+r^{2}=4c-b^{2}\Bigg{\\}}$

How we will describe having infinitely many solutions?
For Geometric view,
Consider

$|x|^{2}=\frac{(-b)^{2}}{4}+\frac{\alpha^{2}}{4}+\frac{\beta^{2}}{4}+\frac{\gamma^{2}}{4}+\frac{\delta^{2}}{4}+\frac{p^{2}}{4}+\frac{q^{2}}{4}+\frac{r^{2}}{4}=\frac{b^{2}+4c-b^{2}}{4}=c$

Conclusion: Solutions are set of all points on $S^{7}$ ($7$ \- sphere) in
8-dimension with norm equal to $\sqrt{c}$ (Note that $c\in\mathbb{R}).$

Case 3. $b\in\mathbb{R}$ and $c\notin\mathbb{R}.$ Let

$x=x_{0}+x_{1}\textbf{i}+x_{2}\textbf{j}+x_{3}\textbf{k}+x_{4}\textbf{l}+x_{5}\textbf{il}+x_{6}\textbf{jl}+x_{7}\textbf{kl}$

and

$c=c_{0}+c_{1}\textbf{i}+c_{2}\textbf{j}+c_{3}\textbf{k}+c_{4}\textbf{l}+c_{5}\textbf{il}+c_{6}\textbf{jl}+c_{7}\textbf{kl}$

Then $x^{2}+bx+c=0$ becomes the real system

$(x_{0}^{2}-x_{1}^{2}-x_{2}^{2}-x_{3}^{2}-x_{4}^{2}-x_{5}^{2}-x_{6}^{2}-x_{7}^{2})+bx_{0}+c_{0}=0$
$\Big{(}x_{0}^{2}+\frac{b}{2}\Big{)}^{2}-x_{1}^{2}-x_{2}^{2}-x_{3}^{2}-x_{4}^{2}-x_{5}^{2}-x_{6}^{2}-x_{7}^{2}\big{)}=\frac{b^{2}}{4}-c_{0}$
$(2x_{0}+b)x_{i}=-c_{i}\ \ where\ \ i=1,2,3,4,5,6,7$

Since $c$ is non-real $(2x_{0}+b)$ is non-zero

$(2x_{0}+b)^{2}-4x_{1}^{2}-4x_{2}^{2}-4x_{3}^{2}-4x_{4}^{2}-4x_{5}^{2}-4x_{6}^{2}-4x_{7}^{2}=(b^{2}-4c_{0})$
$(2x_{0}+b)^{4}-4c_{1}^{2}-4c_{2}^{2}-4c_{3}^{2}-4c_{4}^{2}-4c_{5}^{2}-4c_{6}^{2}-4c_{7}^{2}=(2x_{0}+b)^{2}(b^{2}-4c_{0})$
$(2x_{0}+b)^{4}-(b^{2}-4c_{0})(2x_{0}+b)^{2}+4(c_{1}^{2}+c_{2}^{2}+c_{3}^{2}+c_{4}^{2}+c_{5}^{2}+c_{6}^{2}+c_{7}^{2})=0$
$(2x_{0}+b)^{2}=\frac{(b^{2}-4c_{0})\pm\sqrt{(b^{2}-4c_{0})^{2}+16(c_{1}^{2}+c_{2}^{2}+c_{3}^{2}+c_{4}^{2}+c_{5}^{2}+c_{6}^{2}+c_{7}^{2}})}{2}$
$(2x_{0}+b)=\pm\sqrt{\frac{\Big{(}b^{2}-4c_{0}+\sqrt{(b^{2}-4c_{0})^{2}+16(c_{1}^{2}+c_{2}^{2}+c_{3}^{2}+c_{4}^{2}+c_{5}^{2}+c_{6}^{2}+c_{7}^{2}})\Big{)}}{2}}$
$x_{0}=\frac{(-b\pm\rho)}{2}\ \ where\ \rho\neq 0$
$x=\frac{-b}{2}\pm\frac{\rho}{2}\mp\frac{c_{1}}{\rho}\textbf{i}\mp\frac{c_{2}}{\rho}\textbf{j}\mp\frac{c_{3}}{\rho}\textbf{k}\mp\frac{c_{4}}{\rho}\textbf{l}\mp\frac{c_{5}}{\rho}\textbf{il}\mp\frac{c_{6}}{\rho}\textbf{jl}\mp\frac{c_{7}}{\rho}\textbf{kl}$

Case 4. $b\notin\mathbb{R}.$ Rewrite the equation $x^{2}+bx+c=0$ as

$y^{2}+b^{\prime}y+c^{\prime}=0,$

where $y=x+({\rm Re}\,b)/2$, $b^{\prime}=b-\textrm{Re}b\notin\mathbb{R}$ and
$c^{\prime}=c-({\rm Re}\,b/2)\big{(}b-({\rm Re}\,b/2)\big{)}$. By [2], we
observe that the solution of the quadratic equation
$y^{2}+b^{\prime}y+c^{\prime}=0$ also satisfies

$y^{2}-Ty+N=0,$

where $N=\bar{y}y\geq 0$ and $T=y+\bar{y}\in\mathbb{R}.$ Hence,
$(b^{\prime}+T)y+(c^{\prime}-N)=0,$ and so

$y=-(b^{\prime}+T)^{-1}(c^{\prime}-N)$

because $T\in\mathbb{R}$ and $b^{\prime}\notin\mathbb{R}$ implies that
$b^{\prime}+T\neq 0$ To solve for $T$ and $N$, we substitute $y$ back into the
definitions $T=y+\bar{y}$ and $N=\bar{y}y$ and simplify to obtain the real
system

$N^{2}-(B+T^{2})N+E=0,$ $T^{3}+(B-2N)T+D=0,$

where
$B=b^{\prime}\bar{b^{\prime}}+c^{\prime}+\bar{c^{\prime}}=|b^{\prime}|^{2}+2{\rm
Re}\,c^{\prime},E=c^{\prime}\bar{c^{\prime}}=|c^{\prime}|^{2},D=\bar{b^{\prime}}c^{\prime}+\bar{c^{\prime}}b^{\prime}=2{\rm
Re}\,\bar{b^{\prime}}c^{\prime}$ are real numbers. Note that
$E=|c^{\prime}|^{2}\geq 0.$ if $B<0,$ then $c^{\prime}+\bar{c^{\prime}}<0$ and
$B^{2}-4E=|b^{\prime}|^{2}B+|b^{\prime}|^{2}(c^{\prime}+\bar{c^{\prime}})+(c^{\prime}-\bar{c^{\prime}})^{2}\leq
0$ because of the face that $(c^{\prime}-\bar{c^{\prime}})^{2}\leq 0.$ It
follows that $B^{2}-4E<0,$ otherwise $B^{2}-4E=0$ and so
$|b^{\prime}|^{2}B=|b^{\prime}|^{2}(c^{\prime}+\bar{c^{\prime}})=(c^{\prime}-\bar{c^{\prime}}=0,$
i.e., $b^{\prime}=0\in\mathbb{R},$ a contradiction. Hence, by lemma $(2.2)$,
such system can be solved explicitly as claimed. Consequently, $x=(-{\rm
Re}\,b)/2-(b^{\prime}+T)^{-1}(c^{\prime}-T).$ ∎

Example 1. Consider the equation $x^{2}+5x+6=0$, Find solutions
$x\in\mathbb{O}$
Solution:
$b=5$ and $c=6$, therefore, $b^{2}>4c.$ From theorem 2.3, case 1.

$x=\frac{-5\pm\sqrt{5^{2}-4\times 6}}{2}=\frac{-5\pm 1}{2}=-3\hskip
5.69054pt\textrm{or}-2$

Example 2. Consider the equation $x^{2}+1=0$, Find solutions $x\in\mathbb{O}$
Solution:
$b=0$ and $c=1$, therefore, $b^{2}<4c.$ From theorem 2.3, case 2.

$x=\alpha\textbf{i}+\beta\textbf{j}+\gamma\textbf{k}+\delta\textbf{l}+p\textbf{il}+q\textbf{jl}+r\textbf{kl}$
$\hskip 28.45274pt\textrm{where}\ \
\alpha^{2}+\beta^{2}+\gamma^{2}+\delta^{2}+p^{2}+q^{2}+r^{2}=1\hskip
5.69054pt\textrm{and}\hskip
5.69054pt\alpha,\beta,\gamma,\delta,p,q,r\in\mathbb{R}$

so there are infinitely many solutions to the equation $x^{2}+1=0$ in
$\mathbb{O}$.
Example 3. Consider the equation $x^{2}-x+\textbf{l}=0$, Find solutions
$x\in\mathbb{O}$
Solution:
$b=-1$ and $c=\textbf{l}$, From theorem 2.3, case 3.

$x=\frac{1}{2}\pm\frac{\rho}{2}\mp\frac{\textbf{l}}{\rho}$

where

$\rho=\sqrt{\frac{1+\sqrt{17}}{2}}$

Example 4. Consider the equation $x^{2}+\textbf{l}x+\textbf{j}=0$, Find
solutions $x\in\mathbb{O}$
Solution:
$b=\textbf{l}\notin\mathbb{R}$ and $c=\textbf{j}$, From theorem 2.3, case 4

$x=-\frac{{\rm Re}\,b}{2}-(b^{\prime}+T)^{-1}(c^{\prime}-N)$

since ${\rm Re}\,b=0$ we have $x=-(b^{\prime}+T)^{-1}(c^{\prime}-N)$ where
note that $B^{2}<4E$
Therefore, $T=\pm\sqrt{2\sqrt{E}-B}=\pm 1$ and $N=\sqrt{E}=1$
Thus, $x=-(\textbf{l}+1)^{-1}(\textbf{j}-1)$ or
$x=-(\textbf{l}-1)^{-1}(\textbf{j}-1)$
Finally,

$x=\frac{(\textbf{l}-1)(1-\textbf{j})}{-2}=\frac{\textbf{l}+\textbf{jl}-1+\textbf{j}}{-2}=\frac{1-\textbf{j}-\textbf{l}-\textbf{jl}}{2}$

or

$x=\frac{(\textbf{l}+1)(1-\textbf{j})}{-2}=\frac{\textbf{l}+\textbf{jl}+1-\textbf{j}}{-2}=\frac{-1+\textbf{j}-\textbf{l}-\textbf{jl}}{2}$

##  3 Application

###  3.1 Left spectrum of $2\times 2$ octonionic matrix

We can find left spectrum of $2\times 2$ octonionic matrices by solving
corresponding octonionic quadratic equation. Let’s begin with following lemma

######  Lemma 3.1.

For $p$ and $q\in\mathbb{O},\ \sigma_{l}(pI+qA)=\\{p+qt:t\in\sigma_{l}(A)\\}$,
where $I$ is the identity matrix.

###### Proof.

Let $t$ be a left eigenvalue of $A$ with eigenvector $v$,

$Av=tv$

consider

$(pI+qA)v=pIv+qAv=pv+qtv=(p+qt)v$

Therefore, $p+qt$ where $t\in\sigma_{l}(A)$ represent the set of left spectrum
of $(pI+qA).$ ∎

######  Theorem 3.2.

Let $A=\begin{bmatrix}a&b\\\ c&d\end{bmatrix}$, where $a,b,c,\in\mathbb{O}$.
(i) if $bc=0$, then $\sigma_{l}(A)=\\{a,d\\}$
(ii) if $bc\neq 0$, then
$\sigma_{l}(A)=\\{a+bt:t^{2}+b^{-1}(a-d)t-b^{-1}c=0\\}$

###### Proof.

(i) $A$ is a triangular matrix, then results follows.
(ii) Using Lemma 3.1, we have

$\sigma_{l}(A)=\sigma_{l}\begin{bmatrix}aI+b\begin{bmatrix}0&1\\\
b^{-1}c&b^{-1}(d-a)\end{bmatrix}\end{bmatrix}$

Let $t$ be any left eigenvalue of $\begin{bmatrix}0&1\\\
b^{-1}c&b^{-1}(d-a)\end{bmatrix}$, then there exists non-zero vector
$v=\begin{bmatrix}x\\\ y\end{bmatrix}$ such that

$\begin{bmatrix}0&1\\\ b^{-1}c&b^{-1}(d-a)\end{bmatrix}\begin{bmatrix}x\\\
y\end{bmatrix}=t\begin{bmatrix}x\\\ y\end{bmatrix}$ $y=tx,$ (3.1)
$b^{-1}cx+b^{-1}(d-a)y=ty$ (3.2)

From equations (3.1) and (3.2), we have

$t^{2}+b^{-1}(a-d)t-b^{-1}c=0.$ (3.3)

We can use Theorem $(3.2)$ to find roots of $(3.3)$ and explicitly find left
spectrum of given $2\times 2$ octonionic Matrix $A$. ∎

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