The Moore-Penrose inverses of split quaternions

TL;DR

This paper introduces the Moore-Penrose inverse for split quaternions and applies it to solve specific linear equations, also characterizing similarity and consimilarity of split quaternions.

Contribution

It extends the concept of Moore-Penrose inverse to split quaternions and provides solutions to linear equations involving them, along with conditions for similarity and consimilarity.

Findings

Solutions to equations $axb=d$, $xa=bx$, $xa=bar{x}$ using Moore-Penrose inverse

Necessary and sufficient conditions for similarity of split quaternions

Necessary and sufficient conditions for consimilarity of split quaternions

Abstract

In this paper, we find the roots of lightlike quaternions. By introducing the concept of the Moore-Penrose inverse in split quaternions, we solve the linear equations $axb=d$, $xa=bx$ and $xa=b\bar{x}$. Also we obtain necessary and sufficient conditions for two split quaternions to be similar or consimilar.

Equations

$${\mathbb{H}}_s=\{q=q_0+q_1\mathbf{i}+q_2\mathbf{j}+q_3\mathbf{k},q_i\in \mathbb{R},i=0,1,2,3\},$$

$${\mathbf{i}}^2=-{\mathbf{j}}^2=-{\mathbf{k}}^2=-1,\mathbf{i}\mathbf{j}=\mathbf{k}=-\mathbf{j}\mathbf{i},\mathbf{j}\mathbf{k}=-\mathbf{i}=-\mathbf{k}\mathbf{j},\mathbf{k}\mathbf{i}=\mathbf{j}=-\mathbf{i}\mathbf{k}.$$

$$K(q)=\Im (q{)}^2=-q_1^2+q_2^2+q_3^2.$$

$$q=(q_0+q_1\mathbf{i})+(q_2+q_3\mathbf{i})\mathbf{j}=z_1+z_2\mathbf{j},z_1,z_2\in \mathbb{C}.$$

$$\overrightarrow{qx}=L_q\overrightarrow{x},\overrightarrow{xq}=R_q\overrightarrow{x},$$

$$L(q)=\left(\begin{array}[]{cccc}q_{0}&-q_{1}&q_{2}&q_{3}\\ q_{1}&q_{0}&q_{3}&-q_{2}\\ q_{2}&q_{3}&q_{0}&-q_{1}\\ q_{3}&-q_{2}&q_{1}&q_{0}\end{array}\right),\ \ R(q)=\left(\begin{array}[]{cccc}q_{0}&-q_{1}&q_{2}&q_{3}\\ q_{1}&q_{0}&-q_{3}&q_{2}\\ q_{2}&-q_{3}&q_{0}&q_{1}\\ q_{3}&q_{2}&-q_{1}&q_{0}\end{array}\right).$$

$$Z({\mathbb{H}}_s)=\{q\in {\mathbb{H}}_s:I_q=0\};$$

$$N({\mathbb{H}}_s)=\{q\in {\mathbb{H}}_s:\text{mbox}thereisanintegern\text{mbox}suchthat{q}^n=0\};$$

$$I({\mathbb{H}}_s)=\{q\in {\mathbb{H}}_s:q^2=q\}.$$

$$q^n=(2\Re (q){)}^{n-1}q,q\in Z({\mathbb{H}}_s).$$

$$q_0^2-q_1^2+q_2^2+q_3^2=q_0,2q_0{q}_1=q_1,2q_0{q}_2=q_2,2q_0{q}_3=q_3.$$

$$q=r(e^{\mathbf{i}\alpha }+e^{\mathbf{i}\beta }\mathbf{j}),r\ge 0,\alpha ,\beta \in [0,2\pi ).$$

$$(2r_1\cos {\alpha }_1{)}^{n-1}{r}_1(e^{\mathbf{i}{\alpha }_1}+e^{\mathbf{i}{\beta }_1}\mathbf{j})=r(e^{\mathbf{i}\alpha }+e^{\mathbf{i}\beta }\mathbf{j}).$$

$$(2r_1\cos {\alpha }_1{)}^{n-1}{r}_1\cos {\alpha }_1=r\cos \alpha ,(2r_1\cos {\alpha }_1{)}^{n-1}{r}_1\sin {\alpha }_1=r\sin \alpha ,$$

$$(2r_1\cos {\alpha }_1{)}^{n-1}{r}_1\cos {\beta }_1=r\cos \beta ,(2r_1\cos {\alpha }_1{)}^{n-1}{r}_1\sin {\beta }_1=r\sin \beta .$$

$$AXA=A,XAX=X,(AX{)}^{\ast }=AX,(XA{)}^{\ast }=XA,$$

$$x=A^{+}b+(E_n-A^{+}A)y,\forall y\in {\mathbb{R}}^n.$$

$$a^{+}=\left\{\begin{array}[]{ll}0,&\hbox{provided a=0;}\\ \frac{\bar{c_{1}}-c_{2}{\bf j}}{|c_{1}|^{2}-|c_{2}|^{2}}=\frac{\bar{a}}{I_{a}},&\hbox{provided $|c_{1}|^{2}-|c_{2}|^{2}\neq 0$;}\\ \frac{\bar{c_{1}}+c_{2}{\bf j}}{4|c_{1}|^{2}},&\hbox{provided $|c_{1}|^{2}-|c_{2}|^{2}=0$.}\\ \end{array}\right.$$

$$a{a}^{+}a=a,a^{+}a{a}^{+}=a^{+},a{a}^{+}=\frac{1}{2}(1+\frac{c_2}{\overline{c_1}}\mathbf{j}),a^{+}a=\frac{1}{2}(1+\frac{c_2}{c_1}\mathbf{j}).$$

$$L(a)L(a^{+})L(a)=L(a),L(a^{+})L(a)L(a^{+})=L(a^{+}),L(a)L(a^{+})=\big{(}L(a)L(a^{+})\big{)}^{*},L(a^{+})L(a)=\big{(}L(a^{+})L(a)\big{)}^{*};$$

$$R(a)R(a^{+})R(a)=R(a),R(a^{+})R(a)R(a^{+})=R(a^{+}),R(a)R(a^{+})=\big{(}R(a)R(a^{+})\big{)}^{*},R(a^{+})R(a)=\big{(}R(a^{+})R(a)\big{)}^{*}.$$

$$L(a)^{+}=L(a^{+}),\ R(b)^{+}=R(b^{+}),\ \big{(}L(a)R(b)\big{)}^{+}=L(a^{+})R(b^{+}).$$

$$(1+\frac{c_2}{\overline{c_1}}\mathbf{j})d(1+\frac{u_2}{u_1}\mathbf{j})=4d,$$

$$x=\frac{\overline{c_1}d\overline{u_1}}{4|c_1{|}^2|u_1{|}^2}+y-\frac{1}{4}(1+\frac{c_2}{c_1}\mathbf{j})y(1+\frac{u_2}{\overline{u_1}}\mathbf{j}),\forall y\in {\mathbb{H}}_s.$$

$$L(a)R(b)\big{(}L(a)R(b)\big{)}^{+}\overrightarrow{d}=\overrightarrow{d}.$$

$$\overrightarrow{x}=\big{(}(L(a)R(b)\big{)}^{+}\overrightarrow{d}+(E_{4}-\Big{(}\big{(}L(a)R(b)\big{)}^{+}L(a)R(b)\Big{)}\overrightarrow{y},\forall y\in{\mathbb{H}}_{s}.$$

$$x=a^{+}d{b}^{+}+(y-a^{+}ayb{b}^{+}),\forall y\in {\mathbb{H}}_s.$$

$$x=\frac{(\overline{c_1}+c_2\mathbf{j})d(\overline{u_1}+u_2\mathbf{j})}{16|c_1{|}^2|u_1{|}^2}+y-\frac{1}{4}(1+\frac{c_2}{c_1}\mathbf{j})y(1+\frac{u_2}{\overline{u_1}}\mathbf{j}),\forall y\in {\mathbb{H}}_s.$$

$$x=(1-a^{+}a)y=\frac{1}{2}(1-\frac{c_2}{c_1}\mathbf{j})y,\forall y\in {\mathbb{H}}_s.$$

$$x=a^{+}d+(1-a^{+}a)y=\frac{\overline{c_1}+c_2\mathbf{j}}{4|c_1{|}^2}d+\frac{1}{2}(1-\frac{c_2}{c_1}\mathbf{j})y,\forall y\in {\mathbb{H}}_s.$$

Full text

The Moore-Penrose inverses of split quaternions

Wensheng Cao, Zhenhu Chang

School of Mathematics and Computational Science,

Wuyi University, Jiangmen, Guangdong 529020, P.R. China

e-mail: wenscao@aliyun.com

Abstract    In this paper, we find the roots of lightlike quaternions. By introducing the concept of the Moore-Penrose inverse in split quaternions, we solve the linear equations $axb=d$, $xa=bx$ and $xa=b\bar{x}$. Also we obtain necessary and sufficient conditions for two split quaternions to be similar or consimilar.

Keywords and phrases:   Roots of lightlike quaternion, Moore-Penrose inverse, similar, consimilar

MR(2000) Subject Classifications:  15A33; 11R52

1 Introduction

Let ${\mathbb{R}}$ and ${\mathbb{C}}$ be the field of real numbers and complex numbers, respectively. The split quaternions are elements of a 4-dimensional associative algebra introduced by James Cockle [1] in 1849. Split quaternions can be represented as

[TABLE]

where $1,{\bf i},{\bf j},{\bf k}$ are basis of ${\mathbb{H}}_{s}$ satisfying the equalities:

[TABLE]

Let $\bar{q}=q_{0}-q_{1}{\bf i}-q_{2}{\bf j}-q_{3}{\bf k}$ be the conjugate of $q$ and $I_{q}=\bar{q}q=q\bar{q}=q_{0}^{2}+q_{1}^{2}-q_{2}^{2}-q_{3}^{2}.$ A split quaternion is spacelike, timelike or lightlike if $I_{q}<0,I_{q}>0$ or $I_{q}=0$, respectively. We define $\Re(q)=(q+\bar{q})/2=q_{0}$ to be the real part of $q$ and $\Im(q)=(q-\bar{q})/2=q_{1}{\bf i}+q_{2}{\bf j}+q_{3}{\bf k}$ to be the imaginary part of $q$. It is easy to verify that $I_{pq}=I_{p}I_{q}.$ Let $q^{\prime}=q_{0}-q_{1}{\bf i}+q_{2}{\bf j}+q_{3}{\bf k}$ be the prime of $q$. Then $I_{q^{\prime}}=I_{q}$. For $q=q_{0}+q_{1}{\bf i}+q_{2}{\bf j}+q_{3}{\bf k}$, we define

[TABLE]

Obviously ${\mathbb{C}}={\mathbb{R}}+{\mathbb{R}}{\bf i}\subset{\mathbb{H}}_{s}$ and ${\bf j}z=\bar{z}{\bf j}$ for $z\in{\mathbb{C}}.$ A split quaternion can be written as

[TABLE]

Denote $\overrightarrow{x}=(x_{0},x_{1},x_{2},x_{3})^{T}$ for $x=x_{0}+x_{1}{\bf i}+x_{2}{\bf j}+x_{3}{\bf k}\in{\mathbb{H}}_{s}$, where T denotes the transpose of a matrix. Using the multiplication rule (1), we have the following formulas:

[TABLE]

where

[TABLE]

Unlike the Hamilton quaternion algebra, the split quaternions contain nontrivial zero divisors, nilpotent elements, and idempotents. Accordingly we define the following sets of split quaternions:

[TABLE]

[TABLE]

[TABLE]

In this paper we mainly concentrate on the properties of lightlike split quaternions or some singular matices induced by split quaternions. In Section 2, we will obtain the characteristics of the sets $Z({\mathbb{H}}_{s}),N({\mathbb{H}}_{s}),I({\mathbb{H}}_{s})$ and find the roots of lightlike quaternions. In Section 3, we will obtain some properties of the Moore-Penrose inverse and solve the linear equation $axb=d$. In Section 4, we solve the linear equations $xa=bx$ and $xa=b\bar{x}$. As applications, we obtain some necessary and sufficient conditions for two split quaternions to be similar or consimilar.

2 The root of a split quaternion in $Z({\mathbb{H}}_{s})$

Proposition 2.1**.**

$q^{2}=2\Re(q)q-I_{q},q\in{\mathbb{H}}_{s}$* and*

[TABLE]

Proof.

It is obvious that $q^{2}=q(2\Re(q)-\bar{q})=2\Re(q)q-I_{q}$. If $q\in Z({\mathbb{H}}_{s})$ then $I_{q}=0$, and therefore $q^{n}=(2\Re(q))^{n-1}q.$ ∎

Proposition 2.2**.**

$N({\mathbb{H}}_{s})=\{q=q_{1}{\bf i}+q_{2}{\bf j}+q_{3}{\bf k}\ \mbox{with}\ q_{1}^{2}-q_{2}^{2}-q_{3}^{2}=0\}.$**

Proof.

Let $q=q_{0}+q_{1}{\bf i}+q_{2}{\bf j}+q_{3}{\bf k}\in N({\mathbb{H}}_{s})$. Then $I_{q}=0$ and therefore $q\in Z({\mathbb{H}}_{s})$. By proposition 2.1, $\Re(q)=0$ which implies that $I_{q}=q_{1}^{2}-q_{2}^{2}-q_{3}^{2}=0$. ∎

Proposition 2.3**.**

$I({\mathbb{H}}_{s})=\{0,1\}\cup\{\frac{1}{2}+q_{1}{\bf i}+q_{2}{\bf j}+q_{3}{\bf k}\ \mbox{with}\ \frac{1}{4}+q_{1}^{2}-q_{2}^{2}-q_{3}^{2}=0\}.$**

Proof.

Let $q=q_{0}+q_{1}{\bf i}+q_{2}{\bf j}+q_{3}{\bf k}\in I({\mathbb{H}}_{s})$. By $q^{2}=q$, we have the following equations:

[TABLE]

If $q_{1}^{2}+q_{2}^{2}+q_{3}^{2}\neq 0$ then $q_{0}=\frac{1}{2}$ and $\frac{1}{4}+q_{1}^{2}-q_{2}^{2}-q_{3}^{2}=0$. If $q_{1}^{2}+q_{2}^{2}+q_{3}^{2}=0$ then $q$ is [math] or $1$. ∎

Let $q=z_{1}+z_{2}{\bf j}\in Z({\mathbb{H}}_{s})$. Then $I_{q}=|z_{1}|^{2}-|z_{2}|^{2}=0$, and therefore $q$ can be written as

[TABLE]

Theorem 2.1**.**

Let $q=r(e^{{\bf i}\alpha}+e^{{\bf i}\beta}{\bf j}),r>0,\alpha,\beta\in[0,2\pi)$ and $n\geq 2$. Then the equation $w^{n}=q$ has solution in the following cases:

(1) $w=\sqrt[n]{\frac{r}{(2\cos\alpha)^{n-1}}}(e^{{\bf i}\alpha}+e^{{\bf i}\beta}{\bf j})$ if $\cos\alpha>0$; $\mbox{or}\ \cos\alpha<0,n\ \mbox{is odd}$;

(2) $w=\sqrt[n]{\frac{r}{(2\cos\alpha)^{n-1}}}(-e^{{\bf i}\alpha}-e^{{\bf i}\beta}{\bf j})$ if $\cos\alpha>0,n\ \mbox{is even}$.

Proof.

Let $w$ be a root of the equation $w^{n}=q$. Then $(I_{w})^{n}=I_{q}=0$ and therefore $w\in Z({\mathbb{H}}_{s})$. So $w$ can be written as $w=r_{1}(e^{{\bf i}\alpha_{1}}+e^{{\bf i}\beta_{1}}{\bf j}),r_{1}>0,\alpha_{1},\beta_{1}\in[0,2\pi)$. It follows from Proposition 2.1 that

[TABLE]

That is

[TABLE]

[TABLE]

It is obvious that there is no solution if $\cos\alpha=0$. So we only need to consider the case $\cos\alpha\neq 0$. Thus $\tan\alpha_{1}=\tan\alpha.$ In this case, the above four equations are solvable if and only if one of the following conditions holds:

• (1)

$\alpha_{1}=\alpha,\beta_{1}=\beta,r_{1}=\sqrt[n]{\frac{r}{(2\cos\alpha)^{n-1}}}\ \mbox{provided}\ \cos\alpha>0;$

• (2)

$\alpha_{1}=\alpha\pm\pi\in[0,2\pi),\beta_{1}=\beta\pm\pi\in[0,2\pi),r_{1}=\sqrt[n]{\frac{r}{(2\cos\alpha)^{n-1}}}\ \mbox{provided}\ \cos\alpha>0,n\ \mbox{is even};$

• (3)

$\alpha_{1}=\alpha,\beta_{1}=\beta,r_{1}=\sqrt[n]{\frac{r}{(2\cos\alpha)^{n-1}}}\ \mbox{provided}\ \cos\alpha<0,n\ \mbox{is odd}.$

The above three cases conclude the proof of Theorem 2.1. ∎

Remark 2.1**.**

Proposition 2.2, Theorem 2.1 and the results in [4] show how to find the roots of any split quaternions.

3 The Moore-Penrose inverse of elements in ${\mathbb{H}}_{s}$

We recall that the Moore-Penrose inverse of a complex matrix $A$ is the unique complex matrix $X$ satisfying the following equations:

[TABLE]

where ∗ is the conjugate of a matrix. We denote the Moore-Penrose inverse of $A$ by $A^{+}$. The following lemma is well known.

Lemma 3.1**.**

Let $A\in{\mathbb{R}}^{m\times n},b\in{\mathbb{R}}^{m}$ and $E_{n}$ the identity matrix of order $n$. Then the linear equation $Ax=b$ has a solution if and only if $AA^{+}b=b$, furthermore the general solution is

[TABLE]

Definition 3.1**.**

The Moore-Penrose inverse of $a=c_{1}+c_{2}{\bf j}\in{\mathbb{H}}_{s}$ is defined to be

[TABLE]

For $0\neq a=c_{1}+c_{2}{\bf j}\in Z({\mathbb{H}}_{s})$, we have the following facts:

[TABLE]

Since $L(a)R(b)=R(b)L(a)$ for any $a,b\in{\mathbb{H}}_{s}$, We can verify directly the following proposition.

Proposition 3.1**.**

Let $a=a_{0}+a_{1}{\bf i}+a_{2}{\bf j}+a_{3}{\bf k}\in{\mathbb{H}}_{s}$. Then

[TABLE]

[TABLE]

[TABLE]

Proposition 3.1 implies that the concept of Moore-Penrose inverse of split quaternions is well defined.

Theorem 3.1**.**

Let $0\neq a=c_{1}+c_{2}{\bf j}\in Z({\mathbb{H}}_{s})$ and $0\neq b=u_{1}+u_{2}{\bf j}\in Z({\mathbb{H}}_{s})$. Then the equation $axb=d$ is solvable if and only if

[TABLE]

in which case all solutions are given by

[TABLE]

Proof.

It is obvious that $axb=d$ is equivalent to $L(a)R(b)\overrightarrow{x}=\overrightarrow{d}$. By Lemma 3.1 $axb=d$ is solvable if and only if

[TABLE]

Returning to quaternion form by Proposition 3.1, we have $aa^{+}db^{+}b=d$. Using (8), we can rewrite $aa^{+}db^{+}b=d$ as (10). By Lemma 3.1, the general solution is

[TABLE]

Hence the general solution can be expressed as

[TABLE]

That is

[TABLE]

By (10) we have $\frac{(\bar{c_{1}}+c_{2}{\bf j})d(\bar{u_{1}}+u_{2}{\bf j})}{16|c_{1}|^{2}|u_{1}|^{2}}=\frac{\bar{c_{1}}d\bar{u_{1}}}{4|c_{1}|^{2}|u_{1}|^{2}}$. ∎

In similar way, we have the following corollaries.

Corollary 3.1**.**

Let $0\neq a=c_{1}+c_{2}{\bf j}\in Z({\mathbb{H}}_{s})$. Then the equation $ax=0$ has solutions:

[TABLE]

Corollary 3.2**.**

Let $0\neq a=c_{1}+c_{2}{\bf j}\in Z({\mathbb{H}}_{s})$. Then the equation $ax=d$ is solvable if and only if $aa^{+}d=\frac{1}{2}(1+\frac{c_{2}}{\bar{c_{1}}}{\bf j})d=d$, in which case all solutions are given by

[TABLE]

Corollary 3.3**.**

Let $0\neq a=c_{1}+c_{2}{\bf j}\in Z({\mathbb{H}}_{s})$. Then the equation $xa=d$ is solvable if and only if $da^{+}a=\frac{1}{2}d(1+\frac{c_{2}}{c_{1}}{\bf j})=d$, in which case all solutions are given by

[TABLE]

4 Similarity and consimilarity

Definition 4.1**.**

We say that two split quaternions $a,b\in{\mathbb{H}}_{s}$ are similar (resp. consimilar) if and only if there exists a $q\in{\mathbb{H}}_{s}-Z({\mathbb{H}}_{s})$ such that $qa=bq$ (resp. $aq=\bar{q}b$).

It is obvious that $xa=bx$ is equivalent to

[TABLE]

Noting the linear equations studied in Section 3, we assume that $a,b\in{\mathbb{H}}_{s}-{\mathbb{R}}$ for $xa=bx$ in this section.

We can verify the following proposition directly.

Proposition 4.1**.**

Let $T(a,b)=R(a)-L(b)$. Then the eigenvalues of $T(a,b)$ are

[TABLE]

and

[TABLE]

$\det(T(a,b))=0$* if only if one of the following conditions holds:*

• (1)

$a_{0}=b_{0}$* and $K(a)=K(b)$; in this case $\mathrm{rank}(T(a,b))=2$.*

• (2)

$a_{0}-b_{0}\neq 0$* and $\det(T(a,b))=0$; in this case $K(a),K(b)\geq 0$ and $\mathrm{rank}(T(a,b))=3$.*

Example 4.1**.**

Two examples of the $\det(T(a,b))=0$: (1) $a=1+3{\bf i}+2{\bf j}+{\bf k},b=1+3{\bf i}+{\bf j}+2{\bf k}$, $K(a)=K(b)=-4$, $\mathrm{rank}(T(a,b))=2$; (2) $a=2+{\bf i}+{\bf k},b=1+{\bf k}$, $K(a)=0,K(b)=1$, $\mathrm{rank}(T(a,b))=3$.

Lemma 4.1**.**

Let $a,b\in{\mathbb{H}}_{s}-{\mathbb{R}}$ and $T=R(a)-L(b)$. If $a_{0}=b_{0},K(a)=K(b)$ then

[TABLE]

Proof.

It follows from $a,b\in{\mathbb{H}}_{s}-{\mathbb{R}}$ that $|\Im(a)|^{2}+|\Im(b)|^{2}\neq 0$. Let $S=R(a^{\prime})-L(b^{\prime})$. Since $a_{0}=b_{0}$, we have S=R\big{(}\Im(a^{\prime})\big{)}-L\big{(}\Im(b^{\prime})\big{)} and T=R\big{(}\Im(a)\big{)}-L\big{(}\Im(b)\big{)}. Note that

[TABLE]

[TABLE]

[TABLE]

Hence

[TABLE]

Similarly we can verify that $STS=2(|\Im(a)|^{2}+|\Im(b)|^{2})S$, $ST$ and $TS$ are symmetric matrices. ∎

Theorem 4.1**.**

Let $a,b\in{\mathbb{H}}_{s}-{\mathbb{R}}$ and $a_{0}=b_{0},K(a)=K(b)$. Then the general solution of linear equation $xa=bx$ is

[TABLE]

Proof.

It follows from Lemma 4.1 that the general solution of linear equation $xa=bx$ is $\overrightarrow{x}=(E_{4}-T^{+}T)\overrightarrow{y}$. Returning to quaternion form by Proposition 3.1, we get the formula (13). ∎

Theorem 4.2**.**

Let $a=c_{1}+c_{2}{\bf j},b=u_{1}+u_{2}{\bf j}\in{\mathbb{H}}_{s}-{\mathbb{R}}$, $a_{0}-b_{0}\neq 0$ and $\det(T(a,b))=0$. Then the general solution of linear equation $xa=bx$ is

[TABLE]

Proof.

It follows from Proposition 4.1 that \det\big{(}R(a)-L(b)\big{)}=\det\big{(}R(a)-L(\bar{b})\big{)}. Let

[TABLE]

Then $I_{p}=\det(T(a,b))=0,$ which implies that $0\neq p\in Z({\mathbb{H}}_{s})$. Note that

[TABLE]

Let $x_{1}$ be the general solution of the equation $xp=0$. Then \big{(}R(a)-L(b)\big{)}\big{(}R(a)-L(\bar{b})\big{)}\overrightarrow{x_{1}}=\overrightarrow{0}. By Corollary 3.3,

[TABLE]

It follows from $\mathrm{rank}(T(a,b))=3$ that the general solution of $xa=bx$ is $x=x_{1}a-\bar{b}x_{1}$, which is just (14). ∎

Remark 4.1**.**

Kula and Yayli [3, Theorem 5.5] didn’t treat the case $a_{0}=b_{0},K(a)=K(b)=0$; for example $a=1+5{\bf i}+3{\bf j}+4{\bf k},b=1+13{\bf i}+12{\bf j}+5{\bf k}$. Also they overlooked the case of Theorem 4.2, therefore [3, Proposition 5.3 (ii)] should be amended. For example, $a=1+5{\bf i}+5{\bf j}+2{\bf k},b=2+{\bf i}+{\bf j}+3{\bf k}$; in this case, the eigenvalues of $R(a)-L(b)$ are $-6,-2,0,4$ and $rank(R(a)-L(b))=3$. The solutions of $xa=bx$ can be represented by

[TABLE]

We mention that taking $y=1$ leads to $r=2$ in the above formula.

Lemma 4.2**.**

Let $a=a_{0}+a_{1}{\bf i}+a_{2}{\bf j}+a_{3}{\bf k}\in{\mathbb{H}}_{s}-{\mathbb{R}}$. Then there exists a $q\in{\mathbb{H}}_{s}-Z({\mathbb{H}}_{s})$ such that

[TABLE]

Proof.

Kula and Yayli [3, Propositions 5.1 and 5.2] treated the case $K(a)\neq 0$. We only need to consider the case $K(a)=0$. We at first show that we can conjugate $a$ to $a_{0}+a_{1}{\bf i}-a_{1}{\bf j}$.

If $a_{3}=0$ then $a=a_{0}+a_{1}{\bf i}+a_{2}{\bf j}$ and $|a_{1}|=|a_{2}|$. Note that

[TABLE]

Hence we can find a $p\in{\mathbb{H}}_{s}-Z({\mathbb{H}}_{s})$ such that $pap^{-1}=a_{0}+a_{1}{\bf i}-a_{1}{\bf j}$.

If $a_{3}\neq 0$ then $a_{1}^{2}=a_{2}^{2}+a_{3}^{2}>a_{2}^{2}$. In Theorem 4.1, letting $b=a_{0}+a_{1}{\bf i}-a_{1}{\bf j}$ and taking $y=2a_{1}(1+{\bf k})$ in (13), we get $p=a_{1}-a_{2}+a_{3}{\bf i},I_{p}=2a_{1}(a_{1}-a_{2})\neq 0.$ That is $pap^{-1}=a_{0}+a_{1}{\bf i}-a_{1}{\bf j}$.

If $a_{1}=1$ or $a_{1}=-1$, then by (16), we can find a $q$ such that $qaq^{-1}=a_{0}+{\bf i}+{\bf j}$. Otherwise, letting $a=a_{0}+a_{1}{\bf i}-a_{1}{\bf j}$ and $b=a_{0}+{\bf i}+{\bf j}$ and taking $y=\frac{2(a_{1}^{2}+1){\bf i}}{a_{1}+1}$ in (13), we get $q=(1+a_{1}){\bf i}+(1-a_{1}){\bf j}$ such that $q(a_{0}+a_{1}{\bf i}-a_{1}{\bf j})q^{-1}=a_{0}+{\bf i}+{\bf j}$. ∎

The fact that a real number is similar to itself, together with Lemma 4.2, implies the following theorem.

Theorem 4.3**.**

Two split quaternions $a,b\in{\mathbb{H}}_{s}-{\mathbb{R}}$ are similar if and only if

[TABLE]

Note that $xa=b\bar{x}$ is equivalent to \big{(}R(a)-L(b)F\big{)}\overrightarrow{x}=0. We can verify the following proposition directly.

Proposition 4.2**.**

Let $F=diag(1,-1,-1,-1)$ and $S(a,b)=R(a)-L(b)F$. Then the eigenvalues of $S(a,b)$ are

[TABLE]

[TABLE]

and

[TABLE]

$\det(S(a,b))=0$* if and only if one of the following conditions holds:*

• (1)

$I_{a}=I_{b}$* and $\bar{a}+b=0$; in this case $\mathrm{rank}(S(a,b))=1$.*

• (2)

$I_{a}=I_{b}$* and $I_{\bar{a}+b}\neq 0$; in this case $\mathrm{rank}(S(a,b))=3$.*

• (3)

$I_{a}=I_{b}$* and $0\neq\bar{a}+b\in Z({\mathbb{H}}_{s})$; in this case $\mathrm{rank}(S(a,b))=3$.*

• (4)

$I_{a}\neq I_{b}$* and $0\neq\bar{a}+b\in Z({\mathbb{H}}_{s})$; in this case $\mathrm{rank}(S(a,b))=3$.*

Example 4.2**.**

Four examples of $\det(S(a,b))=0$: (1) $a=1+2{\bf i}+3{\bf j}+4{\bf k},b=-1+2{\bf i}+3{\bf j}+4{\bf k}$, $\bar{a}+b=0$, $\mathrm{rank}(S(a,b))=1$; (2) $a=1+2{\bf i}+3{\bf j}+4{\bf k},b=2+{\bf i}+3{\bf j}+4{\bf k}$, $I_{\bar{a}+b}=10$, $I_{a}=I_{b}=-20,\mathrm{rank}(S(a,b))=3$; (3) $a=1+2{\bf i}+3{\bf j}+4{\bf k},b=-2+{\bf i}+4{\bf j}+3{\bf k}$, $I_{\bar{a}+b}=0$, $I_{a}=I_{b}=-20,\mathrm{rank}(S(a,b))=3$; (4) $a=1+2{\bf i}+3{\bf j}+4{\bf k},b=2+{\bf i}+3{\bf k}$, $I_{a}=-20,I_{b}=-4,I_{a}\neq I_{b}$, $I_{\bar{a}+b}=0,\mathrm{rank}(S(a,b))=3$.

Theorem 4.4**.**

Two split quaternions $a,b\in{\mathbb{H}}_{s}-{\mathbb{R}}$ are consimilar if and only if one of the following two conditions holds:

[TABLE]

Proof.

If $a,b$ are consimilar then there exists an $x\in{\mathbb{H}}_{s}-Z({\mathbb{H}}_{s})$ such that $xa=b\bar{x}$. Thus $xa\overline{xa}=b\bar{x}x\bar{b}$. That is $I_{x}(I_{a}-I_{b})=0$, which implies that $I_{a}=I_{b}$. It is obvious that $x=\bar{a}+b$ is a solution to $xa=b\bar{x}$. If $\bar{a}+b=0$ then the following three split quaternions $x=a_{3}{\bf i}+a_{1}{\bf k}$, $x=a_{2}{\bf i}+a_{1}{\bf j}$ and $x=a_{1}+a_{0}{\bf i}$ are solutions to $xa=b\bar{x}$, among them there is at least one number in ${\mathbb{H}}_{s}-Z({\mathbb{H}}_{s})$. If $x=\bar{a}+b\neq 0$ then it is a basis of the solution space of $xa=b\bar{x}$. The above observation and Proposition 4.2 imply Theorem 4.4. ∎

Acknowledgements. This work is supported by Natural Science Foundation of China (no:11871379 ) and the Innovation Project of Department of Education of Guangdong Province.