1 Introduction

In 1878, Clifford algebras having n number of generators were discovered. Various useful and important results related with Clifford algebras have been studied up to recently.

Let \(\mathbb {R}^{0,n}\) be the standard n-dimensional pseudo-Euclidean space endowed with the quadratic form of signature (0, n). Furthermore, let \(C\ell _{0,n}\) be the corresponding real Clifford algebra of \(\mathbb {R}^{0,n}\).

Some different matrix representations of \(C\ell _{0,n}\) have been established and \(C\ell _{0,n}\) is isomorphic to a matrix algebra [1, 2, 4, 8]. The \(2^{n}\)-dimensional matrix algebra \(L_{2^{n}}(\mathbb {R})\) is proved to be isomorphic to \(C\ell _{0,n}\) [2]. Throughout this paper, we will denote \(L_{2^{n}}(\mathbb {R})\) by \(L_{0, n}(\mathbb {R})\).

The vector generators of \(L_{p,q}(\mathbb {R})\) that is isomorphic to \(C\ell _{p,q}\) are obtained using the set of vector generators \(\left\{ g_{2}, g_{3}, g_{7}, \ldots , g_{2^{n}-1}\right\} \) of \(L_{0, n}(\mathbb {R})\) by multiplying special diagonal matrices to the vector generators \(g_{2}, g_{3}, g_{7}, \ldots , g_{2^{n}-1}\) [5]. Thus, algebraic properties of \(C\ell _{p,q}\) can be derived from the properties of \(C\ell _{0,n}\).

The Cayley–Dickson construction is a well-known process to extend number system. Using the Cayley–Dickson construction, one can obtain complex numbers, quaternions, octonions and sedenions from real numbers in the sequel. A lot of works associated with Cayley–Dickson process have been done [1, 3, 7].

In Sect. 2, We establish a method to construct \(L_{0, n}(\mathbb {R})\) using \(L_{0, n-1}(\mathbb {R})\) and so one can obtain a sequence of algebras basically different from the algebras induced by the Cayley–Dickson construction. Since \(L_{0, 0}(\mathbb {R})\) is isomorphic to \(\mathbb {R}\) and \(L_{0, 1}(\mathbb {R})\) is isomorphic to \(\mathbb {C}\), likewise, we can extend the number system by using \(L_{0, n}(\mathbb {R})\).

In [6], Tian present some properties of quaternion matrices and investigate the solution of some linear equations over real quaternion algebra \(\mathbb {H}\).

In Sect. 3, we investigate the existence of solutions for some linear equations such as \(ax=0, \,\,xa=0,\,\, ax=xa,\,\,ax-xb=c,\,\,axa=a\) over the Clifford algebra \(C\ell _{0,3}\) by means of matrix representation.

2 Extension of Number System by Means of \(L_{0,n}(\mathbb {R})\)

Recall that, the algebra \(L_{0, n}(\mathbb {R})\) is constructed through the following process [2]. Let

$$\begin{aligned} \sigma _{0}=\left( \begin{array}{c@{\quad }c} 1 &{} 0 \\ 0 &{} 1 \end{array} \right) , \sigma _{1}=\left( \begin{array}{c@{\quad }c} 0 &{} 1 \\ 1 &{} 0 \end{array} \right) ,\,\,\, \sigma _{2}=\left( \begin{array}{c@{\quad }c} 0 &{} -\,i \\ i &{} 0 \end{array} \right) ,\,\,\, \sigma _{3}=\left( \begin{array}{c@{\quad }c} 1 &{} 0 \\ 0 &{} -\,1 \end{array} \right) \end{aligned}$$

be the Pauli matrices. Then, we first define \(A^{(1)}=A_1^{1}\sigma _{0}+A_{2}^{1}\sigma _{2}(-i),\,\,\, A_1^{1}, A_{2}^{1}\in {\mathbb {R}}\) and define \(A^{(n)}\) by using \(A^{(n-1)}\in L_{0,n-1}(\mathbb {R})\) as follows. Let \(A^{(n-1)} \in L_{0,n-1}(\mathbb {R})\) and let the first column of \(A^{(n-1)}\) be \((A_1^{n-1} , A_2^{n-1} , \cdots , A_{2^{n-1} }^{n-1})^{\mathrm{T}}\). If i is odd, replace \(A_i^{n-1}\) by \( \left( \begin{array}{cc} A_{2i-1}^{n} &{} -A_{2i}^{n} \\ A_{2i}^{n} &{} A_{2i-1}^{n} \end{array} \right) \). If i is even, replace \(A_i^{n-1}\) by \(\left( \begin{array}{cc} A_{2i-1}^{n} &{} A_{2i}^{n} \\ A_{2i}^{n} &{} -A_{2i-1}^{n} \end{array} \right) \). Through this process, the \(2^{n} \times 2^{n}\) matrix \(A^{(n)}\) and \(L_{0, n}(\mathbb {R}) = \{ A^{(n)} \in M_{2^{n}}(\mathbb {R}) \,| \, A_{i}^{n} \in \mathbb {R} \,\, , \,\, i= 1 ,2, \ldots , 2^{n}-1 \}\) are obtained. Moreover, \(g_i \) is a \(2^n \times 2^n\) matrix in \(L_{0,n} (\mathbb {R})\) such that (i, 1)-th entry is 1, and (j, 1)-th entry is 0 for \(j \not = i\). Thus, \(g_{1}=I_{2^{n}}\) and \(L_{0,0}(\mathbb {R})=\left\{ a_{1}g_{1} | a_{1}\in \mathbb {R}\right\} \),   \(L_{0,1}(\mathbb {R})=\left\{ a_{1}g_{1}+a_{2}g_{2} | a_{1}, a_{2}\in \mathbb {R}\right\} \). For \(n\ge 2\), \(g_{2}, g_{3}, g_{7}, \ldots , g_{2^{n}-1}\) are vector generators of \(L_{0,n}(\mathbb {R})\) and

$$\begin{aligned} \mathcal {B} =\left\{ g_{1}, g_{2}, g_{3},\ldots ,g_{2^{n}-1},\, g_{2}g_{3},\ldots ,\, g_{2}g_{2^{n}-1},\, \ldots ,\, g_{2}g_{3}\ldots g_{2^{n}-1} \right\} \end{aligned}$$

is a vector space basis of \(L_{0,n}(\mathbb {R})\). In what follows, A in \(L_{0,n-1}(\mathbb {R})\) will be identified with the block matrix \( \left( \begin{array}{c@{\quad }c} A &{} O \\ O &{} A \end{array}\right) \) in \(L_{0,n}(\mathbb {R}).\)

Obviously, \(A \in L_{0,1}(\mathbb {R})\) can be expressed uniquely by \(A = a_{1}g_{1}+(b_{1}g_{1})g_{2}\) for some \(a_{1},b_{1} \in \mathbb {R}\). Furthermore, we can generalize the expression of elements in \(L_{0,n}(\mathbb {R})\) as following theorem.

Theorem 2.1

Let \(A \in L_{0,n}(\mathbb {R})\) for \(n\ge 2\). Then, A can be expressed uniquely as \(A = v_{1}+v_{2}g_{2^{n}-1}\) for some \(v_{i} \in L_{0,n-1}(\mathbb {R})\),  \(i=1,2\).

Proof

Since \(\mathcal {B}\) is a vector space basis of \(L_{0,n}(\mathbb {R})\), \(A\in span(\mathcal {B})\), the spanned subspace of \(\mathcal {B}\). Thus, the terms in A can be expressed by the sum of terms without the factor of \(g_{2^{n}-1}\) and terms with the factor of \(g_{2^{n}-1}\). Let \(v_{1}\) be the sum of all the terms without the factor of \(g_{2^{n}-1}\) and let w be the sum of all the terms with the factor of \(g_{2^{n}-1}\). Then, \(w=v_{2}g_{2^{n}-1}\) for some \(v_{2}\), where \(v_{2}\) has no terms with factor of \(g_{2^{n}-1}\). Thus, \(v_{i} \in L_{0,n-1}(\mathbb {R})\), for all \(i=1,2\) and A can be expressed as \(A = v_{1}+v_{2}g_{2^{n}-1}\).

For the uniqueness, suppose that \(A = v_{1}+v_{2}g_{2^{n}-1}= w_{1}+w_{2}g_{2^{n}-1}\) for some \(v_{i}, w_{i} \in L_{0,n-1}(\mathbb {R})\),    \(i=1,2\). Then, \(v_{1}=w_{1}\) and \(v_{2}g_{2^{n}-1}=w_{2}g_{2^{n}-1}\). Since \(g_{2^{n}-1}\) is invertible, \(v_{2}=w_{2}\) and so the uniqueness is proved. \(\square \)

Example 2.2

Let \(A \in L_{0,3}(\mathbb {R})\). Then, \(A=a_{1}g_{1}+a_{2}g_{2}+a_{3}g_{3}+a_{4}g_{2}g_{3} +b_{1}g_{7}+b_{2}g_{2}g_{7}+b_{3}g_{3}g_{7} +b_{4}g_{2}g_{3}g_{7}\) for some \(a_{i}, b_{i}\in \mathbb {R}\), \(i=1,2,3,4\) and so we can write \(A = v_{1}+v_{2}g_{7}\), where \(v_{1}=a_{1}g_{1}+a_{2}g_{2}+a_{3}g_{3}+a_{4}g_{2}g_{3}\) and \(v_{2}=b_{1}g_{1}+b_{2}g_{2}+b_{3}g_{3}+b_{4}g_{2}g_{3} \in L_{0,2}(\mathbb {R})\). \(\Box \)

To simplify the expression of the multiplication of elements in \(L_{0, n}(\mathbb {R})\), we define \(\overline{A}\), the conjugate of A, as follows.

Definition 2.3

Let \(A \in L_{0, n}(\mathbb {R})\). Then, we define the conjugate \(\overline{A}\) of A according to n as follows.

  1. (1)

    For \(n=0\), \(\overline{A} = A\).

  2. (2)

    For \(n=1\), \(\overline{A} = \overline{v_{1}}-\overline{v_{2}}g_{2}\), where \(A = v_{1}+ v_{2}g_{2}\), \(v_{i}\in L_{0, 0}(\mathbb {R})\), \(i=1,2\).

  3. (3)

    For \(n\ge 2\), \(\overline{A} = \overline{v_{1}}-\overline{v_{2}}g_{2^{n}-1}\), where \(A = v_{1}+ v_{2}g_{2^{n}-1}\), \(v_{i} \in L_{0, n-1}(\mathbb {R})\), \(i=1,2\).

Note that the changes of signs of the terms of the conjugate of A in the above definition are not the same as those of the conjugate of elements obtained by the Cayley–Dickson construction.

Example 2.4

Let \(A = (a_{1}g_{1}+a_{2}g_{2})+(b_{1}g_{1}+b_{2}g_{2})g_{3}\in L_{0, 2}(\mathbb {R})\), \(a_{i}, b_{i} \in \mathbb {R}\), \(i=1,2\). Then,

$$\begin{aligned} \begin{array}{lll} \overline{A}&{}=&{}\overline{(a_{1}g_{1}+a_{2}g_{2})}- \overline{(b_{1}g_{1}+b_{2}g_{2})}g_{3} =(a_{1}g_{1}-a_{2}g_{2})-(b_{1}g_{1}-b_{2}g_{2})g_{3}\\ &{}=&{} a_{1}g_{1}-a_{2}g_{2}-b_{1}g_{3}+b_{2}g_{2}g_{3} \end{array} \end{aligned}$$

The definition of conjugate yields the following properties.

Theorem 2.5

Let \(A, B \in L_{0, n}(\mathbb {R})\) and \(\alpha \in \mathbb {R}\). Then,

  1. (1)

    \(\overline{\overline{A}}=A\).

  2. (2)

    \(\overline{A+B}=\overline{A} + \overline{B}\).

  3. (3)

    \(\overline{\alpha A}=\alpha \overline{A}\).

Proof

For \(n=0\), it is obvious. Now, assume that \(n\ge 1\). Let \(A=v_{1}+v_{2}g\) and \(B=w_{1}+w_{2}g\) for some \(v_{i}, w_{i}\in L_{0,n-1}(\mathbb {R})\), \(i=1,2\), where \(g=g_{2}\) for \(n=1\) and \(g=g_{2^{n}-1}\) for \(n\ge 2\). We will prove the theorem by the mathematical induction.

  1. (1)

    If \(n=1\), then \(\overline{\overline{v_{i}}}=v_{i}\),  \(i=1,2\)  and

    $$\begin{aligned} \overline{\overline{A}}=\overline{\overline{v_{1}+ v_{2}g_{2}}} = \overline{\overline{v_{1}}- \overline{v_{2}}g_{2}} = \overline{\overline{v_{1}}} + \overline{\overline{v_{2}}}g_{2} = v_{1}+ v_{2}g_{2}=A. \end{aligned}$$

    Now, assume that \(\overline{\overline{A}}=A\) for all \(A\in L_{0,k}(\mathbb {R})\). Then, for \(n=k+1\), \(A=v_{1}+ v_{2}g_{2^{k+1}-1}\) for some \(v_{1}, v_{2}\in L_{0,k}{(\mathbb R)}\) and by the mathematical induction hypothesis \(\overline{\overline{v_{i}}}=v_{i}\),  \(i=1,2\). Thus,

    $$\begin{aligned} \begin{array}{lll} \overline{\overline{A}}&{}=&{}\overline{\overline{v_{1}+ v_{2}g_{2^{k+1}-1}}} \,=\, \overline{\overline{v_{1}}- \overline{v_{2}}g_{2^{k+1}-1}} \,= \,\overline{\overline{v_{1}}} + \overline{\overline{v_{2}}}g_{2^{k+1}-1} \\ &{}=&{} v_{1}+ v_{2}g_{2^{k+1}-1}\,=\,A. \end{array} \end{aligned}$$
  2. (2)

    If \(n=1\), then \(\overline{v_{i}+w_{i}}=\overline{v_{i}}+\overline{w_{i}}\),   \(i=1,2\)  and

    $$\begin{aligned} \begin{array}{lll} \overline{A+B}&{}=&{}\overline{(v_{1}+w_{1})+ (v_{2}+w_{2})g_{2}} = \overline{(v_{1}+w_{1})}- \overline{(v_{2}+w_{2})}g_{2} \\ &{}=&{} (\overline{v_{1}}+\overline{w_{1}}) - (\overline{v_{2}}+\overline{w_{2}})g_{2} = (\overline{v_{1}}-\overline{v_{2}}g_{2}) + (\overline{w_{1}}-\overline{w_{2}}g_{2}) \\ &{}=&{}\overline{A}+\overline{B}. \end{array} \end{aligned}$$

    Assume that \(\overline{A+B}=\overline{A}+\overline{B}\) for \(A, B\in L_{0,k}(\mathbb {R})\). For \(n=k+1\), \(A=v_{1}+v_{2}g_{2^{k+1}-1}\) and \(B=w_{1}+w_{2}g_{2^{k+1}-1}\) for some \(v_{i}, w_{i}\in L_{0,k}(\mathbb {R})\), \(i=1,2\). Then, \(\overline{v_{i}+w_{i}}=\overline{v_{i}}+\overline{w_{i}}\),   \(i=1,2\) by the mathematical induction hypothesis. Thus,

    $$\begin{aligned} \begin{array}{lll} \overline{A+B}&{}=&{}\overline{(v_{1}+w_{1})+ (v_{2}+w_{2})g_{2^{k+1}-1}} = \overline{(v_{1}+w_{1})}- \overline{(v_{2}+w_{2})}g_{2^{k+1}-1} \\ &{}=&{} (\overline{v_{1}}+\overline{w_{1}}) - (\overline{v_{2}}+\overline{w_{2}})g_{2^{k+1}-1} = (\overline{v_{1}}-\overline{v_{2}}g_{2^{k+1}-1}) + (\overline{w_{1}}-\overline{w_{2}}g_{2^{k+1}-1}) \\ &{}=&{}\overline{A}+\overline{B}. \end{array} \end{aligned}$$
  3. (3)

    If \(n=1\), then \(\overline{\alpha v_{i}}=\alpha \overline{v_{i}}\),   \(i=1,2\)  and

    $$\begin{aligned} \overline{\alpha A}=\overline{\alpha v_{1}+ \alpha v_{2}g_{2}} = \overline{\alpha v_{1}}- \overline{\alpha v_{2}}g_{2} = \alpha \overline{v_{1}} - \alpha \overline{v_{2}}g_{2}=\alpha \overline{A}. \end{aligned}$$

    Now, assume that \(\overline{\alpha A}=\alpha \overline{A}\) for all \(A\in L_{0,k}(\mathbb {R})\). For \(n=k+1\), \(A=v_{1}+ v_{2}g_{2^{k+1}-1}\) for some \(v_{1}, v_{2}\in L_{0,k}{(\mathbb R)}\). Then, \(\overline{\alpha v_{i}}=\alpha \overline{v_{i}}\),   \(i=1,2\) by the mathematical induction hypothesis and hence

    $$\begin{aligned} \begin{array}{lll} \overline{\alpha A}&{}=&{}\overline{\alpha v_{1}+ \alpha v_{2}g_{2^{k+1}-1}} = \overline{\alpha v_{1}}- \overline{\alpha v_{2}}g_{2^{k+1}-1} = \alpha \overline{v_{1}} - \alpha \overline{v_{2}}g_{2^{k+1}-1} \\ &{}=&{} \alpha (\overline{v_{1}}-\overline{v_{2}}g_{2^{k+1}-1})=\alpha \overline{A}. \end{array} \end{aligned}$$

\(\Box \)

Theorem 2.6

If \(w=ag_{1}\) for some \(a\in \mathbb {R}\), then \(g_{2}w=\overline{w}g_{2}\). Furthermore, if \(w\in L_{0,n-1}(\mathbb {R})\), then \(g_{2^{n}-1}w=\overline{w}g_{2^{n}-1}\) for \(n\ge 2\).

Proof

If \(w=ag_{1}\), then \(\overline{w}=\overline{ag_{1}}=ag_{1}\) and so \(g_{2}w=\overline{w}g_{2}\). For the case of \(n\ge 2\), we will prove by the mathematical induction. For \(n=2\), \(w=u_{1}+u_{2}g_{2}\) for some \(u_{1}=ag_{1}\) and \(u_{2}=bg_{1}\), \(a, b \in \mathbb {R}\). Thus, \(g_{3}w=g_{3}(u_{1}+u_{2}g_{2})=u_{1}g_{3}+u_{2}g_{3}g_{2}=(u_{1}-u_{2}g_{2})g_{3}= \overline{w}g_{3}\) and the theorem is true for \(n=2\). Now, assume that \(g_{2^{k}-1}{w}=\overline{w}g_{2^{k}-1}\) for \(w \in L_{0, k-1}(\mathbb {R})\). Note that \(g_{2^{k+1}-1}{u}=\overline{u}g_{2^{k+1}-1}\) for \(u \in L_{0, k-1}(\mathbb {R})\) since \(g_{i}g_{j}=-g_{j}g_{i}\) for \(i\ne j \ge 2\). If \(w = u_{1}+ u_{2}g_{2^{k}-1}\in L_{0, k}(\mathbb {R})\) for some \(u_{1}, u_{2} \in L_{0, k-1}(\mathbb {R})\), then \(\overline{w} = \overline{u_{1}} - \overline{u_{2}}g_{2^{k}-1}\) and

$$\begin{aligned} \begin{array}{lll} g_{2^{k+1}-1}w &{}=&{} g_{2^{k+1}-1}(u_{1}+ u_{2}g_{2^{k}-1})= g_{2^{k+1}-1}u_{1}+g_{2^{k+1}-1}u_{2}g_{2^{k}-1} \\ &{}=&{}\overline{u_{1}}g_{2^{k+1}-1} +\overline{u_{2}}g_{2^{k+1}-1}g_{2^{k}-1} =\overline{u_{1}}g_{2^{k+1}-1}- \overline{u_{2}}g_{2^{k}-1}g_{2^{k+1}-1} \\ &{}=&{} (\overline{u_{1}}- \overline{u_{2}}g_{2^{k}-1})g_{2^{k+1}-1} = \overline{w}g_{2^{k+1}-1}. \end{array} \end{aligned}$$

Thus, it satisfies for the case of \(n=k+1\) and the theorem is proved. \(\Box \)

The multiplication AB of A and B in \(L_{0, n}(\mathbb {R})\) can be expressed by using the conjugate as follows:

Corollary 2.7

Let \(A, B \in L_{0, n}(\mathbb {R})\) and let \(A = v_{1}+ v_{2}g_{2^{n}-1}\), \(B = w_{1}+ w_{2}g_{2^{n}-1}\) for some \(v_{i}, w_{i} \in L_{0, n-1}(\mathbb {R})\), \(i=1,2\). Then, \(AB=(v_{1}w_{1}-v_{2}\overline{{w}_{2}})+(v_{1}w_{2} +v_{2}\overline{{w}_{1}})g_{2^{n}-1}\).

Proof

Note that

$$\begin{aligned} \begin{array}{lll} AB&{}=&{} (v_{1}+ v_{2}g_{2^{n}-1})(w_{1}+ w_{2}g_{2^{n}-1}) \\ &{}=&{} (v_{1}w_{1}+v_{2}g_{2^{n}-1}{w}_{2}g_{2^{n}-1}) +(v_{1}w_{2}g_{2^{n}-1}+v_{2}g_{2^{n}-1}{w}_{1}). \end{array} \end{aligned}$$

But,

$$\begin{aligned} g_{2^{n}-1}{w}_{2}= \overline{w_{2}}g_{2^{n}-1},\,\,\,\, g_{2^{n}-1}{w}_{1}=\overline{w_{1}}g_{2^{n}-1},\,\,\,\, \end{aligned}$$

and so \(AB=(v_{1}w_{1}-v_{2}\overline{w}_{2})+ (v_{1}w_{2}+v_{2}\overline{w}_{1})g_{2^{n}-1}.\)\(\Box \)

Corollary 2.8

Let \(A, B \in L_{0, n}(\mathbb {R})\). Then, \(\overline{AB}=\overline{A}\,\, \overline{B}\).

Proof

For \(n=0\), it is obvious. Now, assume that \(n\ge 1\). Let \(A=v_{1}+v_{2}g\) and \(B=w_{1}+w_{2}g\) for some \(v_{i}, w_{i}\in L_{0,n-1}(\mathbb {R})\), \(i=1,2\), where \(g=g_{2}\) for \(n=1\) and \(g=g_{2^{n}-1}\) for \(n\ge 2\). We will prove the theorem by the mathematical induction.

If \(n=1\), then \(\overline{v_{i}}=v_{i}, \overline{w_{i}}=w_{i}\), \(\overline{v_{i}w_{j}}=v_{i}w_{j}\),  \(i=1,2,\,\,j=1,2\) and so

$$\begin{aligned} \begin{array}{lll} \overline{AB}&{}=&{}\overline{(v_{1}w_{1}-v_{2}\overline{w_{2}})+ (v_{1}w_{2}+v_{2}\overline{w_{1}})g_{2}} = \overline{(v_{1}w_{1}-v_{2}\overline{w_{2}})}- \overline{(v_{1}w_{2}+v_{2}\overline{w_{1}})}g_{2} \\ &{}=&{}(\overline{v_{1}w_{1}}-\overline{v_{2}w_{2}})- (\overline{v_{1}w_{2}}+\overline{v_{2}w_{1}})g_{2}= (v_{1}w_{1}-v_{2}w_{2})-(v_{1}w_{2}+v_{2}w_{1})g_{2} \\ &{}=&{}(v_{1}w_{1}-v_{2}\overline{w_{2}})- (v_{1}w_{2}+v_{2}\overline{w_{1}})g_{2} =(v_{1}-v_{2}g_{2})(w_{1}-w_{2}g_{2})=\overline{A}\,\, \overline{B}. \end{array} \end{aligned}$$

Now, assume that \(\overline{AB}=\overline{A}\,\, \overline{B}\) for all \(A, B\in L_{0,k}(\mathbb {R})\).

For \(n=k+1\), \(A=v_{1}+v_{2}g_{2^{k+1}-1}\) and \(B=w_{1}+w_{2}g_{2^{k+1}-1}\) for some \(v_{i}, w_{i}\in L_{0,k}(\mathbb {R})\), \(i=1,2\). Then, \(\overline{v_{i}w_{j}}=\overline{v_{i}}\,\, \overline{w_{j}}\),   \(i=1,2,\,\,j=1,2\) by the mathematical induction hypothesis and so

$$\begin{aligned} \begin{array}{lll} \overline{AB}&{}=&{}\overline{(v_{1}w_{1}-v_{2}\overline{w_{2}})+ (v_{1}w_{2}+v_{2}\overline{w_{1}})g_{2^{k+1}-1}} \\ &{}=&{}\overline{(v_{1}w_{1}-v_{2}\overline{w_{2}})}- \overline{(v_{1}w_{2}+v_{2}\overline{w_{1}})}g_{2^{k+1}-1} \\ &{}=&{}(\overline{v_{1}w_{1}}-\overline{v_{2}\overline{w_{2}}})- (\overline{v_{1}w_{2}}+\overline{v_{2}\overline{w_{1}}})g_{2^{k+1}-1}\\ &{}=&{} (\overline{v_{1}}\,\, \overline{w_{1}}- \overline{v_{2}}\,\, \overline{\overline{w_{2}}})- (\overline{v_{1}}\,\, \overline{w_{2}}+\overline{v_{2}}\,\, \overline{\overline{w_{1}}})g_{2^{k+1}-1} \\ &{}=&{}(\overline{v_{1}}-\overline{v_{2}}g_{2^{k+1}-1}) (\overline{w_{1}}-\overline{w_{2}}g_{2^{k+1}-1})\\ &{}=&{}\overline{A}\,\, \overline{B}. \end{array} \end{aligned}$$

\(\Box \)

Remark 2.9

Note that, if we identify \(A = v_{1}+ v_{2}g_{2^{n}-1}\) and \(B = w_{1}+ w_{2}g_{2^{n}-1}\) with \(A = (v_{1},v_{2})\) and \(B = (w_{1}, w_{2})\), respectively, then we may write

$$\begin{aligned} AB=(v_{1}w_{1}-v_{2}\overline{w_{2}}, v_{1}w_{2}+v_{2}\overline{w_{1}}). \end{aligned}$$

Obviously, the multiplication is basically different from the multiplication in the Cayley–Dickson construction.

It is well known that \(\mathbb {R} \,\,\rightarrow \,\, \mathbb {C} \,\,\rightarrow \,\, \mathbb {H} \,\,\rightarrow \,\, \cdots \) is an extension of number system through the Cayley–Dickson construction. Since

$$\begin{aligned} L_{0,0}(\mathbb {R})\cong \mathbb {R}\,\,\subseteq \,\, L_{0,1}(\mathbb {R})\cong \mathbb {C}\,\,\subseteq \,\, L_{0,2}(\mathbb {R})\,\,\subseteq \,\, \cdots , \end{aligned}$$

the following sequence of algebra extension is considered as another extension of number system obtained naturally:

$$\begin{aligned} L_{0, 0}(\mathbb {R}) \,\,\rightarrow \,\, L_{0, 1}(\mathbb {R}) \,\,\rightarrow \,\, L_{0, 2}(\mathbb {R}) \,\,\rightarrow \,\, \cdots \,\,\rightarrow \,\, L_{0, n-1}(\mathbb {R}) \,\,\rightarrow \,\, L_{0, n}(\mathbb {R}) \,\,\rightarrow \,\, \cdots \end{aligned}$$

If we let \(e_{1}, e_{2}, \ldots , e_{n}\) be the vector generators of \(C\ell _{0, n}\), then we may assume that

$$\begin{aligned} C\ell _{0, n}=\left\{ u_{1}+u_{2}e_{n} \, |\, u_{1}, u_{2} \in C\ell _{0, n-1} \right\} \end{aligned}$$

since \(L_{0, n}(\mathbb {R})\cong C\ell _{0, n}\).

Theorem 2.10

The Clifford algebra \(C\ell _{0, n}\) can be constructed from \(C\ell _{0, n-1}\) for all n. Thus, the following sequence of Clifford algebras can be considered as an extension of number system naturally.

$$\begin{aligned} C\ell _{0, 0} \,\,\rightarrow \,\, C\ell _{0, 1} \,\,\rightarrow \,\, C\ell _{0, 2} \,\,\rightarrow \,\, \cdots \,\,\rightarrow \,\, C\ell _{0, n-1} \,\,\rightarrow \,\, C\ell _{0, n} \,\,\rightarrow \,\, \cdots \end{aligned}$$

3 Solutions for Some Linear Equations in \(C\ell _{0,3}\)

In this section, we will consider the linear equations such as \(ax=0, \,\,xa=0,\,\, ax=xa,\,\,ax-xb=c,\,\,axa=a\) over the Clifford algebra \(C\ell _{0,3}\). Recall that \(C\ell _{0,3}\) is isomorphic to the matrix algebra \(L_{0, 3}(\mathbb {R})\) and so it is enough to establish the existence of solutions of the linear equations over the matrix algebra \(L_{0, 3}(\mathbb {R})\).

Let \(A\in L_{0, 3}(\mathbb {R})\). Then, \(A=v_{1}+v_{2}g_{7}\) for some \(v_{1}, v_{2}\in L_{0, 2}(\mathbb {R})\). Now, let \(v_{1}=a_{1}g_{1}+a_{2}g_{2}+a_{3}g_{3}+a_{4}g_{2}g_{3},\,\, v_{2}=b_{1}g_{1}+b_{2}g_{2}+b_{3}g_{3}+b_{4}g_{2}g_{3}\) for some \(a_{i}, b_{i}\in \mathbb {R}\). Define \(\overrightarrow{A}\) by

$$\begin{aligned} \overrightarrow{A}=(a_{1}, a_{2}, a_{3}, a_{4}, b_{1}, b_{2}, b_{3}, b_{4})^{\mathrm{T}}. \end{aligned}$$

Moreover, define \(\phi _{i}(v_{j})\) and \(\psi _{i}(A)\) as follows.

$$\begin{aligned} \phi _{1}(v_{1})= & {} \left( \begin{array}{c@{\quad }c@{\quad }c@{\quad }c} a_{1} &{} -\,a_{2} &{} -\,a_{3} &{} -\,a_{4} \\ a_{2} &{} a_{1} &{} -\,a_{4} &{} a_{3} \\ a_{3} &{} a_{4} &{} a_{1} &{} -\,a_{2} \\ a_{4} &{} -\,a_{3} &{} a_{2} &{} a_{1} \end{array} \right) , \,\,\,\,\, \phi _{1}(v_{2}) = \left( \begin{array}{c@{\quad }c@{\quad }c@{\quad }c} b_{1} &{} -\,b_{2} &{} -\,b_{3} &{} -\,b_{4} \\ b_{2} &{} b_{1} &{} -\,b_{4} &{} b_{3} \\ b_{3} &{} b_{4} &{} b_{1} &{} -\,b_{2} \\ b_{4} &{} -\,b_{3} &{} b_{2} &{} b_{1} \end{array} \right) ,\\ \phi _{2}(v_{1})= & {} \left( \begin{array}{c@{\quad }c@{\quad }c@{\quad }c} a_{1} &{} a_{2} &{} a_{3} &{} -\,a_{4} \\ a_{2} &{} -\,a_{1} &{} a_{4} &{} a_{3} \\ a_{3} &{} -\,a_{4} &{} -\,a_{1} &{} -\,a_{2} \\ a_{4} &{} a_{3} &{} -\,a_{2} &{} a_{1} \end{array} \right) ,\,\,\,\,\, \phi _{2}(v_{2}) = \left( \begin{array}{c@{\quad }c@{\quad }c@{\quad }c} b_{1} &{} b_{2} &{} b_{3} &{} -\,b_{4} \\ b_{2} &{} -\,b_{1} &{} b_{4} &{} b_{3} \\ b_{3} &{} -\,b_{4} &{} -\,b_{1} &{} -\,b_{2} \\ b_{4} &{} b_{3} &{} -\,b_{2} &{} b_{1} \end{array} \right) ,\\ \phi _{3}(v_{1})= & {} \left( \begin{array}{c@{\quad }c@{\quad }c@{\quad }c} a_{1} &{} -\,a_{2} &{} -\,a_{3} &{} -\,a_{4} \\ a_{2} &{} a_{1} &{} a_{4} &{} -\,a_{3} \\ a_{3} &{} -\,a_{4} &{} a_{1} &{} a_{2} \\ a_{4} &{} a_{3} &{} -\,a_{2} &{} a_{1} \end{array} \right) , \,\,\,\,\, \phi _{3}(v_{2}) = \left( \begin{array}{c@{\quad }c@{\quad }c@{\quad }c} b_{1} &{} -\,b_{2} &{} -\,b_{3} &{} -\,b_{4} \\ b_{2} &{} b_{1} &{} b_{4} &{} -\,b_{3} \\ b_{3} &{} -\,b_{4} &{} b_{1} &{} b_{2} \\ b_{4} &{} b_{3} &{} -\,b_{2} &{} b_{1} \end{array} \right) ,\\ \phi _{4}(v_{1})= & {} \left( \begin{array}{c@{\quad }c@{\quad }c@{\quad }c} a_{1} &{} a_{2} &{} a_{3} &{} -\,a_{4} \\ -\,a_{2} &{} a_{1} &{} a_{4} &{} a_{3} \\ -\,a_{3} &{} -\,a_{4} &{} a_{1} &{} -\,a_{2} \\ a_{4} &{} -\,a_{3} &{} a_{2} &{} a_{1} \end{array} \right) , \,\,\,\,\, \phi _{4}(v_{2}) = \left( \begin{array}{c@{\quad }c@{\quad }c@{\quad }c} b_{1} &{} b_{2} &{} b_{3} &{} -\,b_{4} \\ -\,b_{2} &{} b_{1} &{} b_{4} &{} b_{3} \\ -\,b_{3} &{} -\,b_{4} &{} b_{1} &{} -\,b_{2} \\ b_{4} &{} -\,b_{3} &{} b_{2} &{} b_{1} \end{array} \right) .\\ \psi _{1}(A)= & {} \left( \begin{array}{c@{\quad }c} \phi _{1}(v_{1}) &{} -\,\phi _{2}(v_{2}) \\ \phi _{2}(v_{2}) &{} \phi _{1}(v_{1}) \end{array} \right) , \,\,\,\,\, \psi _{2}(A) = \left( \begin{array}{c@{\quad }c} \phi _{3}(v_{1}) &{} -\,\phi _{4}(v_{2}) \\ \phi _{3}(v_{2}) &{} \phi _{4}(v_{1}) \end{array} \right) . \end{aligned}$$

By the straightforward calculations, the following proposition is obtained.

Proposition 3.1

Let \(A=v_{1}+v_{2}g_{7}\) and \(X \in L_{0,3}(\mathbb {R})\), where \(v_{1}=a_{1}g_{1}+a_{2}g_{2}+a_{3}g_{3}+a_{4}g_{2}g_{3},\,\, v_{2}=b_{1}g_{1}+b_{2}g_{2}+b_{3}g_{3}+b_{4}g_{2}g_{3}\) for some \(a_{i}, b_{i}\in \mathbb {R}\). Then,

  1. (1)

    \(\overrightarrow{AX}=\psi _{1}(A)\overrightarrow{X}\)

  2. (2)

    \(\overrightarrow{XA}=\psi _{2}(A)\overrightarrow{X}\).

By Proposition 3.1, we have the following properties.

Theorem 3.2

Let \(A, B \in L_{0,3}(\mathbb {R})\). Then,

  1. (1)

    If \(A=g_{1}\), then \(\psi _{i}(A)=A\),  \(i=1,2\).

  2. (2)

    \(A=B\) if and only if \(\psi _{i}(A)=\psi _{i}(B)\),  \(i=1,2\).

  3. (3)

    A is invertible if and only if \(\psi _{i}(A)\) is invertible,  \(i=1,2\). Furthermore,

\(\psi _{i}(A^{-1})=(\psi _{i}(A))^{-1}\).

Proof

Let \(X \in L_{0,3}(\mathbb {R})\). Then, (1) \(\overrightarrow{AX}=\psi _{1}(A)\overrightarrow{X}\) and \(\overrightarrow{AX}=\overrightarrow{X}=A\overrightarrow{X}\). Thus, \(\psi _{1}(A)\overrightarrow{X}=A\overrightarrow{X}\) for all \(X \in L_{0,3}(\mathbb {R})\) and so \(\psi _{1}(A)=A\). Also, \(\overrightarrow{XA}=\psi _{2}(A)\overrightarrow{X}\) and \(\overrightarrow{XA}=\overrightarrow{X}=A\overrightarrow{X}\). Thus, \(\psi _{2}(A)\overrightarrow{X}=A\overrightarrow{X}\) for all \(X \in L_{0,3}(\mathbb {R})\) and so \(\psi _{2}(A)=A\). (2) Obviously, \(A=B\) implies that \(\psi _{i}(A)=\psi _{i}(B)\),  \(i=1,2\). Conversely, if \(\psi _{1}(A)=\psi _{1}(B)\), then \(\psi _{1}(A)\overrightarrow{X}=\psi _{1}(B)\overrightarrow{X}\) and so \(\overrightarrow{AX}=\overrightarrow{BX}\). Thus, \(\overrightarrow{(A-B)X}=O\) for all \(X \in L_{0,3}(\mathbb {R})\) and hence \(A=B\). Also, if \(\psi _{2}(A)=\psi _{2}(B)\), then \(\psi _{2}(A)\overrightarrow{X}=\psi _{2}(B)\overrightarrow{X}\) and so \(\overrightarrow{XA}=\overrightarrow{XB}\). Thus, \(\overrightarrow{X(A-B)}=O\) for all \(X \in L_{0,3}(\mathbb {R})\) and hence \(A=B\). (3) \(\psi _{1}(AA^{-1})=\psi _{1}(A)\psi _{1}(A^{-1})\) and \(\psi _{2}(AA^{-1})=\psi _{2}(A^{-1})\psi _{2}(A)\). But, \(\psi _{i}(AA^{-1})=\psi _{i}(g_{1})=g_{1}\), \(i=1,2\) and hence \(\psi _{i}(A^{-1})=\psi _{i}(A)^{-1}\), \(i=1,2\). \(\Box \)

Theorem 3.3

Let \(A, B \in L_{0,3}(\mathbb {R})\) and \(a\in \mathbb {R}\). Then,

  1. (1)

    \(\psi _{i}(aA)=a\psi _{i}(A)\),     \(i=1,2\).

  2. (2)

    \(\psi _{i}(A+B)=\psi _{i}(A)+\psi _{i}(B)\),     \(i=1,2\).

  3. (3)

    \(\psi _{1}(AB)=\psi _{1}(A)\psi _{1}(B)\),     \(\psi _{2}(AB)=\psi _{2}(B)\psi _{2}(A)\).

  4. (4)

    \(\psi _{1}(A)\psi _{2}(B)=\psi _{2}(B)\psi _{1}(A)\).

  5. (5)

    \(\psi _{i}\) preserves the similar matrix. That is, if \(B=CAC^{-1}\) for some invertible \(C\in L_{0,3}(\mathbb {R})\), then \(\psi _{1}(B)=\psi _{1}(C)\psi _{1}(A)\psi _{1}(C)^{-1}\)  and   \(\psi _{2}(B)=\psi _{2}(C)^{-1}\psi _{2}(A)\psi _{2}(C)\).

Proof

Let \(X \in L_{0,3}(\mathbb {R})\). Then,

  1. (1)

    According to the result of Proposition 3.1, \(\overrightarrow{aAX}=\psi _{1}(aA)\overrightarrow{X}\) and \(a\overrightarrow{AX}=a\psi _{1}(A)\overrightarrow{X}\). Since \(\overrightarrow{aAX}=a\overrightarrow{AX}\), \(\psi _{1}(aA)\overrightarrow{X}=a\psi _{1}(A)\overrightarrow{X}\) for all \(X \in L_{0,3}(\mathbb {R})\) and so \(\psi _{1}(aA)=a\psi _{1}(A)\). Also, \(\overrightarrow{X(aA)}=\psi _{2}(aA)\overrightarrow{X}\) and \(a\overrightarrow{XA}=a\psi _{2}(A)\overrightarrow{X}\). Since \(\overrightarrow{X(aA)}=a\overrightarrow{XA}\), \(\psi _{2}(aA)\overrightarrow{X}=a\psi _{2}(A)\overrightarrow{X}\) for all \(X \in L_{0,3}(\mathbb {R})\) and so \(\psi _{2}(aA)=a\psi _{2}(A)\).

  2. (2)

    \(\overrightarrow{(A+B)X}=\psi _{1}(A+B)\overrightarrow{X}\) and \(\overrightarrow{AX}+\overrightarrow{BX}= (\psi _{1}(A)+\psi _{1}(B))\overrightarrow{X}\). Since \(\overrightarrow{(A+B)X}=\overrightarrow{AX}+\overrightarrow{BX}\), \(\psi _{1}(A+B)\overrightarrow{X}=(\psi _{1}(A)+\psi _{1}(B))\overrightarrow{X}\) for all \(X \in L_{0,3}(\mathbb {R})\) and so \(\psi _{1}(A+B)=\psi _{1}(A)+\psi _{1}(B)\). Also, \(\overrightarrow{X(A+B)}=\psi _{2}(A+B)\overrightarrow{X}\) and \(\overrightarrow{XA}+\overrightarrow{XB}=(\psi _{2}(A)+\psi _{2}(B))\overrightarrow{X}\). Since \(\overrightarrow{X(A+B)}=\overrightarrow{XA}+\overrightarrow{XB}\), \(\psi _{2}(A+B)\overrightarrow{X}=(\psi _{2}(A)+\psi _{2}(B))\overrightarrow{X}\) for all \(X \in L_{0,3}(\mathbb {R})\) and so \(\psi _{2}(A+B)=\psi _{2}(A)+\psi _{2}(B)\).

  3. (3)

    \(\overrightarrow{(AB)X}=\psi _{1}(AB)\overrightarrow{X}\) and \(\psi _{1}(A)\overrightarrow{BX}=\psi _{1}(A)\psi _{1}(B)\overrightarrow{X}\). Since \(\overrightarrow{(AB)X}=\psi _{1}(A)\overrightarrow{BX}\), \(\psi _{1}(AB)\overrightarrow{X}=\psi _{1}(A)\psi _{1}(B)\overrightarrow{X}\) for all \(X \in L_{0,3}(\mathbb {R})\). Hence \(\psi _{1}(AB)=\psi _{1}(A)\psi _{1}(B)\). Also, \(\overrightarrow{X(AB)}=\psi _{2}(AB)\overrightarrow{X}\) and \(\psi _{2}(B)\overrightarrow{XA}=\psi _{2}(B)\psi _{2}(A)\overrightarrow{X}\). Since \(\overrightarrow{X(AB)}=\psi _{2}(B)\overrightarrow{XA}\), \(\psi _{2}(AB)\overrightarrow{X}=\psi _{2}(B)\psi _{2}(A)\overrightarrow{X}\) for all \(X \in L_{0,3}(\mathbb {R})\) and thus, \(\psi _{2}(AB)=\psi _{2}(B)\psi _{2}(A)\).

  4. (4)

    \(\overrightarrow{AXB}=\psi _{1}(A)\overrightarrow{XB}= \psi _{1}(A)\psi _{2}(B)\overrightarrow{X}\) and \(\overrightarrow{AXB}=\psi _{2}(B)\overrightarrow{AX}= \psi _{2}(B)\psi _{1}(A)\overrightarrow{X}\). Thus, \(\psi _{1}(A)\psi _{2}(B)\overrightarrow{X}=\psi _{2}(B)\psi _{1}(A)\overrightarrow{X}\) for all \(X \in L_{0,3}(\mathbb {R})\) and thus, \(\psi _{1}(A)\psi _{2}(B)=\psi _{2}(B)\psi _{1}(A)\).

  5. (5)

    \(\psi _{1}(B)=\psi _{1}(CAC^{-1})=\psi _{1}(C)\psi _{1}(A)\psi _{1}(C)^{-1}\) and \(\psi _{2}(B)=\psi _{2}(CAC^{-1})=\psi _{2}(C)^{-1}\psi _{2}(A)\psi _{2}(C)\). \(\Box \)

Using the properties, we investigate the solutions of the following linear equations. The zero square matrix of any size will be denoted by O for the convenience of expression.

Theorem 3.4

Let \(A=(a_{1}g_{1}+a_{2}g_{2}+a_{3}g_{3}+a_{4}g_{2}g_{3})+ (b_{1}g_{1}+b_{2}g_{2}+b_{3}g_{3}+b_{4}g_{2}g_{3})g_{7} \in L_{0,3}(\mathbb {R})\), \(a_{i}, b_{i}\in \mathbb {R}\), \(i=1,2,3,4\). Then, the equation \(AX=O\) has a nontrivial solution over \(L_{0,3}(\mathbb {R})\) if and only if \((a_{1}=-b_{4}, a_{2}=b_{3}, a_{3}=-b_{2}, a_{4}=b_{2})\) or \((a_{1}=b_{4}, a_{2}=-b_{3}, a_{3}=b_{2}, a_{4}=-b_{2})\).

Proof

Since \(\overrightarrow{AX}=\overrightarrow{O}\) and \(\overrightarrow{AX}=\psi _{1}(A)\overrightarrow{X}\),

$$\begin{aligned} \psi _{1}(A)\overrightarrow{X}=\left( \begin{array}{cc} \phi _{1}(v_{1}) &{} -\phi _{2}(v_{2}) \\ \phi _{2}(v_{2}) &{} \phi _{1}(v_{1}) \end{array} \right) \overrightarrow{X}=\overrightarrow{O}. \end{aligned}$$

By the straightforward calculations,

$$\begin{aligned} \psi _{1}(A)\psi _{1}(A)^{\mathrm{T}}=\left( \begin{array}{c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c} c &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} d \\ 0 &{} c &{} 0 &{} 0 &{} 0 &{} 0 &{} -\,d &{} 0 \\ 0 &{} 0 &{} c &{} 0 &{} 0 &{} d &{} 0 &{} 0 \\ 0 &{} 0 &{} 0 &{} c &{} -\,d &{} 0 &{} 0 &{} 0 \\ 0 &{} 0 &{} 0 &{} -\,d &{} c &{} 0 &{} 0 &{} 0 \\ 0 &{} 0 &{} d &{} 0 &{} 0 &{} c &{} 0 &{} 0 \\ 0 &{} -\,d &{} 0 &{} 0 &{} 0 &{} 0 &{} c &{} 0 \\ d &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} c \end{array} \right) , \end{aligned}$$

where \(c=\sum _{i=1}^{i=4}(a_{i}^{2}+b_{i}^{2}), \,\,\, d=2\sum _{i=1}^{i=4}(-1)^{i+1}a_{i}b_{5-i}.\) Thus, if we let

$$\begin{aligned} J_{4}=\left( \begin{array}{c@{\quad }c@{\quad }c@{\quad }c} 0 &{} 0 &{} 0 &{} 1 \\ 0 &{} 0 &{} -\,1 &{} 0 \\ 0 &{} 1 &{} 0 &{} 0 \\ -\,1 &{} 0 &{} 0 &{} 0 \end{array} \right) , \end{aligned}$$

then

$$\begin{aligned} \begin{array}{lll} (\hbox {det}(\psi _{1}(A)))^{2}&{}=&{} \hbox {det}(\psi _{1}(A)\psi _{1}(A)^{\mathrm{T}})= \hbox {det}\left( \begin{array}{c@{\quad }c} cI_{4} &{} dJ_{4} \\ -\,dJ_{4} &{} cI_{4} \end{array} \right) \\ \\ &{}=&{}\hbox {det} \left( \begin{array}{c@{\quad }c} cI_{4} &{} dI_{4} \\ O &{} cI_{4}+dJ_{4}c^{-1}dJ_{4} \end{array} \right) \\ \\ &{}=&{} \hbox {det}(cI_{4})\hbox {det}(cI_{4}+c^{-1}d^{2}J_{4}^{2}) \\ \\ &{}=&{} \hbox {det}(c^{2}I_{4}-d^{2}I_{4}) \\ \\ &{}=&{} (c^{2}-d^{2})^{4}. \end{array} \end{aligned}$$

Note that \(\hbox {det}(\psi _{1}(A))=0\) is equivalent to \(c^{2}-d^{2}=0\) which is equivalent to \((a_{1}=-b_{4}, a_{2}=b_{3}, a_{3}=-b_{2}, a_{4}=b_{2})\) or \((a_{1}=b_{4}, a_{2}=-b_{3}, a_{3}=b_{2}, a_{4}=-b_{2})\). Thus, the theorem is proved. \(\Box \)

Example 3.5

Let \(A=3g_{2}-g_{3}+g_{2}g_{3}-g_{7}+g_{2}g_{3}g_{7} \in L_{0,3}(\mathbb {R})\). Then, \(v_{1}=3g_{2}-g_{3}+g_{2}g_{3},\,\,\,v_{2}= -g_{1}+g_{2}g_{3}\). Since \(a_{1}=0,\,\,b_{4}=1\) and so the equation \(AX=O\) has only the trivial solution \(X=O\) over \(L_{0,3}(\mathbb {R})\).

Theorem 3.6

Let \(A=(a_{1}g_{1}+a_{2}g_{2}+a_{3}g_{3}+a_{4}g_{2}g_{3})+ (b_{1}g_{1}+b_{2}g_{2}+b_{3}g_{3}+b_{4}g_{2}g_{3})g_{7} \in L_{0,3}(\mathbb {R})\), \(a_{i}, b_{i}\in \mathbb {R}\), \(i=1,2,3,4\). Then, the equation \(XA=O\) has a nontrivial solution over \(L_{0,3}(\mathbb {R})\) if and only if \((a_{1}=-b_{4}, a_{2}=b_{3}, a_{3}=-b_{2}, a_{4}=b_{2})\) or \((a_{1}=b_{4}, a_{2}=-b_{3}, a_{3}=b_{2}, a_{4}=-b_{2})\).

Proof

Since \(\overrightarrow{XA}=\overrightarrow{O}\) and \(\overrightarrow{XA}=\psi _{2}(A)\overrightarrow{X}\),

$$\begin{aligned} \psi _{2}(A)\overrightarrow{X}=\left( \begin{array}{c@{\quad }c} \phi _{3}(v_{1}) &{} -\phi _{4}(v_{2}) \\ \phi _{3}(v_{2}) &{} \phi _{4}(v_{1}) \end{array} \right) \overrightarrow{X}=\overrightarrow{O}. \end{aligned}$$

By the straightforward calculations,

$$\begin{aligned} \psi _{2}(A)\psi _{2}(A)^{\mathrm{T}}=\left( \begin{array}{c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c} c &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} d \\ 0 &{} c &{} 0 &{} 0 &{} 0 &{} 0 &{} -\,d &{} 0 \\ 0 &{} 0 &{} c &{} 0 &{} 0 &{} d &{} 0 &{} 0 \\ 0 &{} 0 &{} 0 &{} c &{} -\,d &{} 0 &{} 0 &{} 0 \\ 0 &{} 0 &{} 0 &{} -\,d &{} c &{} 0 &{} 0 &{} 0 \\ 0 &{} 0 &{} d &{} 0 &{} 0 &{} c &{} 0 &{} 0 \\ 0 &{} -\,d &{} 0 &{} 0 &{} 0 &{} 0 &{} c &{} 0 \\ d &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} c \end{array} \right) , \end{aligned}$$

where \(c=\sum _{i=1}^{i=4}(a_{i}^{2}+b_{i}^{2})\) and \(d=2\sum _{i=1}^{i=4}(-1)^{i+1}a_{i}b_{5-i}.\) Using the similar process of the proof of previous theorem, the equation \(XA=O\) has a nontrivial solution if and only if \(\hbox {det}(\psi _{2}(A))=0\) which is equivalent to \((a_{1}=-b_{4}, a_{2}=b_{3}, a_{3}=-b_{2}, a_{4}=b_{2})\) or \((a_{1}=b_{4}, a_{2}=-b_{3}, a_{3}=b_{2}, a_{4}=-b_{2})\). \(\Box \)

According to Theorems 3.4 and 3.6, we know that the linear equation \(AX=O\) has a nontrivial solution if and only if \(XA=O\) has a nontrivial solution.

Theorem 3.7

Let \(A=v_{1}+v_{2}g_{7} \in L_{0,3}(\mathbb {R})\) for some \(v_{i}\in L_{0,2}(\mathbb {R}), \,\,i=1,2\). Then the equation \(AX=XA\) has a nontrivial solution \(X\in L_{0,3}(\mathbb {R})\).

Proof

Since \(\overrightarrow{AX}=\psi _{1}(A)\overrightarrow{X}\) and \(\overrightarrow{XA}=\psi _{2}(A)\overrightarrow{X}\), the equation \(AX=XA\) implies that \((\psi _{1}(A)-\psi _{2}(A))\overrightarrow{X}=O.\) But,

$$\begin{aligned} \begin{array}{lll} \psi _{1}(A)-\psi _{2}(A)&{}=&{} \left( \begin{array}{c@{\quad }c} \phi _{1}(v_{1})-\phi _{3}(v_{1})&{} -\phi _{2}(v_{2})+\phi _{4}(v_{2})\\ \phi _{2}(v_{2})-\phi _{3}(v_{2})&{} \phi _{1}(v_{1})-\phi _{4}(v_{1}) \end{array} \right) \\ \\ &{}=&{} \left( \begin{array}{c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 \\ 0 &{} 0 &{} -\,2a_{4} &{} 2a_{3} &{} -\,2b_{2} &{} 2b_{1} &{} 0 &{} 0 \\ 0 &{} 2a_{4} &{} 0 &{} -\,2a_{2} &{} -\,2b_{3} &{} 0 &{} 2b_{1} &{} 0 \\ 0 &{} -\,2a_{3} &{} 2a_{2} &{} 0 &{} 0 &{} -\,2b_{3} &{} 2b_{2} &{} 0 \\ 0 &{} 2b_{2} &{} 2b_{3} &{} 0 &{} 0 &{} -\,2a_{2} &{} -\,2a_{3} &{} 0 \\ 0 &{} -\,2b_{1} &{} 0 &{} 2b_{3} &{} 2a_{2} &{} 0 &{} -\,2a_{4} &{} 0 \\ 0 &{} 0 &{} -\,2b_{1} &{} -\,2b_{2} &{} 2a_{3} &{} 2a_{4} &{} 0 &{} 0 \\ 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 \end{array} \right) . \end{array} \end{aligned}$$

Note that all the entries of the first column and eighth column of \(\psi _{1}(A)-\psi _{2}(A)\) are zeros. If we let \(\overrightarrow{X}=(x_{1},x_{2},\ldots ,x_{8})^{\mathrm{T}}\), then \(x_{1}\) and \(x_{8}\) can be chosen arbitrary and hence \(AX=XA\) has a nontrivial solution. \(\Box \)

Example 3.8

Let \(\alpha , \beta , \gamma \in \mathbb {R}\). Then, obviously \(X=\alpha \, g_{1}+\beta A+\gamma \,g_{2}g_{3}g_{7}\) is a solution for the equation \(AX=XA\).

Theorem 3.9

Let \(A, B, C \in L_{0,3}(\mathbb {R})\) and \(A=v_{1}+v_{2}g_{7}, \,\,B=w_{1}+w_{2}g_{7}\) for some \(v_{i}, w_{i}\in L_{0,2}(\mathbb {R}), \,\,i=1,2\). Assume also that \(\hbox {det}(\phi _{1}(v_{1})-\phi _{3}(w_{1}))\ne 0\). Then the equation \(AX-XB=C\) has a unique solution \(X\in L_{0,3}(\mathbb {R})\) if \(\hbox {det}(\phi _{1}(v_{1})-\phi _{3}(w_{1}))\hbox {det}(D)\ne 0\), where \(D=\phi _{1}(v_{1})-\phi _{4}(w_{1})- (\phi _{2}(v_{2})-\phi _{3}(w_{2}))(\phi _{1}(v_{1})- \phi _{3}(w_{1}))^{-1}(-\phi _{2}(v_{2})+\phi _{4}(w_{2}))\).

Proof

Note that \(\overrightarrow{AX}=\psi _{1}(A)\overrightarrow{X}\) and \(\overrightarrow{XB}=\psi _{2}(B)\overrightarrow{X}\), where

$$\begin{aligned} \psi _{1}(A) = \left( \begin{array}{c@{\quad }c} \phi _{1}(v_{1}) &{} -\phi _{2}(v_{2}) \\ \phi _{2}(v_{2}) &{} \phi _{1}(v_{1}) \end{array} \right) , \,\,\,\,\, \psi _{2}(B) = \left( \begin{array}{c@{\quad }c} \phi _{3}(w_{1}) &{} -\phi _{4}(w_{2}) \\ \phi _{3}(w_{2}) &{} \phi _{4}(w_{1}) \end{array} \right) . \end{aligned}$$

Thus, \(AX-XB=C\) implies that \((\psi _{1}(A)-\psi _{2}(B))\overrightarrow{X}=\overrightarrow{C}\). Since

$$\begin{aligned} \psi _{1}(A)-\psi _{2}(B)= \left( \begin{array}{c@{\quad }c} \phi _{1}(v_{1})-\phi _{3}(w_{1})&{} -\phi _{2}(v_{2})+\phi _{4}(w_{2}) \\ \phi _{2}(v_{2})-\phi _{3}(w_{2})&{} \phi _{1}(v_{1})-\phi _{4}(w_{1}) \end{array} \right) \end{aligned}$$

and \(\hbox {det}(\phi _{1}(v_{1})-\phi _{3}(w_{1}))\ne 0\), we have

$$\begin{aligned} \hbox {det}(\psi _{1}(A)-\psi _{2}(B))=\hbox {det}(\phi _{1}(v_{1})-\phi _{3}(w_{1}))\hbox {det}(D), \end{aligned}$$

where \(D=\phi _{1}(v_{1})-\phi _{4}(w_{1})- (\phi _{2}(v_{2})-\phi _{3}(w_{2}))(\phi _{1}(v_{1})- \phi _{3}(w_{1}))^{-1}(-\phi _{2}(v_{2})+\phi _{4}(w_{2}))\). Thus, the linear equation \(AX-XB=C\) has a unique solution if \(\hbox {det}(\phi _{1}(v_{1})-\phi _{3}(w_{1}))\hbox {det}(D)\ne 0\). \(\Box \)

Example 3.10

Let \(A=g_{1}-g_{3} \in L_{0,3}(\mathbb {R})\) and \(B=g_{2}+g_{3} \in L_{0,3}(\mathbb {R})\). Then, \(v_{1}=g_{1}-g_{3}, \,\,w_{1}=g_{2}+g_{3},\,\,v_{2}=O=w_{2}\) and so

$$\begin{aligned} \phi _{1}(v_{1})= & {} \left( \begin{array}{c@{\quad }c@{\quad }c@{\quad }c} 1 &{} 0 &{} 1 &{} 0 \\ 0 &{} 1 &{} 0 &{} -\,1 \\ -\,1 &{} 0 &{} 1 &{} 0 \\ 0 &{} 1 &{} 0 &{} 1 \end{array} \right) ,\,\,\,\,\,\,\,\,\,\, \phi _{3}(w_{1})=\left( \begin{array}{c@{\quad }c@{\quad }c@{\quad }c} 0 &{} -\,1 &{} -\,1 &{} 0 \\ 1 &{} 0 &{} 0 &{} -\,1 \\ 1 &{} 0 &{} 0 &{} 1 \\ 0 &{} 1 &{} -\,1 &{} 0 \end{array} \right) .\\ \phi _{2}(v_{2})= & {} \phi _{2}(w_{2})=O,\,\,\,\,\, \phi _{4}(w_{1})=\left( \begin{array}{c@{\quad }c@{\quad }c@{\quad }c} 0 &{} -\,1 &{} -\,1 &{} 0 \\ 1 &{} 0 &{} 0 &{} -\,1 \\ 1 &{} 0 &{} 0 &{} 1 \\ 0 &{} 1 &{} -\,1 &{} 0 \end{array} \right) . \end{aligned}$$

Thus,

$$\begin{aligned} \hbox {det}(\phi _{1}(v_{1})-\phi _{3}(w_{1}))= \hbox {det}\left( \begin{array}{c@{\quad }c@{\quad }c@{\quad }c} 1 &{} -\,1 &{} 2 &{} 0 \\ -\,1 &{} 1 &{} 0 &{} 0 \\ -\,2 &{} 0 &{} 1 &{} -\,1 \\ 0 &{} 0 &{} 1 &{} 1 \end{array} \right) \ne 0 \end{aligned}$$

and

$$\begin{aligned} \hbox {det}(D)=\hbox {det}(\phi _{1}(v_{1})-\phi _{4}(w_{1}))=\hbox {det}\left( \begin{array}{c@{\quad }c@{\quad }c@{\quad }c} 1 &{} -\,1 &{} 0 &{} 0 \\ 1 &{} 1 &{} 0 &{} -\,2 \\ 0 &{} 0 &{} 1 &{} 1 \\ 0 &{} 2 &{} -\,1 &{} 1 \end{array} \right) \ne 0. \end{aligned}$$

Therefore, for any \(C \in L_{0,3}(\mathbb {R})\), the linear equation \(AX-XB=C\) has a unique solution.

Theorem 3.11

Let \(A \in L_{0,3}(\mathbb {R})\). If \(\hbox {det}(\psi _{1}(A)\psi _{2}(A))\ne 0\), then the equation \(AXA=A\) has a unique solution over \(L_{0,3}(\mathbb {R})\).

Proof

Since \(\overrightarrow{A(XA-I_{n})}=\overrightarrow{O}\) and \(\overrightarrow{A(XA-I_{n})}=\psi _{1}(A)\overrightarrow{XA-I_{n}}=\psi _{1}(A)(\psi _{2}(A)\overrightarrow{X}-(1,0,0,\cdots , 0)^{\mathrm{T}})\), we have \(\psi _{1}(A)\psi _{2}(A)\overrightarrow{X}=Col_{1}(\psi _{1}(A))\), where \(Col_{1}(\psi _{1}(A))\) is the first column of \(\psi _{1}(A)\). By the assumption, \(\overrightarrow{X}=(\psi _{1}(A)\psi _{2}(A))^{-1}Col_{1}(\psi _{1}(A))\) and the solution X exists uniquely. \(\Box \)