Prove a unique real matrix exists to denote complex numbers in the set of Cauchy–Riemann matrices

Question

One model for the complex numbers $\mathbb{C}$ uses the set of Cauchy–Riemann matrices

$CR:=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ such that $a,b,c,d \in \mathbb{R}, a=d, b+c=0$ with matrix addition and matrix multiplication corresponding to $+$ and $×$ in $\mathbb{C}$. We say that a matrix$ \begin{pmatrix}a&b\\c&d\end{pmatrix} \in$ CR is real if $b=c=0$ and imaginary if $a=d=0$, we write $i:=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$

$(a)$ Show that for every imaginary $z ∈$ CR there is a unique real matrix $u ∈$ CR such that $z = u · i.$

$(b)$ Show that for every $z ∈$ CR there are unique real matrices $u, v ∈$ CR such that $z = u + v · i$.

I believe for both parts of the question, I am missing a key aspect of proving uniqueness. Any help would be appreciated

Answer 1

Some intuition on CR matrices may be gained via inspection of the following simple matrix equations (1)-(3), easily verified:

$\begin{bmatrix} a & 0 \\ 0 & a \end{bmatrix} \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & -a \\ a & 0 \end{bmatrix}; \tag 1$

$\begin{bmatrix} 0 & -a \\ a & 0 \end{bmatrix}\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} -a & 0 \\ 0 & -a \end{bmatrix}; \tag 2$

$\begin{bmatrix} 0 & -a \\ a & 0 \end{bmatrix}\begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} = \begin{bmatrix} a & 0 \\ 0 & a \end{bmatrix}; \tag 3$

we have

$i = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}; \tag 4$

$i^2 = -I; \tag 5$

$(-i)i = -i^2 = I; \tag 6$

$i^{-1} = -i. \tag 7$

Note that (1) shows that for every imaginary $z \in CR$ there is a real $u \in CR$ such that

$ui = z; \tag 8$

if

$yi = z, \tag 9$

then in light of (5),

$-y = -yI = yi^2 = zi, \tag{10}$

or

$y = -zi, \tag{11}$

which shows that there is precisely one $u$ satisfying (8); thus is item (a) resolved.

We next observe that the only $w \in CR$ which is both real and imaginary is $0$, for if $w$ is real then

$w = \begin{bmatrix} a & 0 \\ 0 & a \end{bmatrix} \tag{12}$

for some $a \in \Bbb R$, and since $w$ is also imaginary, we have

$w = \begin{bmatrix} 0 & b \\ -b & 0 \end{bmatrix} \tag{13}$

for some $b \in \Bbb R$; equating these two forms of $w$ yields

$\begin{bmatrix} a & 0 \\ 0 & a \end{bmatrix} = \begin{bmatrix} 0 & b \\ -b & 0 \end{bmatrix}; \tag{14}$

comparing entries of these two matrices indicates that

$a = b = 0, \tag{15}$

whence

$w = 0, \tag{16}$

as asserted.

Now any $z \in CR$ may be written

$z = \begin{bmatrix} a & -b \\ b & a \end{bmatrix} \tag{17}$

for some $a, b \in \Bbb R$; we have

$z = \begin{bmatrix} a & 0 \\ 0 & a \end{bmatrix} + \begin{bmatrix} 0 & -b \\ b & 0 \end{bmatrix} = \begin{bmatrix} a & 0 \\ 0 & a \end{bmatrix} + \begin{bmatrix} b & 0 \\ 0 & b \end{bmatrix}\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}; \tag{18}$

we denote the real matrices occurring in this equation by $u$ and $v$:

$u = \begin{bmatrix} a & 0 \\ 0 & a \end{bmatrix}, \tag{19}$

$v = \begin{bmatrix} b & 0 \\ 0 & b \end{bmatrix}, \tag{20}$

then using (4) we have

$z = u + vi; \tag{21}$

if there were two other real matrices $u'$ and $v'$ such that

$z = u' + v'i, \tag{22}$

then

$u' + v'i = u + vi, \tag{23}$

or

$u' - u = vi - v'i = (v - v')i; \tag{24}$

since $u' - u$ and $v - v'$ are both real, $(v - v')i$ is imaginary in light of (1), and thus by what we have just seen in (12)-(18) it follows that

$u' - u = (v - v')i = 0, \tag{25}$

whence

$u' = u, \tag{26}$

and again by virtue of (5),

$v - v' = (v - v')I = -(v - v')i^2 = (-(v - v')i)i = 0, \tag{27}$

and thus

$v = v' \tag{28}$

as well. Thus the uniqueness of $u$, $v$ as in (21) is established, and we have dispensed with item (b).

Answer 2

Since $i^2=-I_2$, For (a) the required $u$ is $$-zi=-\left(\begin{array}{cc} 0 & b_{z}\\ -b_{z} & 0 \end{array}\right)\left(\begin{array}{cc} 0 & -1\\ 1 & 0 \end{array}\right)=\left(\begin{array}{cc} -b_{z} & 0\\ 0 & -b_{z} \end{array}\right).$$For (b) note that$$u+vi=\left(\begin{array}{cc} a_{u} & 0\\ 0 & a_{u} \end{array}\right)+\left(\begin{array}{cc} a_{v} & 0\\ 0 & a_{v} \end{array}\right)\left(\begin{array}{cc} 0 & -1\\ 1 & 0 \end{array}\right)=\left(\begin{array}{cc} a_{u} & -a_{v}\\ a_{v} & a_{u} \end{array}\right),$$which agrees with $\left(\begin{array}{cc} a_{z} & b_{z}\\ -b_{z} & a_{z} \end{array}\right)$ iff $a_{u}=a_{z},\,a_{v}=-b_{z}$.