Complex numbers as Matrices

In this section, we use matrices to give a representation of complex numbers. Indeed, consider the set

\begin{displaymath}{\cal C} = \left\{ \left(\begin{array}{rr}
...\right); \; \mbox{where $a$ and $b$ are real numbers.}\right\}.\end{displaymath}

We will write

\begin{displaymath}M_{a,b} = \left(\begin{array}{rr}

Clearly, the set $\cal C$ is not empty. For example, we have

\begin{displaymath}M_{0,0} = {\cal O},\;\mbox{and}\; M_{1,0} = I_2.\end{displaymath}

In particular, we have

\begin{displaymath}M_{a,b} = M_{c,d}\;\;\mbox{if and only if $a=c$ and $b=d$}\end{displaymath}

for any real numbers a, b, c, and d.

Algebraic Properties of $\cal C$

Addition: For any real numbers a, b, c, and d, we have

Ma,b + Mc,d = Ma+c,b+d.

In other words, if we add two elements of the set $\cal C$, we still get a matrix in $\cal C$. In particular, we have

-Ma,b = M-a,-b.

Multiplication by a number: We have

\begin{displaymath}\lambda M_{a,b} = \lambda \left(\begin{array}{rr}
...da b&\lambda a\\
\end{array}\right) = M_{\lambda a,\lambda b}.\end{displaymath}

So a multiplication of an element of $\cal C$ and a number gives a matrix in $\cal C$.
Multiplication: For any real numbers a, b, c, and d, we have

\begin{displaymath}M_{a,b} M_{c,d} = \left(\begin{array}{rr}
ac - bd&ad + bc\\
-bc-ad &ac - bd\\

In other words, we have

Ma,b Mc,d = Mac-bd, ad+bc.

This is an extraordinary formula. It is quite conceivable given the difficult form of the matrix multiplication that, a priori, the product of two elements of $\cal C$ may not be in $\cal C$ again. But, in this case, it turns out to be true.

The above properties infer to $\cal C$ a very nice structure. The next natural question to ask, in this case, is whether a nonzero element of $\cal C$ is invertible. Indeed, for any real numbers a and b, we have

\end{array}\right) \le...
...+ b^2)\left(\begin{array}{rr}

So, if $a^2 + b^2 \neq 0$, the matrix Ma,b is invertible and

\begin{displaymath}M_{a,b}^{-1} = \frac{1}{a^2 + b^2} \left(\begin{array}{rr}
\end{array}\right)= \frac{1}{a^2 + b^2}M_{a,-b} .\end{displaymath}

In other words, any nonzero element Ma,b of $\cal C$ is invertible and its inverse is still in $\cal C$ since

\begin{displaymath}M_{a,b}^{-1} = M_{\displaystyle \frac{a}{a^2 + b^2},\frac{b}{a^2 + b^2}}\end{displaymath}

In order to define the division in $\cal C$, we will use the inverse. Indeed, recall that

\begin{displaymath}a \div b = \frac{a}{b} = a \cdot \frac{1}{b} = a \cdot b^{-1}\end{displaymath}

So for the set $\cal C$, we have

Ma, b÷Mc, d = Ma, b×Mc, d-1 = Ma, b×M$\scriptstyle {\frac{c}{c^2 + d^2}}$,$\scriptstyle {\frac{-\ d}{c^2 + d^2}}$

which implies

Ma, b÷Mc, d = M$\scriptstyle {\frac{ac+bd}{c^2 + d^2}}$, -  $\scriptstyle {\frac{ad-bc}{c^2 + d^2}}$

The matrix Ma,-b is called the conjugate of Ma,b. Note that the conjugate of the conjugate of Ma,b is Ma,b itself.

Fundamental Equation. For any Ma,b in $\cal C$, we have

Ma,b = a M1,0 + b M0,1 = a I2 + b M0,1.

Note that

M0,1 M0,1 = M-1,0 = - I2.

Remark. If we introduce an imaginary number i such that i2 = -1, then the matrix Ma,b may be rewritten by

a + bi

A lot can be said about $\cal C$, but we will advise you to visit the page on complex numbers.

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Author: M.A. Khamsi

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