How is this a matrix representation of $a+ib$?

Question

Could somebody explain how this matrix represents $a+ib$? Really not sure what I’m missing here.

Answer 1

Consider the $\mathbb R$-linear map $f \colon \mathbb C \to \mathbb C$ given by $z \mapsto (a+bi) \cdot z$.

The matrix representing $f$ with respect to the ordered $\mathbb R$-basis $(1,i)$ is calculated by:

$f(1) = a+bi = a \cdot 1 + b \cdot i$, and $f(i) = ai - b = -b\cdot 1 + a \cdot i$,

hence equal to $\begin{pmatrix}a & -b \\ b & a\end{pmatrix}$. This justifies all the statements made in your pictures.

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Response to the comment: It was asked in the comments why one should consider the map $f$ in the first place. Remember from your calculus course that the derivative of a real differentiable map is linear (a matrix). Since "complex differentiable" implies "real differentiable", we would like to find a criterion such that "complex differentiable" = "real differentiable + criterion". So let us investigate real and complex linear maps $\mathbb C \to \mathbb C$.

If you take any $\mathbb R$-linear map $\mathbb C \to \mathbb C$, you can of course represent it by a matrix $\begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22}\end{pmatrix}$ with respect to the above basis $(1,i)$.

Now, any complex linear map $\mathbb C \to \mathbb C$ is given by multiplication with some complex number $\omega$, i.e, $z \mapsto \omega z$. Hence, a real linear map is complex linear if and only if $a_{12} = -a_{21}$ and $a_{11} = a_{22}$.

But we are not only interested in linear maps $\mathbb C \to \mathbb C$, right? So let us take a differentiable function $g \colon \mathbb C \to \mathbb C$ (differentiable means: differentiable as a map $\mathbb R^2 \to \mathbb R^2$, where we identify $\mathbb C$ and $\mathbb R^2$ as usual). Again, we can ask: when is $g$ holomorphic?

Write $g = u+iv$, where $u,v \colon \mathbb R^2 \to \mathbb R$.

So when is $g$ complex differentiable in $z_0 = x_0+iy_0 \in \mathbb C$? This is the case if and only if

$$g'(z_0) := \lim_{z \to z_0} \frac{g(z)-g(z_0)}{z-z_0}$$

exists. Let $(x,y)$ be coordinates of $\mathbb R^2$. Writing $g'(z_0) = a+bi$, we obtain that the following statements are equivalent to the complex differentiability of $g$ in $z_0$:

$$\lim_{z \to z_0} \bigg \vert \frac{g(z)-g(z_0)-g'(z_0)(z-z_0)}{z-z_0}\bigg \vert = 0 \iff \lim_{z \to z_0} \frac{\bigg \vert\begin{pmatrix}u(x,y) \\ v(x,y)\end{pmatrix} -\begin{pmatrix}u(x_0,y_0) \\ v(x_0,y_0)\end{pmatrix} - \begin{pmatrix}a & -b \\ b & a\end{pmatrix}\begin{pmatrix}x-x_0 \\ y-y_0\end{pmatrix}\bigg \vert}{\bigg \vert\begin{pmatrix}x-x_0 \\ y-y_0\end{pmatrix}\bigg \vert} = 0.$$

The latter means that $\begin{pmatrix}a & -b \\ b & a\end{pmatrix}$ is the derivative of $g$ as a real differentiable function. On the other hand, the Jacobian matrix of $g$ is $Dg = \begin{pmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y}\end{pmatrix}$. So $f$ is complex differentiable in $z_0$ if and only if the Cauchy-Riemann equations hold at $z_0$.

Answer 2

$$\begin{pmatrix}a & -b \\ b & a\end{pmatrix} =a\begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix}+b\begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}=aI+bJ.$$

As you can observe,

$$I^2=I,\\IJ=JI=J,\\J^2=-I$$

which matches

$$1\cdot1=1,\\1\cdot i=i\cdot1=i,\\i^2=-1,$$

and all computation rules on the complex are preserved.

Answer 3

Here's another way to think about this correspondence, using linear algebra. Think of $\Bbb{C}$ as a two-dimensional vector space over $\Bbb{R}$. Then, complex multiplication by a fixed number $a + ib$ is a linear map. That is, if we let $T_{a+ib}(z) = (a + ib)z$, for any $z_1, z_2 \in \Bbb{C}$ and $\lambda \in \Bbb{R}$, we have $T_{a+ib}(z_1 + z_2) = T_{a+ib}(z_1) + T_{a+ib}(z_2)$ and $T_{a+ib}(\lambda z_1) = \lambda T_{a+ib}(z_1)$.

Now, $B = \{1, i\}$ forms the standard ordered basis for $\Bbb{C}$, and we can compute the matrix for $T$ under this basis. We compute \begin{align*} T_{a+ib}(1) &= (a + ib)1 = a + ib \implies [T_{a+ib}(1)]_B = \begin{bmatrix}a \\ b\end{bmatrix} \\ T_{a+ib}(i) &= (a + ib)i = -b + ia \implies [T_{a+ib}(1)]_B = \begin{bmatrix}-b \\ a\end{bmatrix}. \end{align*} Thus, our matrix representation is $$M(a + ib) = \begin{bmatrix} a & -b \\ b & a\end{bmatrix}.$$ Now, note that $$T_{a + ib}T_{c + id}(z) = (a + ib)(c + id)z = T_{(a + ib)(c + id)}(z).$$ We also get $$T_{a + ib}(z) + T_{c + id}(z) = (a + ib)z + (c + id)z = (a + ib + c + id)z = T_{a + ib + c + id}(z).$$ In terms of matrices, we get \begin{align*} M(a + ib) + M(c + id) &= M(a + ib + c + id) \\ M(a + ib)M(c + id) &= M((a + ib)(c + id)). \end{align*}

In short, the matrix $\begin{bmatrix} a & -b \\ b & a\end{bmatrix}$ is the standard matrix for the linear transformation of multiplying by $a + ib$, hence multiplication and addition are respected, and the matrices act identically to the complex numbers.