How are these two sentences related?

Question

Let $T$ be a transformation from $\mathbb{R}^2$ to $\mathbb{R}^2$, represented by the matrix:

$$ T = \left [ { \begin{array}{cc} 0 & -1 \\ 1 & 0 \\ \end{array} } \right ] $$

The eigenvalues of this matrix are $i$ and $-i$.

In this video by 3Blue1Brown, he says that this particular fact has directly related to another fact:

Multiplying a complex number by $i$, is equivalent to rotating the complex plane by $90^0$ and checking where the original complex number (point in complex plane) lands.

How are these two facts related? How could a fact about matrices in $\mathbb{R}^2$ having complex valued eigenvalues be related to a fact about complex planes and a relationship between complex numbers?

Answer 1

This matrix $T = \left [ { \begin{array}{cc} 0 & -1 \\ 1 & 0 \\ \end{array} } \right ]$ corresponds to $i$ in complex number.

Consider $T^2$, it is equal to $- I$, just like $i^2 = -1$ in ordinary number.

Another way to think of it is by change of coordinates, the rotation matrix $T = \left [ { \begin{array}{cc} \cos \theta & - \sin \theta \\ \sin \theta & \cos \theta \\ \end{array} } \right ]$,

plug in $\theta = \pi /2 $, this corresponds to rotation by $\pi /2$, which is equivalent to $i$

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Just added:

Next, by the matrix equation $Av=\lambda v$. Left side means rotation by $\pi /2$, if you relate $v$ to a complex number, this just means $i$ times $v$, so $Av=i v$, equivalent to the fact that eigenvalue is $i$.

Answer 2

Another view: $T$ generalizes to something called an almost complex structure on, say, a vector bundle. The only condition is that $J$ is a linear operator such that $J^2 = - \textrm{Id}$; such structures are valuable because they admit a $\mathbb{C}$-module structure on an otherwise real vector space by defining $i x := Jx$ for $x \in V$, where $V$ is a $\mathbb{R}$-vector space. Since $J$ satisfies its own characteristic polynomial (Cayley-Hamilton), the condition $J^2 = - \textrm{Id}$ implies that there are eigenvalues $\pm i$.

You can construct these on $\mathbb{C}^n$ in a natural way: look at the endomorphism induced by multiplying vectors by $i$ when viewing $\mathbb{C}^n$ as $\mathbb{R}^{2n}$. In your example, consider $\mathbb{C}$ as $\mathbb{R}^2$ in the standard fashion; that is, $x+ i y \mapsto (x,y)$. Then, look at the induced matrix transformation when multiplying by $i$ on each basis vector. We see $$i \cdot (1,0) = i \cdot 1 = i = (0,1)$$ and $$i \cdot (0,1) = i \cdot i = -1 = (-1,0)$$ Thus the induced endomorphism when viewing $i$ as a transformation of $\mathbb{C}$ as a real vector space is precisely the matrix $$T = \begin{pmatrix} 0 & -1 \\ 1 & 0 \\ \end{pmatrix}$$ And, of course, $i = e^{i \pi/2}$ in the complex plane, which means that multiplication by $i$ is the same as rotating a complex number $90$ degrees. Thus we deduce that $T$ as above has the effect of rotating a real vector $90$ degrees in $\mathbb{R}^2$.