How would the most general $2 \times 2$ normal matrix look like?

Question

How would the most general $2 \times 2$ normal matrix look like?

The normal matrix satisfy equation: $A^*A=AA^*$ where $A^*$ denotes conjugate transpose.

I was thinking about the matrix:

$$ \begin{pmatrix} a & -b \\ b & a \\ \end{pmatrix} $$ because its columns are orthogonal to each other and it satisfies the given equation:

$$ \begin{pmatrix} a & -b \\ b & a \\ \end{pmatrix} \begin{pmatrix} a & b \\ -b & a \\ \end{pmatrix} = \begin{pmatrix} a^2 + b^2 & 0 \\ 0 & b^2 + a^2 \\ \end{pmatrix} $$ $$ $$

$$ \begin{pmatrix} a & b \\ -b & a \\ \end{pmatrix} \begin{pmatrix} a & -b \\ b & a \\ \end{pmatrix} = \begin{pmatrix} a^2 + b^2 & 0 \\ 0 & b^2 + a^2 \\ \end{pmatrix} $$ $$ $$

It is true for real matrices, and I suppose for the complex one too. But is this the most general case, or is there something else?

Answer 1

Let $$A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}.$$ For $A$ to be normal we require that $$\begin{bmatrix} \overline{a} & \overline{c} \\ \overline{b} & \overline{d} \end{bmatrix} \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} \overline{a} & \overline{c} \\ \overline{b} & \overline{d} \end{bmatrix},$$ or equivalently $$\begin{bmatrix} |a|^2 + |c|^2 & \overline{a}b + \overline{c}d \\ \overline{b}a + \overline{d}c & |b|^2 + |d|^2 \end{bmatrix} = \begin{bmatrix} |a|^2 + |b|^2 & a\overline{c} + b\overline{d} \\ c\overline{a} + d\overline{b} & |c|^2 + |d|^2 \end{bmatrix}.$$

Firstly notice that by equality of the diagonal entries we must have $$|b|^2 = |c|^2.$$ And furthermore comparing the off-diagonal entries we must have $$ \overline{a}b + \overline{c}d = a\overline{c} + b\overline{d}, $$ $$ \overline{b}a + \overline{d}c = c\overline{a} + d\overline{b}. $$

In the real case the first requirement reduces to $c=|b|$, but for complex matrices we could also have for example $b=1$ and $c=i$. For now let's just consider the real case: the most general case is to fix $b$ and let $c=|b|$. Then we have two cases:

1. If $b$ is nonnegative $a$ and $d$ could be any real values.
2. If $b$ is negative $a=d$.

The second case above corresponds to your example - I wouldn't consider this to be more general than the first case - for real numbers 1 and 2 together is most general.

So now for complex numbers there is a lot more work to be done: you could explore this further by for example setting $a = a_1 + ia_2, b=b_1 + ib_2, \ldots$, fixing some and then working out by the constraints what values the other variables could assume...

Answer 2

You can also characterize a normal matrix $A$ by the fact there exists an unitary matrix such that $U^{-1}AU=D$ where $D$ is diagonal.

Unitary $2x2$ matrices are well characterized, see for instance the wiki page: https://en.wikipedia.org/wiki/Unitary_matrix

$\exists (\alpha,\beta,\xi,\zeta)\in\mathbb R^4$ such that $U=\begin{bmatrix}\cos \alpha & -\sin \alpha\\\sin \alpha & \cos \alpha\end{bmatrix}\begin{bmatrix}e^{i\xi} & 0\\0 & e^{i\zeta}\end{bmatrix}\begin{bmatrix}\cos \beta & -\sin \beta\\\sin \beta & \cos \beta\end{bmatrix}$

$U=R(\alpha).\operatorname{diag}(e^{i\xi},e^{i\zeta}).R(\beta)$

$U^{-1}=R(-\beta).\operatorname{diag}(e^{-i\xi},e^{-i\zeta}).R(-\beta)$

Now you can take $D=\begin{bmatrix}z_1 & 0\\0 & z_2\end{bmatrix}$

Multiplying all these won't probably be very nice looking, but you get to parametrize any normal matrix with $4$ reals and $2$ complexes.

Yet, somehow, this isn't very satisfying for me, the dimension on $\mathbb R$ still appear to be $8$ while C.Haattingh showed that we need $|b|^2=|c|^2$ which can be rewritten $b=re^{it_b}$ and $c=re^{it_c}$ and we reduced the dimension by $1$ already. So I suspect that my $D$ and $(\xi,\zeta)$ are in fact not independent... Is it possible to fix that ?