How to determine all $2 \times 2$ normal matrices?

Question

Determine all $2 \times 2$ normal matrices.

In particular, how would I show that there are normal matrices which are neither unitary, Hermitian, skew-Hermitian, symmetric, nor skew-symmetric. The only thing I do know is $AA^* = A^*A$, but I'm not sure how to proceed. I tried writing in $a,b,c,d$ as entries of $2 \times 2$ matrix, but it seemed to lead nowhere.

Answer 1

Note that $A$ is normal if and only if $AA^* - A^*A = 0$. If we take $$ A = \pmatrix{a&b\\c&d}, $$ then we find $$ AA^* - A^*A = \pmatrix{|b|^2 - |c|^2 & -b \bar a + a \bar c + b \bar d - d \bar c\\ \overline{-b \bar a + a \bar c + b \bar d - d \bar c} & |c|^2 - |b|^2}. $$ $A$ will be normal if and only if all the above entries are zero. In particular, we see that this means that $A$ is normal if and only if $$ |c| = |b|, \quad -b \bar a + a \bar c + b \bar d - d \bar c = 0. $$ In order to take advantage of the first equation, write $b,c$ in polar form. That is, let's say $$ b = r_1e^{i\theta}, \quad c = r_1e^{i \phi} $$ where $r_1 \geq 0$ and $\theta,\phi \in \Bbb R$. Substituting these into the second equation yields $$ - r_1e^{i \theta}\bar a + a r_1e^{-i\phi} + re^{i\theta} \bar d - d r_1e^{-i\phi} = 0 \implies\\ e^{i \theta}[\bar d-\bar a] + e^{-i \phi}[a-d] = 0 \implies\\ e^{i \theta}[\overline{d-a}] - e^{-i \phi}[d-a] = 0 \implies\\ e^{i (\theta + \phi)}[\overline{d-a}] - [d-a] = 0. $$ Write $d-a$ in polar form. That is, take $d-a = r_2 e^{i \psi}$. We can write the above as the equation $$ e^{i (\theta + \phi)}r_2 e^{-i\psi} - r_2e^{i\psi} = 0 \implies\\ e^{i(\theta + \phi - \psi)} = e^{i \psi} \implies\\ e^{i(\theta + \phi)} = [e^{i \psi}]^2 \implies\\ \pm \exp\left[i\frac{\theta + \phi}2\right] = e^{i \psi} = \frac{d-a}{r_2} \implies\\ d-a = \pm r_2 \exp\left[i\frac{\theta + \phi}2\right]. $$

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The above analysis leads to the following parameterization of the normal matrices. Select any $a \in \Bbb C$, $r_1,r_2 \geq 0$ and angles $\phi,\psi$. The matrix with this value of $a$ and $$ b = r_1e^{i\theta}, \quad c = r_1e^{i \phi}, \quad d = a \pm r_2 \exp\left[i\frac{\theta + \phi}2\right] $$ is necessarily normal, and every normal matrix can be written in this way. We could also tweak some definition to get the equivalent parameterization $$ a,b = [\text{arbitrary complex}], \quad c = b e^{2i k_1}, \quad d = a + k_2 e^{i k_1} $$ where $k_1,k_2$ are real. This amounts to $$ A = aI + \pmatrix{0&b\\b e^{2i k_1} & k_2 e^{ik_1}}. $$

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As far as the second part goes: it suffices to take a multiple of a unitary matrix. For instance, take the matrix of any rotation by an angle that is not a multiple of $\pi/2$, and multiply it by some $\alpha > 1$.

Alternatively: for any normal $A$, there exists a $\gamma \in \Bbb C$ such that $A + \gamma I$ is neither unitary, Hermitian, nor skew-Hermitian. To see that this is the case, it suffices to look at the expressions for $M^*M$ and $M \pm M^*$, where $M = A + \gamma$ (in terms of $A$ and $\gamma$, not in terms of the entries).

Alternatively: note that every diagonal matrix is normal. However, not every diagonal matrix is Hermitian, skew-Hermitian, or unitary.

Answer 2

I'll give a slightly different take on the other answer.

Start by noticing that the identity component of $A$ doesn't affect it being normal, that is, if $A=\alpha I+\tilde A$ with $\tilde A$ traceless, then $A$ is normal iff $\tilde A$ is. We can thus assume without loss of generality $A$ to be traceless, and write it as $$A = \begin{pmatrix} z & b \\ c & -z\end{pmatrix},\qquad z,b,c\in\mathbb C.$$ The normality condition, $AA^\dagger=A^\dagger A$, translates into the following two conditions on the parameters: $$ \begin{cases}|b|=|c|, \\ c\bar z=z \bar b.\end{cases} $$ Writing the polar representations of the parameters as $z=r_z e^{i\varphi_z}, b=r_b e^{i\varphi_b}$, $c=r_c e^{i\varphi_c}$.

1. Assuming $z\neq0$: these conditions amount to $r_b=r_c$ and $\varphi_c +\varphi_b = 2\varphi_z$. It follows that the generic form of a normal 2x2 complex matrix is $$ A=\alpha I + \begin{pmatrix} r_z e^{i(\varphi_b+\varphi_c)/2} & r_b e^{i\varphi_b}\\ r_b e^{i\varphi_c}& -r_z e^{i(\varphi_b+\varphi_c)/2}\end{pmatrix}, $$ for any $\alpha\in\mathbb C$ and $r_z,r_b,\varphi_b,\varphi_c\in\mathbb R$. We can further simplify this assuming $\varphi_z=0$ (which amounts to collecting a global phase for the matrix). We then get the general solution $$ A=\alpha I + e^{i\varphi_z}\begin{pmatrix} r_z & r_b e^{i\varphi_b}\\ r_b e^{-i\varphi_b}& -r_z\end{pmatrix}, $$ for all $\alpha\in\mathbb C$ and $r_z,r_b,\varphi_b,\varphi_z\in\mathbb R$.
2. For $z=0$, the only constraint is $|b|=|c|$. The general solution is thus $$A = \alpha I +\begin{pmatrix}0 & b e^{i\varphi} \\ b & 0\end{pmatrix},$$ for any $\alpha,b\in\mathbb C$ and $\varphi\in\mathbb R$.