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.

  • $\begingroup$ @hardmath But is normalization a necessary condition? $\endgroup$ – user_9 Jan 18 at 17:41
  • $\begingroup$ I'm suggesting it as a tactic, to discover the $2\times 2$ normal matrices in the most economical way. For example, $A$ is normal if and only if $A- I$ is normal. And $A$ is normal if and only if $\beta A$ is normal for some nonzero real $\beta$. I think we can help ourselves by reframing the problem with less than the four parameters you started with. $\endgroup$ – hardmath Jan 18 at 17:46
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    $\begingroup$ Do you know the spectral theorem for normal matrices? $\endgroup$ – user108903 Jan 18 at 17:58
  • $\begingroup$ Not at all. I've attended about 3 lectures on intro to linear algebra. $\endgroup$ – user_9 Jan 18 at 18:00
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    $\begingroup$ Ok, well that would give you one easy way to describe all normal matrices in terms of the unitary matrices. Maybe you can look it up. For the second part of the problem one approach is to think of a unitary matrix without the last four properties [try a real rotation matrix] and then rescale it to make it non-unitary. $\endgroup$ – user108903 Jan 18 at 18:20

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 $$ e^{i (\theta + \phi)}r_2 e^{-i\psi} - r_2e^{i\psi} = 0 \implies\\ e^{i(\theta + \phi - \psi)} = e^{i \psi} \implies\\ e^{i(\theta + \phi)} = [e^{i \psi}]^2 \implies\\ \pm \exp\left[i\frac{\theta + \phi}2\right] = e^{i \psi} = \frac{d-a}{r_2} \implies\\ d-a = \pm r_2 \exp\left[i\frac{\theta + \phi}2\right]. $$

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 $$ A = aI + \pmatrix{0&b\\b e^{2i k_1} & k_2 e^{ik_1}}. $$

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, skew-Hermitian, symmetric, nor skew-symmetric. To see that this is the case, it suffices to look at the expressions for $M^*M$, $M \pm M^*$, and $M \pm M^T$ where $M = A + \gamma$ (in terms of $A$ and $\gamma$, not in terms of the entries).


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