2×2 normal matrices are multiples of unitary matrices

hbghlyj

Among 2×2 matrices, normal matrices are multiples of unitary matrices.
Proof:
$A=\pmatrix{a&b\\c&d}$ is normal$\Leftrightarrow A^{\rm H}A=AA^{\rm H}$
\[\Leftrightarrow\left(
\begin{array}{cc}
a a^*+c c^* & b a^*+d c^* \\
a b^*+c d^* & b b^*+d d^* \\
\end{array}
\right)=\left(
\begin{array}{cc}
a a^*+b b^* & a c^*+b d^* \\
c a^*+d b^* & c c^*+d d^* \\
\end{array}
\right)\]
What should I do next

hbghlyj

$aa^*+cc^*=aa^*+bb^*\Rightarrow cc^*=bb^*$. Let $k=\sqrt{c^*\over b}$, then $|k|=1$, so $k^*=\frac1k$. We have
$$kb=k^*c^*$$
and
$$kc=k^*b^*$$
Now we show $kA$ is self-adjoint.
$$kA=k\pmatrix{a&b\\c&d}=\pmatrix{ka&k^*c^*\\k^*b^*&kd}$$
and$$(kA)^*=\pmatrix{k^*a^*&kb\\kc&k^*d^*}$$
To establish $kA=(kA)^*$, it suffices to show both $ka$ and $kd$ are real.
$ba^*+dc^*=ac^*+bd^*\Rightarrow b(a-d)^*=c^*(a-d)\Rightarrow k(a-d)=k^*(a-d)^*\Rightarrow k(a-d)\in\Bbb R$
We've used all the equations, but cannot show both $ka$ and $kd$ are real.

hbghlyj

I suspect this proposition is false, so I wrote a counterexample in Wikipedia talk page
The matrix $A=\pmatrix{1&0\\0&2}$ is Hermitian, so $A$ is normal, but eigenvalues of $A$ are $1$ and $2$, so $A$ is not a multiple of a unitary matrix.

Czhang271828

In fact, each complex matrix $A\in \mathbb C^{n\times n}$ is a linear combination of at most $2$ unitary matrices. Consider the SVD decomposition $A=U\Sigma V$ and write $\Sigma$ as a linear combination of $2$ unitary matrices.

E.g. Write $\begin{pmatrix}4&0\\3&-5\end{pmatrix}$ into linear combination of $2$ unitary matrices.

Step I. Find the SVD decomposition ($A=U\Sigma V$), i.e.,
$$
\begin{pmatrix}4&0\\3&-5\end{pmatrix}=\begin{pmatrix}\frac{-1}{\sqrt 5}&\frac{2}{\sqrt 5}\\\frac{-2}{\sqrt 5}&\frac{-1}{\sqrt 5}\end{pmatrix}\begin{pmatrix}2\sqrt{10}&0\\0&\sqrt{10}\end{pmatrix}\begin{pmatrix}\frac{-1}{\sqrt 2}&\frac{1}{\sqrt 2}\\\frac{1}{\sqrt 2}&\frac{1}{\sqrt 2}\end{pmatrix}.
$$
Step II. Write $\Sigma $ as a linear combination of $2$ unitary matrices, e.g.,
$$
\begin{pmatrix}2\sqrt{10}&0\\0&\sqrt{10}\end{pmatrix}=\begin{pmatrix}\frac{2\sqrt{10}+3i}{2}&0\\0&\frac{\sqrt{10}+i\sqrt{39}}{2}\end{pmatrix}+\begin{pmatrix}\frac{2\sqrt{10}-3i}{2}&0\\0&\frac{\sqrt{10}-i\sqrt{39}}{2}\end{pmatrix}.
$$
Step III. Recover $A=U\Sigma_1V+U\Sigma_2V$, i.e.,
$$
\begin{align*}
A&=\frac{1}{20}\left(\begin{array}{cc}40 +3i \sqrt{10}+2 i\sqrt{390} & 2i \sqrt{390}-3i \sqrt{10} \\ 30 +6i \sqrt{10}-i\sqrt{390} & -50 -6i \sqrt{10}-i\sqrt{390}\end{array}\right)\\
&\quad\,+\frac{1}{20}\left(\begin{array}{cc}40 -3i \sqrt{10}-2 i\sqrt{390} & -2i \sqrt{390}+3i \sqrt{10} \\ 30 -6i \sqrt{10}+i\sqrt{390} & -50 +6i \sqrt{10}+i\sqrt{390}\end{array}\right).
\end{align*}
$$

hbghlyj

Matrix Analysis, Horn & Johnson (1985), p. 109
2.5 Normal matrices Exercise 13

$$A\text{ is normal iff }A^* = AV\text{ for some unitary matrix }V$$

Czhang271828

hbghlyj 发表于 2023-6-9 04:09
Matrix Analysis, Horn & Johnson (1985), p. 109
2.5 Normal matrices Exercise 13

$\Longleftarrow$: Since\[[A^\ast =AV]\Leftrightarrow [A=V^\ast A^\ast]\Leftrightarrow [VA=A^\ast],\]we see that $A$ and $V$ commutes. Hence\[A^\ast A=(AV)A=A(VA)=AA^\ast.\]
$\Longrightarrow$: If $A$ is normal, then we have the decomposition $A=U^\ast \Lambda U$ for unitary matrix $U$ and diagonal matrix
\begin{align*}
\Lambda&=\mathrm{diag}(\lambda_1,\ldots ,\lambda_n)\\
&=\mathrm{diag}(|\lambda_1|,\ldots |\lambda_n|)\cdot \mathrm{diag}(e^{i\theta_1},\ldots, e^{i\theta_n})\\
&=:R\cdot \Theta.
\end{align*}
Hence $A=U^\ast R \Theta U$, and \[ A^\ast =U^\ast \Theta^\ast RU=U^\ast R\Theta UU^\ast (\Theta^\ast)^2 U=A(U^\ast \Theta^\ast \Theta^\ast U).\]
Therefore, $V=U^\ast \Theta^\ast \Theta^\ast U$.