# Can we always find this kind of unitary matrix to diagonalize this special form hermitian matrix?

Suppose we have a hermitian matrix $$M$$ of form $$\left( \begin{matrix} \alpha& -\beta ^*\\ \beta& -\alpha ^*\\ \end{matrix} \right)$$ where $$\alpha$$ is a hermitian matrix and $$\beta$$ is an antisymmetric matrix with equal dimension, and I use the star $$*$$ for complex conjugate. Since $$M$$ is hermitian, then it can be diagonalized into $$TMT^{\dagger}=\left( \begin{matrix} d& 0\\ 0& -d\\ \end{matrix} \right)$$ where $$d$$ is diagonal, and I used the special form of $$M$$ to deduce that the eigenvalues are real and appear in matched pairs $$\pm \lambda$$.

But I want to say more about the unitary matrix $$T$$, i.e., can $$T$$ always be chosen to be of form $$\left( \begin{matrix} \gamma& \mu\\ \mu ^*& \gamma ^*\\ \end{matrix} \right)$$ where $$\gamma$$ and $$\mu$$ are two matrices with equal dimensions?

Edit

The question is found in this notes(from the bottom of page 6 to 7) for introduction of Jordan-Wigner transformation. The author states the above rules while I really can't see how to show that. I quoted the closely related part for your convenience:

One way of obtaining a rigorous proof is to find a $$T$$ satisfying $$T M T^{\dagger}=\left[\begin{array}{cc} d & 0 \\ 0 & -d \end{array}\right]$$ and then to apply the cosine-sine (or CS) decomposition from linear algebra, which provides a beautiful way of representing block unitary matrices, and which, in this instance, allows us to obtain a $$T$$ of the desired form with just a little more work.

Let $$\dim(M)=2n$$.
$$\newcommand{\b}[1]{\begin{bmatrix}#1\end{bmatrix}}$$ Let $$S=\b{0 &I_n\\I_n &0}$$. Then $$S^{-1}=S^*=S$$. $$S^{-1}MS=-M^*.$$

Recall that $$\Bbb C^{2n\times1}$$ is a Hermitian space with inner product defined by $$\langle x, y\rangle=x^\dagger y$$.
Let $$\kappa:\Bbb C^{2n\times1}\to\Bbb C^{2n\times1}$$, $$\kappa(x)=Sx^*$$, i.e., $$\kappa\left(\b{a\\b}\right)=\b{b^*\\a^*}.$$
$$\kappa$$ is an antiunitary operator and $$\kappa^{-1}=\kappa$$.

Let $$\{0\}\not=M_\lambda=\{x\in\Bbb C^{2n\times1}\mid Mx=\lambda x\}$$, the $$\lambda$$-eigenspace of $$M$$.
Since $$M$$ is Hermitian, $$\lambda\in\Bbb R$$ and $$\Bbb C^{2n\times1}=\bigoplus M_{\lambda}$$.
Suppose $$x\in M_{\lambda}$$. Since $$MSx^*=S(S^{-1}MS)x^*=S(-M^*)x^*=-S(Mx)^*=-S(\lambda x)^*=-\lambda Sx^*,$$ we have $$\kappa(x)\in M_{-\lambda}$$. Hence $$\kappa$$ induces an antiunitary isomorphism between $$M_{\lambda}$$ and $$M_{-\lambda}$$.

Let $$\lambda_i$$, where $$1\le i\le k$$ and $$k\ge0$$ be all positive eigenvalues of $$M$$.
Let $$B_{\lambda_i}$$ be an orthonormal basis of $$M_{\lambda_i}$$.
Let $$B_{-\lambda_i}=\left\{\kappa(x)\mid x\in B_{\lambda_i}\right\}$$.
We can see that $$B_{-\lambda_i}$$ is an orthonormal basis of $$M_{-\lambda_i}$$.

Note that $$\kappa$$ induces an antiunitary operator on $$M_0$$. Let us choose a "$$\kappa$$-paired" orthonormal basis of $$M_0$$.

Let $$R_1=M_0$$. Starting from $$t=1$$, repeat the following procedure as long as $$R_t$$ is not $$\{0\}$$. We will keep $$R_t$$ as an even-dimensional $$\kappa$$-invariant $$\Bbb C$$-linear subspace of $$M_0$$.

• Suppose we can find $$x\in R_t$$ such that $$x$$ and $$\kappa(x)$$ are $$\Bbb C$$-linear independent.
Let $$\mathscr Z_t$$ be the $$\Bbb C$$-linear space spanned by $$x$$ and $$\kappa(x)$$. Let $$R_{t+1}$$ be the $$\Bbb C$$-linear subspace of $$R_t$$ that is orthogonal to $$\mathscr Z_t$$. $$\dim(R_{t+1})$$ is even since it is $$\dim(R_{t})-2$$.
Verify that $$\kappa$$ induces an antiunitary operator on $$\mathscr Z_t$$ and $$R_{t+1}$$.
Thanks to the linear independence of $$x$$ and $$\kappa(x)$$, we can find $$0\not=y\in\mathscr Z_t$$ such that $$y$$ and $$\kappa(y)$$ is orthogonal. Let $$z_t=y/||y||$$.
$$\{z_t, \kappa(z_t)\}$$ is an orthonormal basis of $$\mathscr Z_t$$.
Add $$1$$ to $$t$$.
• Otherwise for all $$x\in R_t$$ and $$x$$ and $$\kappa(x)$$ are $$\Bbb C$$-linear dependent.
We claim that if $$0\not=x$$ and $$\kappa(x)$$ are $$\Bbb C$$-linear dependent, then $$x=\b{u\\e^{i\Theta}u}$$ for some $$u\in\Bbb C^{n\times1}$$ and $$\Theta\in \Bbb R$$. The proof is straightforward.
Since $$R_t$$ is even-dimensional, there are $$x,y\in R_t$$ that are $$\Bbb C$$-linear independent. Then either $$x+y$$ and $$\kappa(x+y)=\kappa(x)+\kappa(y)$$ are not $$\Bbb C$$-linear dependent, or $$x-iy$$ and $$\kappa(x-iy)=\kappa(x)-i\kappa(y)$$ are not $$\Bbb C$$-linear dependent.
This contradicts the assumption of this case. Hence, this case does not exist.

Since the dimension of $$R_t$$ decreases by $$2$$ at each iteration, the procedure above will run $$\dim(M_0)/2$$ times.
We obtain that $$M_0$$, as an orthogonal sum of $$\mathscr Z_t$$, $$t=1,\cdots, \dim(M_0)/2$$, has an orthonormal basis $$z_1, \cdots, z_{\dim(M_0)/2}, \kappa(z_1), \cdots, \kappa(z_{\dim(M_0)/2})$$.

Let $$\cup B_{\lambda_i}=\{w_1, \cdots, w_{n-\dim(M_0)/2}\}$$. Let $$T$$ be the following unitary matrix $$\b{w_1, \cdots, w_{n-\dim(M_0)/2}, z_1, \cdots, z_{\dim(M_0)/2}, \kappa(w_1), \cdots, \kappa(w_{n-\dim(M_0)/2}), \kappa(z_1), \cdots, \kappa(z_{\dim(M_0)/2})}$$ Then $$T$$ is of the required form $$\b{\gamma &\mu\\\mu^* &\gamma^*}$$ and $$MT=\b{\Lambda & 0 & 0 & 0\\0 & 0 & 0 & 0\\0&0 &-\Lambda& 0\\0 & 0 & 0 &0}T=\b{d &0 \\ 0 & -d}$$ where $$\Lambda$$ is a diagonal real matrix with $$\lambda_i$$'s on the diagonal, and $$d$$ is the $$n$$-dimensional diagonal real matrix $$\b{\Lambda & 0\\ 0 & 0 }$$.

Since $$T^\dagger M(T^\dagger)^\dagger=T^\dagger MT=\b{d &0 \\ 0 & -d}$$, and $$T^\dagger$$ is an unitary matrix of the form $$\b{\gamma &\mu\\\mu^* &\gamma^*}$$ as well, we have done.

• Thanks for the answer, but I think I need sometime to digest the answer and added some references to my original post. And $\beta$ is antisymmetric so a number might not be a choice, I think.. Commented Sep 1, 2022 at 5:41
• @Sherlock You are right. Since $\beta$ is antisymmetric, a one-dimensional $\beta$ must be $0$. My answer was completely wrong. Please check my updated answer. It takes quite some time for me to figure out how we can handle the $0$-eigenspace of $M$. Hopefully, my answer is good enough for you to accept. Commented Sep 12, 2022 at 3:02

First all blocks are considered matrices of equal size. Set $$M=\begin{pmatrix}A&B\\-B&-A\end{pmatrix}$$ a $$2n\times 2n$$ complexe matrix with $$A^*=A$$ and $$B^*=-B$$. Let $$S=\frac{1}{\sqrt{2}}\begin{pmatrix}I&-I\\I&I\end{pmatrix}$$, $$X=A+B$$ so $$X^*=A-B$$. Here take the polar decomposition of $$X$$ as $$X=U|X|$$ with $$H^*|X|H=D$$ where $$D$$ is a diagonal matrix and $$H$$ unitary. Now we apply the following $$K:=(LSRS)M(LSRS)^*$$ where $$R=\begin{pmatrix}U^*&0\\0&I\end{pmatrix}$$ and $$L=\begin{pmatrix}H^*&0\\0&H^*\end{pmatrix}$$ are unitaries. We check that $$K=\begin{pmatrix}-D&0\\0&D\end{pmatrix}$$ and that $$LSRS=\begin{pmatrix}E&F\\-F&-E\end{pmatrix}$$. This is different than the form presented for $$T$$, in dimension $$2$$ ($$E$$ and $$F$$ are scalars) this is almost invalid for $$b\neq 0$$, $$M=\begin{pmatrix}a&ib\\-ib&-a\end{pmatrix}$$, the two eigenvectors are up to a factor: $$\begin{pmatrix}\frac{-i(\sqrt{a^2+b^2}-a)}{b}\\1\end{pmatrix}$$ and $$\begin{pmatrix}\frac{i(\sqrt{a^2+b^2}+a)}{b}\\1\end{pmatrix}$$...

• Sorry, I want to ask do you use $*$ for complex conjugate? I use $*$ as complex conjugate instead of conjugate transpose. Commented Aug 16, 2022 at 12:50
• Hey it is for $A^*=\bar{A}^T$ (conjugate transpose as adjoint). Commented Aug 16, 2022 at 13:06
• I'm sorry that I use $A^*$ to mean $\bar A$. I just get used to use $A^\dagger$ stands for $\bar A^T$... Sorry about the ambiguity. Commented Aug 16, 2022 at 13:13