# Block diagonalize skew symmetric matrix with orthogonal transformation

I have a real-valued, antisymmetric matrix (A and B real):

$$H=\begin{bmatrix} 0 & A & 0 \\ -A & 0 & B \\ 0 & -B & 0 \\ \end{bmatrix}$$

which can be diagonalized with a unitary transformation:

$$S= \frac{1}{\sqrt{A^2+B^2}} \begin{bmatrix} B & \frac{-A}{\sqrt{2}} & \frac{-A}{\sqrt{2}} \\ 0 & \frac{-i}{\sqrt{2}}\sqrt{A^2+B^2} & \frac{i}{\sqrt{2}}\sqrt{A^2+B^2} \\ A & \frac{B}{\sqrt{2}} & \frac{B}{\sqrt{2}} \\ \end{bmatrix}$$

such that

$$D=S^\dagger H S = \begin{bmatrix} 0 & 0 & 0 \\ 0 & i\lambda & 0 \\ 0 & 0 & -i\lambda \\ \end{bmatrix}$$

where $\lambda=\sqrt{A^2+B^2}$

It is a known property of skew symmetric matrices that they can be brought to the block diagonal form containing the same eigenvalues as follows:

$$\Sigma=Q^T H Q = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & \lambda \\ 0 & -\lambda & 0 \\ \end{bmatrix}$$

where $Q$ is a real orthogonal matrix. How do you generally construct the matrix $Q$? I have found two unitary transformations $$W_{1} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \frac{-i}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ 0 & \frac{1}{\sqrt{2}} & \frac{-i}{\sqrt{2}} \\ \end{bmatrix}$$

and

$$W_{2} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \frac{1}{\sqrt{2}} & \frac{-i}{\sqrt{2}} \\ 0 & \frac{i}{\sqrt{2}} & \frac{-1}{\sqrt{2}} \\ \end{bmatrix}$$

which transform $D$ to $\Sigma$ as $\Sigma=W_{j}DW_{j}^\dagger$ for $j=1,2$. Then we have $H=TDT^\dagger$ where $T=SW_{j}^\dagger$. I would expect $T$ to be the real orthogonal matrix $Q$ I am looking for (or its transpose), but instead it is a complex-valued matrix (unitary by construction). How do I construct the real orthogonal matrix $Q$ in this example (and in general)?

Similar questions on stackexchange that I couldn't figure out the answer from are:

In general, if $u\pm iv$ are eigenvectors to a conjugate pair of eigenvalues $a\pm ib$ of a real normal matrix $H$, we must have $\|u\|=\|v\|$ and $u\perp v$. Therefore, with respect to the ordered basis $\{u,v\}$, the matrix representation of the linear map $x\mapsto Mx$ when restricted on the linear span of $u,v$ is given by $\pmatrix{a&-b\\ b&a}$.
In your case, the conjugate pair of eigenvectors are $u\pm iv$, where $u=\frac1{\sqrt{2(A^2+B^2)}}(-A,0,B)^\top$ and $v=\frac1{\sqrt{2(A^2+B^2)}}(0,\sqrt{A^2+B^2},0)^\top$. The matrix $H$ also has an eigenvector $w=\frac1{A^2+B^2}(B,0,A)^\top$ in its null space. Therefore, if you take $Q=\pmatrix{\frac{w}{\|w\|},\frac{u}{\|u\|},\frac{v}{\|v\|}}=\pmatrix{w,\sqrt{2}u,\sqrt{2}v}$, then $Q$ is real orthogonal and $Q^THQ$ will give you the desired block-diagonal form $\Sigma$.
• A small correction: $\|u\|$ and $\|v\|$ do not have to equal one another (take $b = 0$ for instance). But if they do, then $u \perp v$ follows. Commented Jun 4, 2021 at 0:27
• @JohnJiang Sorry for the late reply. When we write “a conjugate pair of eigenvalues $a\pm ib$”, it is understood that we are talking about non-real eigenvalues. Hence $b$ is nonzero. Commented Mar 9, 2022 at 20:52