Transform a 2x2 matrix into an anti-symmetric matrix

[dRic2]

Hi,
I have a 2x2 hermitian matrix like:
$$
A = \begin{bmatrix}
a && b \\
-b && -a
\end{bmatrix}
$$
(b is imaginary to ensure that it is hermitian). I would like to find an orthogonal transformation M that makes A skew-symmetric:
$$
\hat A = \begin{bmatrix}
0 && c \\
-c && 0
\end{bmatrix}
$$
Is it possible, or I need to constrain my problem more? I need M to be orthogonal and with det(M) = 1. I was thinking maybe there are some tricks involving Pauli matrices.

Best,
Ric

 

[pasmith]

This is possible since the minimal polynomials of $A$ and $\hat A$ are $m_A(\lambda) = \lambda^2 + b^2 - a^2$ and $m_{\hat A}(\lambda) = \lambda^2 + c^2$, so if $c^2 = b^2 - a^2$ they will have the same minimal polynomial and the same Jordan normal form (which in this case is diagonal).

However, I don't think $M$ will be orthogonal unless the eigenvectors of $A$ are orthogonal, which does not appear to be the case in general: the eigenvectors are $(b, a \mp \lambda)$ and their inner product is $$|a|^2 + |b|^2 + 2\operatorname{Im}(a\bar{\lambda}) - |\lambda|^2$$ where $\lambda^2 = a^2 - b^2$.

 

[dRic2]

Question: since A is hermitian $\lambda$ are reals, so if $a$ is real, then $\text{Im}(a\lambda) = 0$
and
$$
|a|^2 - |b|^2 - (a^2 - b^2) = |b|^2 + b^2 = 0
$$
because $b$ is purely imaginary. Right?

 

[Maarten Havinga]

No, you have $|b|^2$ with the wrong sign on the left.

 

[dRic2]

No, you have $|b|^2$ with the wrong sign on the left.

Sorry it was a typo. It should read $+|b|^2$ and it should be correct.

 

[Maarten Havinga]

It is correct with that addition