# Transform a 2x2 matrix into an anti-symmetric matrix

• I
• dRic2

#### dRic2

Gold Member
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

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 $\lambda^2 = a^2 - b^2$.

• topsquark, PeroK and 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?

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

Maarten Havinga said:
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

• dRic2