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

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dRic2
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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.Ric
 
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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.
 
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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.
 
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It is correct with that addition
 
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