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My question here is: is the next matrix an hermitian matrix?

When you transpose and conjugate it, you obtain the same matrix, but i'm not really sure, because i found only $2$ eigenvectors, and if this is an hermitian matrix, i should to find $3$ eigenvectors, cause it has two eigenvalues $(0,2)$ with multiplicities $(2,1)$. (Remember that an hermitian matrix is a diagonalizable matrix).

\begin{bmatrix}1&0&-i\\0&0&0\\i&0&1\end{bmatrix}

Thanks!!

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  • $\begingroup$ What is the dimension of the eigenspace of 0? $\endgroup$ – MattG88 Dec 1 '16 at 2:11
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    $\begingroup$ This is diagonalizable; for the eigenvalue $0$ you have eigenvectors $(0,1,0)$ and $(1,0,-\mathrm{i})$. $\endgroup$ – Morgan Rodgers Dec 1 '16 at 2:11
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The Hermitian conjugate of that matrix is equal to the matrix: Transpose the matrix bringing a $+i$ to the upper right, then complex conjugate it, and you see you get back the original. So this matrix is Hermitian.

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