Given $ \begin{equation*} Y = \begin{pmatrix} y & 0 & 0 \\ 0 & 0 & -iy \\ 0 & iy & 0 \end{pmatrix} \end{equation*}$
Given another matrix $X$ which has two degenerate eigenvalues and commutes with $Y$, that is $XY = YX$. Can we claim anything about $X$ in general, more specifically if $X$ is Hermitian or not. I used commutative property to find $X$ as
$ \begin{equation*} X = \begin{pmatrix} a & ci & c \\ -mi & t & -n \\ m & n & t \end{pmatrix} \end{equation*}$
But I can' use the property that $X$ has two degenerate eigenvalues. Help would be appreciated.
Thanks in advance.
No $X$ need not be Hermitian. Given any Hermitian matrix $X$ satisfying the constraints in the question, we define $X' = i X$. Then $X'$ also satisfies all constraints in the question but is not Hermitian.
But let's investigate $X$ further. Working in the orthonormal basis in which $Y$ is diagonal we have $$ Y = \begin{pmatrix} y & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & -y \end{pmatrix} $$ Then by the commutation constraint $X$ needs to be of the form $$ X = \begin{pmatrix} x_{00} & x_{01} & 0 \\ x_{10} & x_{11} & 0 \\ 0 & 0 & x_{22} \end{pmatrix} $$
But now take the top left block to be any $2\times 2$ matrix that has two distinct eigenvalues. If we pick $x_{22}$ equal to one of those eigenvalues then we satisfy the constraints in the question. This shows that $X$ need not even be normal as we can pick the top left block to be a non-normal matrix.