showing that $\sigma(\mathbf{XYX}^{−1}) = \sigma(\mathbf{Y})$

Question

I wanted to show the above ($\mathbf{X}$ is nonsingular btw), but I don't think my current proof is complete/quite right, wondering if this is ok:

Let $\mathbf{M} = \mathbf{XYX}^{-1}$. Given that we can represent a matrix $\mathbf{M}$ this way, then we know $\mathbf{M}$ is diagonalizable. Notice that $\mathbf{M}$ and $\mathbf{Y}$ are similar matrices. So, $\mathbf{Y}$ is a diagonal matrix, that shows the eigenvalues of $\mathbf{M}$ on its diagonal.

Taking $\mathbf{Y}$ on its own, we know that one of the properties of a diagonal matrix is that its eigenvalues are on the diagonal. Since $\mathbf{Y}$ is a diagonal matrix, its eigenvalues are on its diagonal. $$ \therefore \sigma(\mathbf{M}) = \sigma(\mathbf{XYX}^{-1}) = \sigma(\mathbf{Y}) $$

(Note: I realize $\mathbf{Y}$ is not always diagonal, how would I circumvent this?)

Also, I think you're making this too complicated: If $Yv=\lambda v$ for some nonzero $v$, show that you can always find a nonzero $u$ such that that $XYX^{-1}u=\lambda u$. — Semiclassical, Feb 11 '21 at 23:37
One can also see this statement as a special case of a more general result: Eigenvalues are invariant under cyclic permutations, i.e., $\sigma(ABC) = \sigma(CAB) = \sigma(BCA)$. — Rammus, Feb 12 '21 at 08:56

score 1 · Accepted Answer · answered Feb 12 '21 at 04:00

I guess $\sigma$ represents the eigenvalues of a matrix.

If that is the case, the eigenvalue $\lambda$ should be an eigen value for both $\mathbf{XYX}^{−1}$ and $\mathbf{Y}$. That means, there are two vectors $u$ and $v$ such that

$$ Yv=\lambda v $$ and $$ XYX^{-1} u= \lambda u \\ \Longrightarrow YX^{-1}u = X^{-1}(\lambda u) = \lambda X^{-1} u $$

So if we define $v$ as $X^{-1} u$ we find the associated eigenvectors for $\mathbf{XYX}^{−1}$ and $\mathbf{Y}$ for the eigenvalue $\lambda$. So for each eigenvalue $\lambda$ (and associated eigen vector $u$) of $\mathbf{XYX}^{−1}$, $\mathbf{Y}$ has the same eigen value $\lambda$ (and associated eigen vector $X^{-1}u$)

showing that $\sigma(\mathbf{XYX}^{−1}) = \sigma(\mathbf{Y})$

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