Assume that there exists an invertible matrix $P$ such that $P^{-1}X^nP$ is a diagonal matrix with distinct eigenvalues, then can I say that $P^{-1}XP$ is also a diagonal matrix with distinct eigenvalues? If so, how do I prove it?
[Math] If $X^n$ is a diagonal matrix with distinct eigenvalues, then is $X$ also a diagonal matrix with distinct eigenvalues
linear algebramatrices
Best Answer
Assume that $M^n$ is the diagonal matrix with diagonal $(a_j)$ of distinct entries, in particular $\prod\limits_{j}(M^n-a_j)=0$. Thus $\prod\limits_j\prod\limits_{\ell=1}^n(M-b_{j,\ell})=0$ where, for each $j$, $x^n-a_j=\prod\limits_{\ell=1}^n(x-b_{j,\ell})$, that is, the $b_{j,\ell}$ are the $n$th roots of $a_j$. Since all the coefficients $(b_{j,\ell})_{j,\ell}$ are distinct, the polynomial $\prod\limits_j\prod\limits_{\ell=1}^n(x-b_{j,\ell})$ has simple roots and is null when evaluated at $M$. Thus $M$ is diagonalizable with a diagonal of distinct entries, say, $M=QDQ^{-1}$. Hence $M^n=QD^nQ^{-1}$ and $D^n$ must have diagonal entries $(a_j)$ since $M^n$ and $D^n$ are similar. In other words, $M^n=RD^nR^{-1}$ where $R$ is a permutation matrix and $(R^{-1}Q)M^n(Q^{-1}R)=M^n$. Thus, $Q=R$, that is, $Q$ was a permutation matrix from the onset and $QDQ^{-1}$ is also a diagonal matrix.
Finally, if $M^n$ is diagonal with distinct elements, so is $M$. Using $M=A$, this answers the question the OP asked in a comment. To answer the one the OP asked in the main post, apply this to $M=P^{-1}XP$.