TITLE: Show a matrix M is invertible if and only if the eigenvalues of A are all distinct
QUESTION [5 upvotes]: Given a matrix $A \in M_n(\mathbb{C})$,
define a matrix $$M=\begin{pmatrix}n&tr({A})&tr({A^2})&\cdots&tr(A^{n-1})\\tr({A}) &tr({A^2})&tr({A^3})&\cdots&tr({A}^n)\\ \vdots&\vdots&\vdots&\cdots&\vdots\\ tr({A}^{n-2}) &tr({A^{n-1}})&tr({A^{n}})&\cdots&tr({A}^{2n-1})\\
tr({A}^{n-1}) &tr({A^n})&tr({A^{n+1}})&\cdots&tr({A}^{2n-2})\end{pmatrix}$$
Can we show that $M$ is invertible if and only if the eigenvalues of $A$ are all distinct? I have a feeling that this relates to the Vandermonde matrix. I also have some work using the linearity of trace and the Cayley-Hamilton theorem. Do we have any other thoughts as to how to approach this problem?
Note that the matrix is complex, so we do not know if it is diagonalizable.

REPLY [3 votes]: Let $\lambda_1, \ldots, \lambda_n$ be the eigenvalues of $A$, and consider the Vandermonde matrix
$$ V = V(\lambda_1, \ldots, \lambda_n) = \begin{bmatrix} 1 & \ldots & 1 \\ \vdots & & \vdots \\ \lambda_1^{n-1} & \ldots & \lambda_n^{n-1} \end{bmatrix} $$
Then
$$VV^T = \begin{bmatrix} n  & \ldots & \sum_i \lambda_i^{n-1} \\ \vdots &  & \vdots \\ \sum_i \lambda_i^{n-1} & \ldots & \sum_i \lambda_i^{2n - 2} \end{bmatrix} = M$$
Therefore $M$ is invertible if and only if $V$ is invertible if and only if the $\lambda_i$'s are distinct.