Let $A$ be the matrix of eigenvectors of an Hermitian $H$. Prove that $HA=AB$ with $B$ diagonal matrix with the eigenvalues of $H$

Question

Suppose we have a hermitian matrix $H$, and a matrix $A$ composed of eigenvectors of $H$, such that

$\langle A_i \mid A_i \rangle =1$, where $A_i$ is the $i$-th column of matrix H.

How to prove that $HA=AB$, where $B$ is a diagonal matrix whose diagonal elements are eigenvalues of $H$?

Thanks

Answer 1

Let $(X_j)$ be $n$ eigenvectors for $H$, possibly linearly dependent or even repeated. So we just have $HX_j=\lambda_j X_j$ for $j=1,\ldots,n$ and no further assumption.

Then consider the matrix $A$ whose $j$th column is $X_j$. When you compute the matrix product $HA$, the $j$th column is $HX_j=\lambda_j X_j$.

Now denote $D$ the diagonal matrix whose diagonal is $\{\lambda_1,\ldots,\lambda_n\}$. When you compute the matrix product $AD$, the $j$th column is $\lambda_jX_j$.

So $HA=AD$ as they have the same columns.

Note: you don't need the assumption that the $X_j$'s have norm one, and more importantly, you don't need $H$ to be hermitian. $H$ could be any matrix, as long as you take eigenvectors of $H$ to form the columns of $A$.