Let $\mathbf{C}$ be an $n\times n$ Hermitian matrix. Let $\dagger$ indicate a matrix conjugate-transpose.
Let $\mathbf{V}\mathbf{D}\mathbf{V}^\dagger$ be the eigendecomposition of $\mathbf{C}$, where $\mathbf{V}$ is a unitary matrix whose columns comprise an orthonormal basis of eigenvectors of $\mathbf{C}$, and $\mathbf{D}$ is a real diagonal matrix whose main diagonal contains the corresponding eigenvalues.
The following observations show up (numerically):
Let $\vec{N}:=(\mathbf{V}\circ\mathbf{V}^\ast)\vec{1}$, where $\ast$ indicates an element-wise complex conjugation (no transpose!), $\circ$ indicates the Hadamard or element-wise matrix product, and $\vec{1}$ is an $n\times 1$ vector of ones.
1) Then $\vec{N}=\vec{1}$.
This is expected since $\mathbf{V}$ is an orthogonal matrix.
Now, let $\vec{d}$ be a column vector whose entries are those across the main diagonal of $\mathbf{D}$, that is, a column vector whose entries are the eigenvalues of $\mathbf{C}$: $d_i=D_{ii}$. Similarly, let $\vec{c}$ be a column vector whose entries are those across the main diagonal of $\mathbf{C}$: $c_i=C_{ii}$.
Let $\vec{M}:=(\mathbf{V}\circ\mathbf{V}^\ast)\vec{d}$.
2) Then $\vec{M}=\vec{c}$. How to prove this?
I'm sure these properties of Hermitian matrices are obvious to the trained eye. Can anyone prove observation #2 to be true?
It turns out the matrix need not even be Hermitian. See the information given in the references in the answer below.
Best Answer
It turns out that this is a well know property of the Hadamard product. See Horn and Johnson, doi: 10.1017/CBO9780511840371, Chapter 5.
The proof is given by Elizabeth Million, Theorem 2.1 found at: http://buzzard.ups.edu/courses/2007spring/projects/million-paper.pdf