Are eigenvalues of sum of two matrices equal to sum of other two matrices with identical eigenvalues?

Question

Consider I want the eigenvalues of $C=AA^T+BB^T$. How can I prove that the non-zero eigenvalues of $C$ is equal to matrix $D$ in which $D=A^TA+B^TB$?

If not possible in general, $A$ and $B$ are i.i.d Gaussain matrices.

I know this claim is true as I simulated it in Matlab.

EDIT: I also Assume that C has no Zero eigenvalues

Answer 1

This isn't true. E.g. $$ A=\pmatrix{0&1\\ 0&0},\ B=\pmatrix{1&0\\ 0&0}, $$ $$ C=AA^T+BB^T=\pmatrix{2&0\\ 0&0},\ D=A^TA+B^TB=\pmatrix{1&0\\ 0&1}. $$ The only nonzero eigenvalue of $C$ is $2$ and it is not an eigenvalues of $D$.

It is true, however, that when all entries of $A$ and $B$ are i.i.d. $N(\mu,\sigma^2)$, we have $$ \mathbb E(C)=\mathbb E(D)=2n(\sigma^2I_n+\mu^2E) $$ where $E$ is the all-one matrix.