Let $A\in\mathbb{R}^{n\times n}$ be an real symmetric matrix with eigenvalues $\lambda_1, \lambda_2, \cdots, \lambda_n$ with some of which be nonzero and repeated, i.e., there exist $\lambda_i \ne 0$ with its algebraic multiplicity larger than 1.

Question:

Is it possible to find a real positive diagonal matrix $D\in\mathbb{R}^{n\times n}$ such that all nonzero eigenvalues in $DAD$ have algebraic multiplicity 1?

Thought:

I am considering deeming this as a $\text{rank}(A)$ perturbation, but it is not trivial to write $DAD$ in matrix addition by $A$. I also think that it is possibly true that if $D$ has distinct diagonal entries, then the eigenvalues of the product $DAD$ are distinct. However, a counter example is the simple case $A = \text{diag}(1, 1, 4, 0)$ and $D = \text{diag}(2, 3, 1, 4)$ resulting in $DAD = \text{diag}(4, 9, 4, 0)$ with new repeated nonzero eigenvalues. I wonder whether this is a rare case if $A$ is not a diagonal matrix and how rare these scenarios are.

I am well aware Ostrowski's inequality as $DAD$ and $A$ are congruent. But that only provides me an upper bound and lower bound of eigenvalues of $DAD$ and does not suggest a way to reduce multiplicities of nonzero eigenvalues.

Any insights, references, or suggestions for further reading would be greatly appreciated.

Related post: [1]