# Conditions for distinct nonzero eigenvalues in product DAD for symmetric matrix A with repeated nonzero eigenvalues and diagonal matrix D

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 currently considering deeming this as a $$\text{rank}(A)$$ perturbation such that $$DAD=A+A\odot (xx^T-11^T)$$ with $$D = \text{diag}(x)$$, though it does not reduce the problem complexity to me since the perturbation still depends on $$A$$.

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 if those nonzero eigenvalues will be distinct.

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

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• What happens when you take something like the Jacobian of the spectrum $\sigma(\text{diag}(x_1,\dots,x_n)\cdot A\cdot \text{diag}(x_1,\dots,x_n))$ at $(x_1,\dots,x_n)=(1,\dots,1)$? Commented Sep 7, 2023 at 2:21

We can do this by (degenerate) first order perturbation theory. Let's take $$D=1+\epsilon C$$, with $$\epsilon\not=0$$ small and $$C$$ also diagonal. Then $$DAD = A +\epsilon(CA+AC) + O(\epsilon^2) .$$ Let's focus on a multiple eigenvalue $$\lambda$$, with eigenvector basis $$v_1,\ldots ,v_k$$. Perturbation theory says that in first order, the $$k$$ copies of $$\lambda$$ will be moved to $$\lambda+\epsilon\mu_j$$. Here, the $$\mu_j$$ are the eigenvalues of $$CA+AC$$, compressed to $$L(v_1,\ldots, v_k)$$. In other words, they are the eigenvalues of the $$k\times k$$ matrix with entries $$\langle v_j, (CA+AC) v_m\rangle = 2\lambda \langle v_j, Cv_m\rangle .$$
Now we only need to make sure that not all the $$\mu_j$$ are equal to one another to at least remove some of the degeneracies. This we can of course easily do, for example by taking $$C$$ as a suitable rank $$1$$ matrix.
We have reduced $$\sum (n(\lambda)-1)$$ (with $$n(\lambda)$$ denoting the multiplicity of $$\lambda$$) by at least $$1$$; notice here that for any fixed $$C$$, our perturbation will never introduce new degeneracies, provided we take $$\epsilon$$ small enough. Finally, repeating this whole step will eventually get us to a matrix with no degeneracies other than possibly $$\lambda=0$$.
Note also that this argument does not work for $$\lambda=0$$, which is exactly as it must be since of course a multiple eigenvalue $$\lambda=0$$ can not be moved by such a multiplicative perturbation.