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.


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?


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.

Related post: [1]

  • $\begingroup$ 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 $(x_1,\dots,x_n)=(1,\dots,1)$? $\endgroup$ Commented Sep 7, 2023 at 2:21

1 Answer 1


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.

  • $\begingroup$ My use of perturbation theory here is done the physicist's way, but fully rigorous treatments are also possible. Standard references are Kato and Reed-Simon, vol. 4. $\endgroup$ Commented Sep 7, 2023 at 15:49

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