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My question is as follows:

Let $A$ be a real symmetric $n\times n$ matrix and $D=\mathrm{diag}(d_1,\ldots,d_n)$ be a diagonal matrix whose diagonal entries are all positive real numbers(denoted as $d_1,\ldots,d_n$). If $A$ has $k$ positive eigenvalues, $m$ zero eigenvalues(i.e. $0$ is an eigenvalue with multiplicity $m$) and $n-m-k$ negative eigenvalues. Then does $DA$ have the same number of positive, zero, negative eigenvalues?

Intuitively, I think this statement is true because, heuristically, if we assume that $A$ is also diagonal, its diagonal entries are exactly the eigenvalues of $A$, and the identity $\mathrm{diag}(d_1,\ldots,d_n)\cdot \mathrm{diag}(a_1,\ldots,a_n)=\mathrm{diag}(a_1d_1,\ldots,a_nd_n)$ directly gives the result. Also it is well-known that the statement is true if $A$ is positive semi-definite. However, I'm not sure that this is also true for the general cases where $A$ is not necessarily positive semi-definite. I tried to find a counterexample, but it also didn't work.

Does anyone have ideas?

Thanks in advance!

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2 Answers 2

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Yes. Since $B=D^{1/2}AD^{1/2}$ is congruent to $A$, they have the same inertia, by Sylvester's law of inertia. However, $B$ is similar to $D^{1/2}BD^{-1/2}=DA$. Therefore $DA$ also has the same number of positive/zero/negative eigenvalues as $A$.

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  • $\begingroup$ Thanks a lot! I should have known the term 'matrix congruence'. $\endgroup$
    – bellcircle
    Commented Mar 22, 2021 at 7:10
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Does Sylvester's law only work for symmetric $A$? I have a similar problem with blocked $A = \begin{bmatrix}L& B\\B& L\end{bmatrix}$ and $D = \begin{bmatrix}D_1& \\& D_2\end{bmatrix}$; where $L$ is symmetric but $B$ is not symmetric. I know $A$ has all nonnegative eigenvalues, does that hold for $DA$?

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