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Let $A$ be a symmetric positive definite matrix, I want to prove that $D^{-1}A$ and $L^{T}D^{-1}L$ have same Eigenvalues where $D=\text{diag}(\text{diag}(A))$ and $L$ is a lower-triangular matrix such that $A=LL^{T}$.

I observed that $D^{-1}A=D^{-1}LL^{T}$ and thus we can write $L^{T}D^{-1}A=(L^{T}D^{-1}L)L^{T}$.How can we proceed from here? I would hope for some hints. I noticed that $D^{-1}$ and $L^{T}D^{-1}L$ are similar given this form. Therefore, they both must have same eigenvalues but how can I show that $D^{-1}$ and $D^{-1}A$ have same eigenvalues to complete this proof by substitution as it appears?

Update: As mentioned in the comments, this does not work since $L^{T}D^{-1}L$ and $D^{-1}$ are congruent and not similar. Therefore, I would really hope for some help in finding an approach to prove the claim.

Remark: $D:=\text{diag}(\text{diag}(A))$ means that $D$ is a diagonal matrix whose diagonal entries are the diagonal entries of $A$. This is based on MATLAB's notation.

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  • $\begingroup$ Regarding the third paragraph: $D^{-1}$ and $L^\top D^{-1} L$ are congruent, not similar. So although they have the same number of positive, negative, and zero eigenvalues, they might not have exactly the same eigenvalues. $\endgroup$
    – angryavian
    Commented Dec 9, 2021 at 21:33
  • $\begingroup$ Ah I see so this breaks down my main approach. Does there exist an alternative approach? @angryavian $\endgroup$ Commented Dec 9, 2021 at 21:35
  • $\begingroup$ Could you clarify what $D=diag(diag(A))$ means? $\endgroup$ Commented Dec 9, 2021 at 22:21
  • $\begingroup$ It means that $D$ is a diagonal matrix whose diagonal entries are the diagonal entries of the matrix $A$ @Golden_Ratio $\endgroup$ Commented Dec 9, 2021 at 22:22
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    $\begingroup$ If $a_1,\ldots,a_n$ are the diagonal entries of $A$, then $\det(D^{-1}A)=1$ and $\det(D^{-1})=\prod_i a^{-1}_i$ so they can have different eigenvalues. Nevertheless that $\det(D^{-1}A)=\det(L^TD^{-1}L)$ and $trace(D^{-1}A)=trace(L^TD^{-1}L)$, which is a good sign. $\endgroup$
    – Surb
    Commented Dec 9, 2021 at 22:57

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$$D^{-1}L L^t = (L^t)^{-1}( L^t D^{-1} L) L^t.$$ So, the two matrices are conjugate (note that $L$ is invertible because $A$ is positive definite).

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