# How to show that $D^{-1}A$ and $L^{T}D^{-1}L$ have same Eigenvalues

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.

• 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. Commented Dec 9, 2021 at 21:33
• Ah I see so this breaks down my main approach. Does there exist an alternative approach? @angryavian Commented Dec 9, 2021 at 21:35
• Could you clarify what $D=diag(diag(A))$ means? Commented Dec 9, 2021 at 22:21
• It means that $D$ is a diagonal matrix whose diagonal entries are the diagonal entries of the matrix $A$ @Golden_Ratio Commented Dec 9, 2021 at 22:22
• 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.
– Surb
Commented Dec 9, 2021 at 22:57

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