linear algebra - Eigenvalue decomposition of $D , A , D$ with $A$ symmetric and $D$ diagonal

Let $A$ be a real, symmetric matrix. It admits the eigenvalue decomposition

$A = U \Lambda U^T$

where the eigenvectors are chosen to be orthogonal. Let $D$ be a diagonal matrix and

$B = D A D = D U \Lambda U^T D = (DU) \Lambda (DU)^T = V \Lambda V^T. \tag{$\ast$}$

Assume none of $A$, $B$, and $D$ is the identity matrix, $I$. In general, we have that

$V^T = U^T D \neq V^{-1} = U^T D^{-1}$ and

$V V^T = D^2 \neq I$.

I would like to clarify my understanding of the situation. Are the following statements correct in general?

1. Eq. ($\ast$) is not an eigenvalue decomposition of $B$.


2. Eq. ($\ast$) is not a diagonalization of $B$.


3. $A$ and $B$ have different sets of eigenvalues, and one can say nothing about the eigenvalues of $B$ based on the knowledge of $\Lambda$.

Thank you.

Regards,
Ivan

Answer

Note that $U^T = U^{-1}$, so the congruence

$$A = U \Lambda U^T = U \Lambda U^{-1}$$

is also a similarity relation. In other words, the eigenvalue decomposition is a unitary similarity of $A$ and $\Lambda$.

Since the same cannot be said about $V := DU$, relation $(*)$ remains congruence that is not a similarity, hence it doesn't preserve the eigenvalues. However, it is a diagonalization by congruence (usually, when we say "diagonalization" without the additional "by something", we assume that it's "by similarity").

However, there is a relation between the eigenvalues of $A$ and $B$, albeit a weaker one. By Sylvester's law of inertia, if $D$ is nonsingular, then $A$ and $B$ have the same number of negative, zero, and positive eigenvalues.