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

  • $\begingroup$ The inequalities below 'So, we have that' are clearly wrong, just let every matrix be the identity matrix and everything is equal. What exactly is an eigenvalue decomposition? $\endgroup$ – Git Gud Dec 7 '13 at 12:57
  • $\begingroup$ @GitGud, sorry, I don't get your point. I assume that none of the matrices is the identity matrix. I wrote "an eigenvalue decomposition" as it is not unique. $\endgroup$ – Ivan Dec 7 '13 at 13:06
  • $\begingroup$ I was just stating a trivial counter example to your statement. I'm pretty sure others can be found. Also nowhere in your question did you say the matrices weren't the identity matrix. As for the second part of my comment, I don't know what an eigenvalue decomposition is and I'm asking you to tell me, please. $\endgroup$ – Git Gud Dec 7 '13 at 13:11
  • $\begingroup$ @GitGud, that is why I used "in general." The definition of the eigenvalue decomposition can be found, e.g., here. $\endgroup$ – Ivan Dec 7 '13 at 13:14
  • $\begingroup$ You only used 'in general' after the inequalities. As for the wikipedia link, it doesn't really help. I can't find the definition of eigendecomposition or eigenvalue decomposition. I can't help you. Edit: I can answer 2. and 3., though: it can surely be a diagonalization of $B$, just let everything be the identity matrix. $\endgroup$ – Git Gud Dec 7 '13 at 13:19

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.

  • $\begingroup$ Thank you for your answer. Regarding diagonalization, according to the definition given in Wikipedia, it should also be a similarity. $\endgroup$ – Ivan Dec 7 '13 at 13:43
  • $\begingroup$ And thank you for the reference! I have not heard about this law. $\endgroup$ – Ivan Dec 7 '13 at 13:45
  • $\begingroup$ You're welcome. As I have written in my answer: "diagonalization" implies "by similarity", but you can also make a "diagonalization by congruence". Here is just one random example of a research paper dealing with the subject. $\endgroup$ – Vedran Šego Dec 7 '13 at 14:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.