1
$\begingroup$

I need your expertise in evaluating the following problem:

Let $D \in \mathbb{R}^{d \times d}$ be a diagonal matrix and let $V \in \mathbb{R}^{d \times d}$ be a orthogonal matrix, then it is know that:

  • $\left| \left| D \right| \right| _2 = \max_{i \in [d]}\left\lbrace \lambda_i\right\rbrace$, where $\lambda_i$ is the $i^{th}$ eigenvalue of $D$ for every $i \in [n]$.
  • $\left| \left| V \right| \right|_2 = 1 $

So what I want to know is the $l_2$ norm of $DV^{\top}$ and the $l_2$ norm of $V^{\top}D^{-1}$?

$\endgroup$
  • $\begingroup$ Notice that the $\lambda_i$ may be negative. You forgot the absolute value. $\endgroup$ – user251257 Oct 2 '16 at 21:16
2
$\begingroup$

The norm is the same as the one of D because the map associated with V (or its inverse) is an isometry.

$\endgroup$
  • $\begingroup$ can you elaborate a little more? $\endgroup$ – user3492773 Oct 2 '16 at 20:42
  • $\begingroup$ So, $$ \left| \left| D V^{\top} \right| \right|_2 = \left| \left| D\right| \right|_2 ??$$ $\endgroup$ – user3492773 Oct 2 '16 at 20:46
  • $\begingroup$ an $l^2$ norm of a matrix $M$ (or its associated transformation) is the supremum of $||Mx||$ taken over vectors with $||x||=1$ - so ask yourself: what is the "longest" image you can produce from a unit vector using your transformation $x \rightarrow DVx$? The $V$ doesn't do anything to the magnitude of a vector, only $D$ has an effect on it - the same as before. The only real difference between D and DV are the eigenvectors. $\endgroup$ – Max Freiburghaus Oct 2 '16 at 20:51
  • $\begingroup$ Yes, you're right. $\endgroup$ – Max Freiburghaus Oct 2 '16 at 20:52

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.