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 $V^{\top}D^{-1}$?

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

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

  • $\begingroup$ 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

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