Out of "Numerical Linear Algebra" by Trefethen and Bau:

Suppose A is an $m\times n$ matrix and $B$ is the $n\times m$ matrix obtained by rotating $A$ ninety degrees clockwise on paper (not exactly a standard mathematical transformation!). Do $A$ and $B$ have the same singular values? Prove that the answer is yes or give a counterexample.

My best guess is ... well, a guess.

How do I approach this?

Note that this is in one of the earlier chapters of the book ("Lecture 4", which introduces the SVD), in which we have not heard of eigenvalues, yet (and won't for quite a while if the index is any indication, which lists them 20+ "Lectures" later).

For now, these are the definitions available:

Let $S$ be the unit sphere in $\mathbb{R}^n$, $A\in\mathbb{R}^{m\times n}$.


singular values of $A$: lenghts $\{\sigma_i\}$ of the principal semiaxes of $AS$ (numbered by convention in decreasing order)


left singular vectors of $A$: unit vectors $\{u_i\}$ oriented in the directions of the principal semiaxes of $AS$ (numbered to match corresponding singular values)


right singular vectors of $A$: unit vectors $\{v_i\}\subset S$ that are the preimages of the principal semiaxes of $AS$ (numbered such that $Av_j = \sigma_ju_j$)


We can describe the $90^\circ$ counter-clockwise rotation as follows: let $K$ denote the $m \times m$ matrix with $1$s on the antidiagonal. So if $n=3$, then $$ K = \pmatrix{0&0&1\\0&1&0\\1&0&0} $$ Then given an $m \times n$ matrix $A$, $A^TK$ is the rotated matrix.

Since $A^T$ has the same singular values as $A$ and $K$ is orthogonal, conclude that the rotated matrix indeed has the same singular values.

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  • $\begingroup$ @DanielXiang apparently you're right. $\endgroup$ – Ben Grossmann Aug 15 '17 at 19:26
  • $\begingroup$ and then would $K$ be $m \times m$? $\endgroup$ – Daniel Xiang Aug 15 '17 at 19:30

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