A and B have same singular values

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Homework Help Overview

The discussion revolves around the singular values of two matrices, A and B, where B is derived from A by a non-standard rotation. The original poster is exploring whether A and B have the same singular values and is seeking a proof or counterexample.

Discussion Character

  • Conceptual clarification, Assumption checking, Mixed

Approaches and Questions Raised

  • The original poster attempts to understand the properties of singular values in relation to matrix operations and is looking for relevant properties that might apply. Some participants question the correctness of the example provided and suggest alternative representations of matrix B.

Discussion Status

Participants are actively engaging with the problem, offering different perspectives on the transformation of matrix B and discussing the implications of isometries in relation to singular values. There is no explicit consensus yet, but several lines of reasoning are being explored.

Contextual Notes

There is a mention of the non-standard nature of the transformation applied to matrix A to obtain B, which may affect the assumptions being made about the singular values. The original poster has not yet provided a definitive proof or counterexample.

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Homework Statement


Suppose A is an m x n matrix and B is the n x 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 the answer is yes or give a counterexample.

The Attempt at a Solution


I have not been able to come up with a counterexample, so I am assuming the answer is yes. But I do not know how to prove that. What properties of Singular Values are there that deal with operations on matrices, like transposing and row-swapping, etc.?

EDIT: Example-

A = [tex]\left(\begin{array}{cc} -1 & 2 \\3 & 1 \\ 2 & -1 \end{array}\right)[/tex]
B = [tex]\left(\begin{array}{ccc} 2 & 3 & -1 \\-1 & 1 & 2 \end{array}\right)[/tex]

It follows that [tex]A^{T}A[/tex] = [tex]BB^{T}[/tex], so naturally the eigenvalues would be the same for these. Furthermore, the singular values of A and B would have to be the same as well. This is a good idea of what I am trying to prove but I cannot think of how this would be proved in the general case for m x n and n x m matrices A & B...
 
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The second column of B is upside down in your example.

I don't know if this helps, but you can say

[tex]B = \left[\begin{pmatrix}-1 & 2 \\3 & 1 \\ 2 & -1 \end{pmatrix} \begin{pmatrix}0 & 1 \\1 & 0 \end{pmatrix}\right]^T[/tex]
 
vela said:
The second column of B is upside down in your example.

I don't know if this helps, but you can say

[tex]B = \left[\begin{pmatrix}-1 & 2 \\3 & 1 \\ 2 & -1 \end{pmatrix} \begin{pmatrix}0 & 1 \\1 & 0 \end{pmatrix}\right]^T[/tex]

Umm, I don't buy into that. . . According to the definition of B it seems like the second column of B is in fact correct in my example. . . I am not seeing what you are referring to.
 
Oh, sorry! I rotated counter-clockwise. How about

[tex]B = A^T\begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{pmatrix}[/tex]?

Again, I just want to point out, I don't know how the proof will work out, but this is a way you might be able to generalize for the mxn case what the transformation is.
 
I'm pretty sure A and B have the same singular values if and only if you can find isometries P and Q such that A = PBQ. The transposing operation and the row swapping operation are both isometries
 

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