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revisiting

singular values of "rotated" matrix

Maybe I've since forgotten - or maybe I never bothered to ask myself - but

Since $A^T$ has the same singular values as $A$ and $K$ is orthogonal, conclude that [ $A^\top K$ ] indeed has the same singular values.

Why does this hold?

Why can we say that multiplying by an orthogonal matrix retains the singular values?

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    $\begingroup$ The (nonzero) singular values of $A$ are related to the eigenvalues of $A^TA$ which are in turn related to the eigenvalues of $AA^T$. Now for $B=A^TK$ you have $BB^T=A^TA$. $\endgroup$
    – A.Γ.
    Aug 21, 2017 at 8:11

2 Answers 2

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We calculate the singular values of $A$ by decomposing it into $A=USV^T$ with diagonal $S$ and orthogonal $U$ and $V$. If we have another orthogonal $K$, we have $AK= USV^TK = U S (V^TK)$ which is the SVD of $AK$.

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    $\begingroup$ I see, and $V^\top K $ is orthogonal because $V^\top K (V^\top K)^\top = V^\top K K^\top V = I $. Thank you. $\endgroup$
    – User1291
    Aug 21, 2017 at 8:19
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Let $A = UDV^T$ be the singular value decomposition.

If you premultiply it with $K$, $KA=(KU)DV^T$

$KU$ is an orthogonal matrix as well. In fact $(KU)DV^T$ is a singular value decomposition for $KA$. The $D$ matrix is not changed.

Similar reasoning for post-multiplication.

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