# How to prove if two matrices are orthogonally equivalent, then they have the same singular values

The question was, "Show that if two matrices are orthogonally equivalent, then they have the same singular values, and there are simple relationships between their singular vectors"

I tried to show like this.

Let $π΄=ππ΅π^β$ for some unitary $π$,

Suppose $π΄=π_π΄ Ξ£_π΄ π_π΄^β$ πππ $π΅=π_π΅ Ξ£_π΅ π_π΅^β$

$π_π΄ Ξ£_π΄ π_π΄^β=ππ_π΅ Ξ£_π΅ π_π΅^β π^β$

$Ξ£_π΄ = π_π΄^β (ππ_π΅ Ξ£_π΅ π_π΅^β π^β)π_π΄$

$πΌΞ£_π΄ πΌ^β=(π_π΄^β ππ_π΅) Ξ£_π΅ (π_π΅^β π^β π_π΄)$

Any tips to prove this ?

• What are orthogonally (or whatever) equivalent matrices? Like similar matrices? – DonAntonio Dec 12 '16 at 12:11
• Orthogonal equivalence implies unitary equivalence and similarity. – Falcon Dec 12 '16 at 12:16
• @F Don't tell me what orthogonal equivalence implies, but what it means . – DonAntonio Dec 12 '16 at 12:19

Let $\mu$ a singular value of $A$. Then there is $x \ne 0$ such that
$A^{\star}Ax= \mu x$. Let $y=Q^{\star}x$. Then $y\ne 0$ and
$\mu x=QB^{\star}Q^{\star}QBQ^{\star}x=QB^{\star}BQ^{\star}x$ thus
$$\mu y=B^{\star}By$$
and $\mu$ is a singular value of $B$