# 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 $𝐴=𝑄𝐵𝑄^∗$ for some unitary $𝑄$,

Suppose $𝐴=𝑈_𝐴 Σ_𝐴 𝑉_𝐴^∗$ 𝑎𝑛𝑑 $𝐵=𝑈_𝐵 Σ_𝐵 𝑉_𝐵^∗$

$𝑈_𝐴 Σ_𝐴 𝑉_𝐴^∗=𝑄𝑈_𝐵 Σ_𝐵 𝑉_𝐵^∗ 𝑄^∗$

$Σ_𝐴 = 𝑈_𝐴^∗ (𝑄𝑈_𝐵 Σ_𝐵 𝑉_𝐵^∗ 𝑄^∗)𝑉_𝐴$

$𝐼Σ_𝐴 𝐼^∗=(𝑈_𝐴^∗ 𝑄𝑈_𝐵) Σ_𝐵 (𝑉_𝐵^∗ 𝑄^∗ 𝑉_𝐴)$

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 $B$