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=𝑄𝐡𝑄^βˆ— $ for some unitary $ 𝑄$,

Suppose $ 𝐴=π‘ˆ_𝐴 Ξ£_𝐴 𝑉_𝐴^βˆ— $ π‘Žπ‘›π‘‘ $ 𝐡=π‘ˆ_𝐡 Ξ£_𝐡 𝑉_𝐡^βˆ— $

$ π‘ˆ_𝐴 Ξ£_𝐴 𝑉_𝐴^βˆ—=π‘„π‘ˆ_𝐡 Ξ£_𝐡 𝑉_𝐡^βˆ— 𝑄^βˆ— $

$ Ξ£_𝐴 = π‘ˆ_𝐴^βˆ— (π‘„π‘ˆ_𝐡 Ξ£_𝐡 𝑉_𝐡^βˆ— 𝑄^βˆ—)𝑉_𝐴 $

$ 𝐼Σ_𝐴 𝐼^βˆ—=(π‘ˆ_𝐴^βˆ— π‘„π‘ˆ_𝐡) Ξ£_𝐡 (𝑉_𝐡^βˆ— 𝑄^βˆ— 𝑉_𝐴) $

Any tips to prove this ?

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


Your Answer

By clicking β€œPost Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.