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I'm trying to prove the following statement: if A and B are unitarily equivalent, then they have the same singular values

so my proof goes like this: unitarily equivalent means $A=QBQ^*$ so if $A=U_A\Sigma_AV_A$ and $B=U_B\Sigma_BV_B$ then $A=U_A\Sigma_AV_A=Q(U_B\Sigma_BV_B)Q^*$ . Multiplying on both sides, we can get $I\Sigma_AI^* = (U_A^*QU_B)\Sigma_A(V_B^*Q^*V_A)$.

From here I am stuck. I want to say that I can use uniqueness of the SVD to help me here but I am not sure how. How do I use the uniqueness of the SVD (and to what extent is it unique?) to help complete this proof?

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  • $\begingroup$ The singular values of $A$ are the square roots of the eigenvalues of $A^*A$. If $A$ and $B$ are unitarily equivalent, then so are $A^*A$ and $B^*B$. Hence, $A^*A$ and $B^*B$ have the same eigenvalues. $\endgroup$
    – amsmath
    Commented Oct 31, 2017 at 14:48
  • $\begingroup$ I know that is true, but how does that relate to my incomplete proof above? @amsmath thanks. $\endgroup$
    – makansij
    Commented Oct 31, 2017 at 14:49
  • $\begingroup$ You should replace $\Sigma_A$ by $\Sigma_B$ on the RHS anyways. $\endgroup$
    – amsmath
    Commented Oct 31, 2017 at 14:50
  • $\begingroup$ $B = QU_A\Sigma_AV_AQ^*$ is an SVD of $B$ since $QU_A$ and $V_AQ^*$ are unitary. Since the SVD is unique, $\Sigma_A = \Sigma_B$. $\endgroup$
    – amsmath
    Commented Oct 31, 2017 at 15:15
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    $\begingroup$ What do you mean by "the entire decomposition"? Of course $U\Sigma V$ is trivially unique because that is just the decomposed matrix. No, the SVD is unique in the sense that when $A = U_1\Sigma_1V_1 = U_2\Sigma_2V_2$, then $\Sigma_1 = \Sigma_2$ (up to permutations of the diagonal entries). $\endgroup$
    – amsmath
    Commented Oct 31, 2017 at 16:30

1 Answer 1

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Suppose $B=U\Sigma V^*$ is a SVD for $B$, with $U$ and $V$ unitary and $\Sigma$ diagonal with decreasing values on the diagonal. If $A=QBQ^*$ with $Q$ unitary, then $$ A=(QU)\Sigma (V^*Q^*) $$ The product of unitary matrices is unitary. Thus this is a SVD for $A$.

The uniqueness for the SVD is only related to the matrix $\Sigma$ (in the canonical diagonal form with decreasing values on the diagonal).

Anyway, this is overkill: the singular values of $A$ are the positive square roots of the nonzero eigenvalues of $A^*A$. If $A$ and $B$ are unitarily similar, then also $A^*A$ and $B^*B$ are.

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