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Suppose there exist matrices $A \in \Bbb C^{m \times n}$ and $B = U_1 A U_2$, where $U_1 \in \Bbb C^{m \times m}$ and $U_2 \in \Bbb C^{n \times n}$ are unitary matrices (not necessarily related to each other). Then, $A$ and $B$ have the same singular values.

The reason for this is because $A^H A$ and $B^H B$ are related the following way ($A^H$ is the Hermitian transpose of $A$): $$B^H B = (U_1 A U_2)^H (U_1 A U_2) = U_2^H A^H U_1^H U_1 A U_2 = U_2^H A^H A U_2 = U_2^{-1} (A^H A) U_2$$ Since $A^H A$ and $B^H B$ are similar matrices (by definition), they share the same eigenvalues. Since the singular values of any matrix $M$ are the positive square-roots of the eigenvalues of $M^T M$, $A$ and $B$ have the same singular values.

Is there a special name relating these types of matrices (just like "similar matrix" relates $A$ and $B$ with the same eigenvalues)? I know that orthogonally equivalent matrices $C$ and $D = U C U^H$ are kind of similar to this, but $C$ and $D$ are always square, and the two unitary matrices are related as the inverse of one another ($U^H = U^{-1}$). In my problem statement, $A$ and $B$ can be rectangular, and the 2 unitary matrices $U_1$ and $U_2$ don't have to be related to each other (in fact, even their dimensions may differ).

5Pack
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3 Answers3

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If $B=UAV$ for some unitary matrices $U$ and $V$ (of possibly different sizes), $A$ and $B$ are said to be unitarily equivalent. Note that unitary equivalence is really an equivalence relation.

However, beware that quite a lot of authors also use the term "unitarily equivalent" to mean "unitarily similar".

user1551
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On page 381 of the solutions manual to "Introduction to Linear Algebra" by Gilbert Strang ($5^{th}$ edition), Strang states,

"The matrices $A$, $Q_1 A Q_2^T$, and $\Sigma$ are all 'isometric' $=$ sharing the same $\Sigma$.

($Q_1$ and $Q_2$ = unitary, $\Sigma$ = diagonal matrix of singular values)

5Pack
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Not really an answer but a suggestion: Your question is obviously very much related to the Singular value decomposition In the SVD a complex matrix $A$ is written as $A = U M V^*$ with $U$ and $V$ unitary. As your definition is equivalent to saying that the two matrices $A$ and $B$ admit SVD with the same $M$ you could say that your two matrices are SVD-equivalent?

H. H. Rugh
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