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Consider the SVD of matrix $A$:

$$A = U \Sigma V^\top$$

If $A$ is a symmetric, real matrix, is there a guarantee that $U = V$?

There is a similar question here that also posits $A$ is positive semi-definite. But I'm wondering whether $U$ would be equal to $V$ if $A$ is symmetric?

Adrian
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  • No: if $A$ fails to be positive semidefinite, then it cannot be the case that $U = V$ (unless we allow for negative diagonal entries in $\Sigma$). – Ben Grossmann Nov 09 '20 at 00:09
  • This is 'almost true' I.e. $A=UD\Sigma U^T$ where $D$ is a diagonal orthogonal matrix. However it's awkward to ask about this since SVD isn't in general unique. (Specializing to invertible $A$ and talking about Polar Decomposition instead of SVD is a nice way to sidestep uniqueness issues.) – user8675309 Nov 09 '20 at 00:59

3 Answers3

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No: if $\Sigma$ is diagonal with non-negative entries, then $U \Sigma U^T$ will necessarily be positive semidefinite. Indeed, we note that for any column vector $x$, we have $$ x^\top(U\Sigma U^\top)x = (U^\top x)^\top \Sigma (U^\top x) \geq 0. $$

Ben Grossmann
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In this question, $A$ is symmetric and therefore must be square.

However, in the general case, if $A$ is not a square matrix, it can be converted into a square matrix $AA^T$ or $A^TA$ and an eigendecomposition can be done on that to get the SVD components. The $V$ are eigenvectors of $A^TA$ and the $U$ are eigenvectors of $AA^T$. The singular values in $\Sigma$ are the square roots of the non-zero eigenvalues of $A$. This (the square root of the eigenvalues) is where the positive semi-definite condition for the SVD to be equivalent to the eigendecomposition comes from. See The SVD and the EVD in http://www.math.kent.edu/~reichel/courses/intr.num.comp.1/fall11/lecture7/svd.pdf for a more detailed explanation.

[Edited to incorporate Joppy's comment.]

nosuchthingasmagic
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  • In the question, $A$ must be a square matrix, since it is symmetric. Also, if $A$ is symmetric but not positive semidefinite, then the right singular vectors could be the eigenvectors of $A$, but the left singular vectors will have to be changed by a sign flip (for the negative eigenvalues of $A$). – Joppy Nov 09 '20 at 00:35
  • Yes, of course. I will edit my answer to make it clearer that this is for SVD in general rather than for the question (where the matrix is implied to be square). And you are right, there would need to be a sign flip in the vectors for negative eigenvalues. Thanks for the comment! – nosuchthingasmagic Nov 09 '20 at 04:34
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If $A=USU^T$ is a SVD, $A$ has to be positive semidefinite, as shown in another answer here.

The converse is not true. If $A$ is positive semidefinite and $A=USV^T$ is a SVD, $U$ is not necessarily equal to $V$. E.g. $A=U0V^T$ is a SVD for any two unitary matrices $U$ and $V$.

However, if $A$ is positive definite, we must have $U=V$ in its SVD, as shown in the answers to Is $U=V$ in the SVD of a symmetric positive semidefinite matrix?

user1551
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