Is there a relationship between SVD and similarity transformation?

Question

Is there a relationship between SVD and similarity transformation? I mean, in articles I read about the SVD method, I came across the equation $A=USV^T$ which seems to me like similarity transformation $A^\prime=QAQ^T$. Is that right? How are these two concepts connected?

Answer 1

Not directly. An SVD of $A$ is related to the orthogonal (unitary in $\mathbb{C}$) similarity diagonalization of a positive definite matrix $A^TA$ (or $AA^T$), because

$$A^TA = (U \Sigma V^T)^T (U \Sigma V^T) = V \Sigma^T \Sigma V^T.$$

If $A$ is a square matrix, then

$$A^TA = V \Sigma^2 V^T.$$

Other than that, (also if $A$ is a square matrix!) you can put

$$A = U \Sigma V^T = U (\Sigma V^T U) U^T,$$

which is a similarity of $A$ and $\Sigma V^T U$, which is a product of a diagonal and orthogonal matrix, but I don't see how would that be useful. Similarly, you could also do

$$A = U \Sigma V^T = V (V^T U \Sigma) V^T.$$

Of course, if $A$ is positive semidefinite, then you can choose $U$ and $V$ such that $U = V$, so the SVD is the same as the Eigenvalue decomposition, but this is just a special case.