Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 1 down vote accepted

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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