Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

The Singular Value Decomposition of a matrix A satisfies

$\mathbf A = \mathbf U \mathbf \Sigma \mathbf V^\top$ The visualization of it would look like enter image description here

But when $\mathbf A$ is symmetric we can do:

$\begin{align*} \mathbf A\mathbf A^\top&=(\mathbf U\mathbf \Sigma\mathbf V^\top)(\mathbf U\mathbf \Sigma\mathbf V^\top)^\top\\ \mathbf A\mathbf A^\top&=(\mathbf U\mathbf \Sigma\mathbf V^\top)(\mathbf V\mathbf \Sigma\mathbf U^\top) \end{align*}$

and since $\mathbf V$ is an orthogonal matrix ($\mathbf V^\top \mathbf V=\mathbf I$), so we have:

$\mathbf A\mathbf A^\top=\mathbf U\mathbf \Sigma^2 \mathbf U^\top$

I have two questions:

  • Is the above statement correct? when Matrix $\mathbf A$ is symmetric and we compute SVD we would get $\mathbf U\mathbf \Sigma^2 \mathbf U^\top$

  • How would the decomposition looks like in a symmetric matrix? As we are getting the eigenvectors and squared eigenvalues in matrices $\mathbf U $ and $\mathbf \Sigma$

share|cite|improve this question
This question seems relevant: Relationship between eigendecomposition and singular value decomposition – littleO Nov 24 '12 at 18:19

Answer1: I think your statement above the two questions is correct. But only when A is square, symmetric, positive definite, we can use Cholesky decomposition so that A= BB^T, then B can be decomposed using SVD: B= U \Sigma U^T, thus A=U\Sigma^2U^⊤

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.