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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 $\mathbf \Sigma$

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

1 Answer 1

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^⊤

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