# Visualization of Singular Value decomposition of a Symmetric Matrix

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

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