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If we have a symmetric matrix $A$ and compute its singular value decomposition $A=U\Sigma V^T$, is it the case that $U$ and $V$ are also symmetric? I know that $U$ and $V$ must be equal since they are the singular vectors of $A$.

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    $\begingroup$ You will have $U=V$, but there is no reason to expect symmetry; they are just generic orthogonal matrices which are not generally symmetric. In fact even if they have a symmetric representative they will still have a different nonsymmetric representative (keep in mind that $U$ and $V$ are not unique!) $\endgroup$ – Ian May 5 '17 at 5:41
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$2\times 2$ symmetric matrices have an SVD where $U$ and $V$ are symmetric. Given a two-by-two $A=U\Sigma U^T,$ the matrix $U$ is either a reflection or rotation. For reflections, $U$ is already symmetric, and for rotations we can multiply by $D=\mathrm{diag}(1,-1)$ to get a reflection: $A=UD\Sigma (UD)^T$ and $UD$ is orthogonal and symmetric.

A counterexample can be constructed from an orthogonal matrix $U$ with $|U_{12}|\neq |U_{21}|$ such as

$$U=\frac 1 3\begin{pmatrix} 2&2&-1\\ -1&2&2\\ 2&-1&2\\ \end{pmatrix}.$$

Take $A=U\Sigma U^T$ with $\Sigma=\mathrm{diag}(3,2,1).$ Any other SVD $A=U_2\Sigma V_2^T$ will satisfy $U_2=UD$ for some diagonal orthogonal $D$ - see "How unique (on non-unique) are U and V in Singular Value Decomposition (SVD)?". This means $U_2$ cannot be symmetric.

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