Decompose a Real symmetric matrix into only 2 matrices

Asked Oct 10 '14 at 05:36

Active Oct 10 '14 at 05:41

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For a real symmetric matrix $A\in\mathbb{S}^n$, we can decompose it into $A=U\Lambda U^{T}$ by Spectral Theorem.
In Matlab, using the command $[L,S,R] = svd(A)$, we obtain $LS R^{T}$ (here, $L$ stands for the $\textbf{L}$eft matrix, and $R$ stands for the $\textbf{R}$ight matrix). It is normally shown that $L\neq R$. This means that $U$ and $L$ are not the same matrix.
My question is here: $\textbf{How to get $U$ from $L$ and $R$ theoretically?}$ Can anyone help me, please?

edited Apr 13 '17 at 12:20

Community

asked Oct 10 '14 at 05:36

pej

So, $U$ is a matrix whose columns are (normalized) eigenvectors of $A$. Do you know a similar description of $L$ and of $R$? – Gerry Myerson Oct 10 '14 at 06:24
Also, while usually $L\ne R$, is that still true in your special case, where $A$ is real symmetric? – Gerry Myerson Oct 10 '14 at 06:25
@GerryMyerson: 1) $UU^{T}=U^{T}U=I$ and $U$ contains eigenvectors of $A$. – pej Oct 10 '14 at 06:36
Yes, but what about $L$ and $R$? – Gerry Myerson Oct 10 '14 at 06:37
@GerryMyerson: 2) $A$ is still real symmetric, but I see that in the formulation $A=LSR^{T}$, $S\neq\Lambda$ and $S$ is just a diagonal matrix. – pej Oct 10 '14 at 06:39
@GerryMyerson: $L$ and $R$ are also unitary matrices. – pej Oct 10 '14 at 06:54
$S$ is not just a diagonal matrix. Its diagonal entries are the "singular values" of $A$. These are the eigenvalues of $AA^t$. But when $A$ is symmetric.... – Gerry Myerson Oct 10 '14 at 08:08

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