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Assuming that $A$ is a diagonalizable matrix, does the singular value decomposition of $A$ coincide with its spectral decomposition?

I think no, because in the spectral decomposition $A = Q^{-1} \Lambda Q$, $Q$ doesn't have to be unitary but the SVD of a matrix $A = UDV^t$ gives unitary matrices $U$ and $V$. However, this could be remedied by diving the columns of $Q$ by their lengths. Right? So, now I think that up to swapping columns and multiplying them by scalars, the SVD and the spectral decomposition of a matrix are the same. Is it correct?

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  • $\begingroup$ No, normalizing the eigenvectors doesn't make them orthogonal (i.e. doesn't turn $Q$ into a unitary matrix). $\endgroup$ – Rahul Nov 6 '18 at 11:55
  • $\begingroup$ @Rahul Yes. But can't we do Gram-Schmidt on the eigenvectors to find a nice orthonormal basis? $\endgroup$ – stressed out Nov 6 '18 at 11:59
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The SVD of $A$ corresponds more closely to the spectral decompositions of $A'A$ and $AA'$. If you assume $A=U\Lambda V'$, you see that $A'A=U\Lambda^2U'$ and $AA'=V\Lambda^2V'$. Repackaged, the spectral decomposition (exercise: work out the detailed formula yourself) of the big matrix $M=\begin{pmatrix}0 &A\\A'&0\end{pmatrix}$ is built out of $U$, $V$, and $\Lambda$; the spectrum of $M$ is the union of that of $\Lambda$ and of $-\Lambda$.

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