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?

  • $\begingroup$ No, normalizing the eigenvectors doesn't make them orthogonal (i.e. doesn't turn $Q$ into a unitary matrix). $\endgroup$ – user856 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

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 $-\Lambda$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.