# Does the singular value decomposition coincide with the spectral decomposition for square matrices?

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?

• No, normalizing the eigenvectors doesn't make them orthogonal (i.e. doesn't turn $Q$ into a unitary matrix). – Rahul Nov 6 '18 at 11:55
• @Rahul Yes. But can't we do Gram-Schmidt on the eigenvectors to find a nice orthonormal basis? – 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 $$-\Lambda$$.