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How does SVD work if the $A^TA$ matrix is singular? Lets say that the matrix looks like $$\begin{matrix}X & 0 \\0 & 0 \\\end{matrix} $$ Where X is an arbitrary real valued number. Is it still possible to obtain $\Sigma$ ? or will it now be in Jordan form? What about the case of repeated Eignevalues? how will this now impact $\Sigma$ along with $U$ and $V$ Matrix?

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  • $\begingroup$ Why do you think there is a problem when $A^\top A$ is singular? If this is the case, then some of the eigenvalues of $A^\top A$ are zero, so $\Sigma$ will have some zero diagonal entries. $\endgroup$ – angryavian Apr 1 '18 at 18:26
  • $\begingroup$ $A^\top A$ is always symmetric. Thus, it is always diagonalizable. No need to bring Jordan forms to the discussion. $\endgroup$ – Rodrigo de Azevedo Jul 15 '18 at 18:52
  • $\begingroup$ The Singular Value Decomposition is possible for any matrix. $\endgroup$ – Yves Daoust Jul 15 '18 at 18:57
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Singular value decomposition works the same whether A$^{\mathrm T}$A is singular or not or whether it has multiple eigenvalues or not. In all cases, $\Sigma$ will be diagonal (not necessarily of the form of the corresponding matrix in the Jordan decomposition).

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