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If $J$ is a singular matrix, then $(J^T J)^{-1}$ is singular too. I'm trying to prove that $J^T J+\lambda I$ is a singular matrix, where $I$ denotes identity matrix. Any suggestions please? Thanks

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    @J.M. Nevermind! I interpreted yo$u$r comment correctly now :) I'm with you, synonymous.2012-08-24

2 Answers 2

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Let $x$ be nonzero such that $Jx=0$. Then $J^\mathrm{T} Jx=0$, so $J^\mathrm{T} J$ is singular, and $(J^\mathrm{T} J)^{-1}$ does not even exist.

As for the second sentence in your question, $J$=the zero matrix and its transpose are certainly singular, but $J^\mathrm{T} J+\lambda I$ will certainly not be if $\lambda\neq 0$.

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    Yes, thanks! :-)2012-08-24
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Let $J^T J +\lambda I = A$. Then, $J^T J = A-\lambda I$.

Since $J^T J$ is not invertible, then $\lambda$ is an eigenvalue of $A$.

If $A$ has an eigenvalue $\lambda = 0$, then your condition that $J^T J + \lambda I$ is singular will be met, because $\det A = \prod_{i=1}^n \lambda_i$.

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    OK, yeah I did not know where you were beginning, but that's consistent.2012-08-24