# Prove that $J^T J+\lambda I$ is a singular matrix

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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Is $\lambda$ intended to be an eigenvalue? – axblount Aug 24 '12 at 16:16
I answered what is written there, but it all seems a little nonsensical. If you decide to change the question a LOT, then consider posting it as a second question. But don't completely delete your original. – rschwieb Aug 24 '12 at 16:20
"if (stuff) is singular then (inverse) is singular too" - isn't "singular" supposed to be synonymous with "non-invertible"? :) – J. M. Aug 24 '12 at 16:30
@J.M. Nevermind! I interpreted your comment correctly now :) I'm with you, synonymous. – rschwieb Aug 24 '12 at 16:39

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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What is x? Is a vector? – Mark Aug 24 '12 at 18:02
@Mark Exactly. You know the characterization that $M$ is singular iff $Mx=0$ for some nonzero vector $x$ right? – rschwieb Aug 24 '12 at 18:10
Yes, thanks! :-) – Mark Aug 24 '12 at 18:28

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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I don't think "since" is what you mean, but I see the idea you want to mention. Basically any vector $x$ such that $J^T Jx=0$ is an eigenvector of $A$ with eigenvalue $\lambda$. – rschwieb Aug 24 '12 at 16:42
I was using the definition of an eigenvalue $\det (A-\lambda I)=0$. If $J^T J$ is nonsingular, then the determinant is not zero, and so $\lambda$ is not an eigenvalue. But since $J^T J$ is singular, then $\lambda$ is an eigenvalue. – Emily Aug 24 '12 at 16:51
I meant that you can't possibly say "since $J^TJ$ is not invertible" because that is not a hypothesis. It is the conclusion, according to the OP's question. – rschwieb Aug 24 '12 at 17:24
I think you're missing what I'm saying. $J$ is singular so naturally $J^T J$ is also singular. Generally speaking, $\lambda$ is an eigenvalue of $B+\lambda I$ if and only if $B$ (square) is singular. Since $J^T J$ is singular, then $\lambda$ is an eigenvalue of $J^T J + \lambda I$. In other words, $\det J = 0$ implies that $\lambda$ is an eigenvalue of $J^TJ + \lambda I$. – Emily Aug 24 '12 at 17:31
OK, yeah I did not know where you were beginning, but that's consistent. – rschwieb Aug 24 '12 at 18:08