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Suppose a square, real and symmetric matrix $G\in\mathbb{R}^{n\times n}$ is given, and it is known to have one zero eigenvalue associated with all ones eingenvector, $1_n$. I'm aware that the (possibly) negative spectrum could be shifted to non-negative, $$(G + c\cdot I)u = Gu + c\cdot Iu = \lambda\cdot u + c\cdot u = (\lambda + c)\cdot u,$$ where $c$ is at least the absolute value of the smallest negative eigenvalues.

a) with the above shifting, are the zero eigenvalue affected (I'm asking because of certain contradictory I found in the literature)?

b) given $J_n=I_n-\frac{1}{n}1_n1_n^T$, what can one say about the spectral decomposition of $$G-c\cdot J_n$$ Is this the same as the above method of shift along the diagonal? I read that that all the eigenvalues are shifted, except for the zero eigenvalue that remains (corresponding to $1_n$), but I wonder how this could be shown.

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up vote 3 down vote accepted

a) Yes, the zero eigenvalue is affected just like any other eigenvalue. If $v$ is an eigenvector of $G$ with eigenvalue $\lambda$, then $(G+cI)v=Gv+cIv=\lambda v+cv=(\lambda +c)v$, so $v$ is an eigenvector of $G+cI$ with eigenvalue $\lambda+c$. Without knowing what you'd tried and what you'd read, my general advice would be to trust more in your own mind and less in written authorities, in the Enlightenment tradition; this is a calculation that's readily done and allows you to decide the question yourself without worrying too much about contradictions in the literature.

b) The matrix $J$ is a projection matrix onto the orthogonal complement of $1_n$. That is, you can write any vector $v$ in $\mathbb R^n$ as $v=c_1 1_n + v^\perp$ with $v^\perp\perp1_n$, and then $Jv=v^\perp$, that is, $J$ orthogonally projects out the $1_n$ component. Since $G$ is real and and symmetric, its eigenvectors are orthogonal. Thus $J v=v$ for all eigevectors of $G$ except $1_n$, for which $J1_n=0$, and it follows by a calculation like the one under a) that subtracting $cJ$ from $G$ indeed shifts all eigenvalues except that of $1_n$ by $c$. Note, however, that this doesn't mean that only non-zero eigenvalues are shifted; all eigenvalues except that of $1_n$ are shifted, including any further zero eigenvalues that $G$ may have.

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Thanks for the advice and the answer. With b): are you implying that centering the vector (subtracting the mean value) is projecting out the $1n$ component? (should be so). And as for the literature, I'm surprised how misleading it $might$ be. – user506901 Mar 21 '12 at 7:37
@user506901: I'm not sure whether I was implying that, since I wasn't thinking in terms of centering; but it's certainly true. – joriki Mar 21 '12 at 12:42
Taken from the literature: "The eigenvectors of $G$ and $G'=G-cJ$ remain the same, but with a change of order. Note that $\lambda_i(G)= 0$ maps into $\lambda_i(G')= 0$ with eigenvector $1_n$,..." I suppose the author made a typo; instead of "$\lambda_i(G')= 0$ with eigenvector $1_n$", one should have "$\lambda_i(G')= 0$ with eigenvector $0_n$", as from your calculations. – user506901 Mar 30 '12 at 11:44
@user506901: No. My calculations don't refer to any vector $0_n$, and you haven't define what that denotes. In case you mean the zero vector (with all entries zero), this is not an eigenvector, since eigenvectors are by definition non-zero; otherwise the zero vector would be an eigenvector of every matrix and every matrix would have a zero eigenvalue. The quote you give from the literature is correct; the eigenvector $1_n$ with eigenvalue $0$ remains an eigenvector with eigenvalue $0$ for the shifted matrix. – joriki Mar 30 '12 at 11:59
That's what I wanted to hear. One could easily misinterpret your "Thus $Jv=v$ for all eigevectors of $G$ except $1_n$, for which $J1_n=0$" – user506901 Mar 30 '12 at 12:25

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