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.
Best Answer
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.