Spectral decomposition with eigenvalue shift

Question

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 eigenvector, $1_n$. 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.

Answer 1

1. When you shift $G$ by $cI$, all eigenvalues are shifted by $c$, including the zero eigenvalue.

2. When you shift by $G$ by $cJ$, and if $G 1=0$ as you assume, all eigenvalues except one on the zero eigenvalues are shifted by $c$. Thus if $G$ has an $n$fold zero eigenvalue, the shifted matrix has $n-1$ eigenvalues $c$.

You can prove these statements by extending $1$ to an orthonormal basis consisting of eigenvectors of $G$. Then your problem has been diagonalized in the new basis and all these properties are trivially verified.