Laplacian matrix eigenvalues

Question

Let $L$ be the laplacian of a connected graph. Is the maximum eigenvalue of $A=\begin{bmatrix} I & 0\\0&0\end{bmatrix} -L$ different than $1$??

Answer 1

Let $J$ be the matrix $\begin{bmatrix}I&0\\0&0\end{bmatrix}$. Let's assume by absurd that the maximum eigenvalues of $A$ is $1$, then \begin{equation}\label{eq} \left ( J-L\right )x_{max,A}=x_{max,A} \end{equation} where $x_{max,A}$ is the maximum eigenvector of $A$ related to $\lambda_{max,A}=1$, that is the maximum eigenvalue of $A$.

Right multiply the first eq by $x_{max,J}^T$ \begin{equation}\label{eq1} x_{max,J}^TJx_{max,A}-x_{max,J}^TLx_{max,A}=x_{max,J}^Tx_{max,A} \end{equation} $J$ is symmetric then $x_{max,J}^TJ=1x_{max,J}^T$ therefore \begin{equation}\label{eq2} x_{max,J}^TLx_{max,A}=0 \end{equation}

But $x_{max,J}^T=[1\,0\,...\,0]$ then $x_{max,A}=[1\,...\,1]$, since L is sum for row null.

But from the first eq $x_{A}$ can not be equal to $[1\,...\,1]$ unless $J$ is the identity matrix.

What do you think?

Answer 2

The statement is equivalent to asking whether $$ A - I = -\left(\pmatrix{0&0\\0&I} + L\right) $$ Is necessarily invertible. My intuition is that this is not the case, but I'm still hunting for a counterexample.