Laplacian matrix eigenvalues 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 $1$??
 A: 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.
A: 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?
