Proof matrices and their eigenvalues

Question

Let $C=A-B$ where $A=\begin{bmatrix}I &0\\ 0&0\end{bmatrix}$ and $B$ is a Laplacian matrix of a connected graph, so sum for rows is null and it doesn't have any zero row(or column). Eigenvalues of B are in [0,1) . $A$, $B$ and $C$ have the same dimension.

I think it's can be proved that $\lambda_i(C)\neq1$.

Answer 1

Counterexample: $$ A = \pmatrix{ 1&0&0\\ 0&0&0\\ 0&0&0 },\; B= \pmatrix{ 0&0&0\\ 0&-1/2&1/2\\ 0&1/2&-1/2 } $$ As for your newest question, note that Laplacian matrices are always positive semidefinite (i.e. have only non-negative eigenvalues). Since the sum of two positive semidefinite matrices is itself positive semidefinite, $C$ will be positive semidefinite and thus cannot have $-1$ as an eigenvalue.

Answer 2

Pretty sure you are wrong. My counterexample: $$B=\begin{bmatrix}0 & 0 & 0 & 0\\ 0 & -1 & 1 & 0\\ 0&0&-1&1\\ 0 & 0 & 0 & 0\end{bmatrix} \\A=\begin{bmatrix}1&0&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&0\end{bmatrix}$$