# Largest eigenvalue of a block matrix less than 1

I have a block matrix in the following form: $$M = \begin{bmatrix} 0 & 0 & 0 & 0 \\ -B_{1,0}A & (I-B_{1,1})A & -B_{1,2}A & -B_{1,3}A \\ -B_{2,0}A & -B_{2,1}A & (I-B_{2,2})A & -B_{2,3}A \\ -B_{3,0}A & -B_{3,1}A & -B_{3,2}A & (I-B_{3,3})A \\ \end{bmatrix}$$ And the following conditions hold:

1) The sum of block elements in each row is $A$, i.e. $\sum_{j=0}^3B_{i,j}=0, \forall i$.

2) The largest eigenvalues of all blocks $(I-B_{i,i})A, \forall i,$ is less than 1, and $I-B_{i,i}$ is positive semidefinite.

Is the largest eigenvalue of matrix $M$ always less than 1? How can I prove it?

No. Counterexample: consider the scalar case, where $A=2$, every $I-B_{ii}$ (with $i>1$) is equal to $\frac14$ and $$M = \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0&\frac12&0&\frac32\\ 0&0&\frac12&\frac32\\ 0&0&\frac32&\frac12 \end{bmatrix}.$$ The eigenvalues of $M$ are $0,\frac12,-1$ and $2$. By continuity, if you perturb the entries of $M$ on the last three rows by a little, you may also obtain a counterexample where $M$ is entrywise nonzero on the last three rows.