Spectral radius and powers of a $2\times 2$ block matrix

The problem I'm struggling is the following:

Let $n$ be a positive integer and let $A=% \begin{pmatrix} B & C\\ C & B \end{pmatrix} \in\mathcal{M}_{2n}(\mathbb{R}_{+})$, where $B,C\in\mathcal{M}_{n} (\mathbb{R}_{+})$. I am interested in putting conditions on $B$ and $C$ such that the spectral radius of $A$ is less than $1$.

I think that the answer is that $B+C$ and $B-C$ have spectral radius less than $1$, but I'm not very familiar working with block matrices and I don't know how to prove it (I came up with this guess by working with the scalar case, when $B$ and $C$ are just nonnegative numbers).

I am also interested in computing the powers of $A$ in terms of the powers of $B$ and $C$, in the case when the spectral radius of $A$ is less than $1$ (is this similar to the case when $n=1$, or is there something fundamentally different?)

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We find that the characteristic polynomial of $A$ is the product of the characteristic polynomials of $B-C$ and $B+C$. – Davide Giraudo Jul 14 '12 at 18:22
Are $B$ and $C$ supposed to be symmetric? If not, then $A$ won't be symmetric. – Geoff Robinson Jul 14 '12 at 19:44
@GeoffRobinson: $B$ and $C$ are not necessarily symmetric. I apologize for the confusion. I edited the title. – digital-Ink Jul 14 '12 at 20:21

We have \begin{align} \det(A-XI)&=\det\pmatrix{B-XI&C\\C&B-XI}\\ &=\det\pmatrix{B-XI&C\\C+B-XI&B-XI+C}\\ &=\det\pmatrix{I&0\\0&B+C-XI}\det\pmatrix{B-XI&C\\I&I}\\ &=\det(B+C-XI)\det\pmatrix{B-C-XI&C\\0&I}\\ &=\det(B+C-XI)\det(B-C-XI), \end{align} hence $\sigma(A)=\sigma(B+C)\cup \sigma(B-C)$. We deduce that the spectral radius of $A$ is $<1$ if and only if so are those of $B+C$ and $B-C$.
Can it be said more, knowing that $B$ and $C$ have non-negative values? – digital-Ink Jul 14 '12 at 22:03
I'm not sure, even when $B$ and $C$ are diagonal. (but it doesn't prove we can't said more, merely that I can't say more). – Davide Giraudo Jul 15 '12 at 14:17
If $B$ and $C$ are diagonal, then the eigenvalues are all nonnegative, since they are on the diagonal, hence $\rho(B-C)<1$, provided that $\rho(B+C)<1$. Is this true in general? – digital-Ink Jul 16 '12 at 7:55
Using the same argument as in math.stackexchange.com/q/172135/22728, I think it follows that $\rho(B-C)\leq\rho(B+C)$, so $\rho(A)<1$ iff $\rho(B+C)<1$. – digital-Ink Jul 18 '12 at 14:49