Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?)

share|improve this question
1  
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
2  
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

1 Answer 1

It's a partial answer, just for the first part of the question.

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 $B-C$.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.