Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $\rho(M)$ denote the spectral radius (modulus of the largest eigenvalue) of a square matrix $M$.

I am looking for a characterization or anything else interesting about the set of matrices $A$ having the following properties:

(i) $A$ is (entrywise) nonnegative;

(ii) $A$ is irreducible;

(iii) $\rho(A) \geq 1$;

(iv) if $B$ is obtained from $A$ by deleting row $i$ and column $i$ (for any $i$), then $\rho(B) < 1$.

share|improve this question
    
What if diagonal entries are $\frac35$ and off diagonal $\frac15$? –  Aaron Meyerowitz Oct 23 '12 at 17:06
    
Shoot - too slow. I was gonna go with diagonal 2/3, off diagonal 1/4. –  Zack Wolske Oct 23 '12 at 17:07
    
If the matrix is irreducible with largest eigenvalue exactly $1$ then the property should hold. An easy way to achieve this is to have the entries on each row add to $1$. –  Aaron Meyerowitz Oct 23 '12 at 17:14
    
Deleted my last comment to fix an error: For 2x2 matrices, if you replace "trace > 1" by "trace > 1 + determinant", then it is a characterization. One still needs the diagonal entries to be less than 1, but there is no upper bound on the off-diagonals. For example, diagonal entries 3/4 and off-diagonal entries 2 satisfies (i)-(iv). –  Zack Wolske Oct 23 '12 at 17:40
add comment

1 Answer

Let's assume the matrix $A$ has all its row sums equal to $\lambda$, the largest eigenvalue. We can rescale the rows and columns of any other nonnegative irreducible matrix by a similarity transformation $A \to D^{-1} A D$ where $D$ is diagonal with positive diagonal entries (namely the entries of the Perron eigenvector of $A$). This makes all row sums equal to $\lambda$, and preserves the eigenvalues of $A$ and of the matrices obtained by removing the $i$'th row and column.

Let $B_k$ be the matrix obtained from $A$ by removing row and column number $k$. The row sums of $B_k$ are $\lambda - a_{ik}$ for $i \ne k$. Thus the largest eigenvalue of $B_k$ is at most $\lambda - \min_{i \ne k} a_{ik}$ and at least $\lambda - \max_{i \ne k} a_{ik}$. So a necessary condition is $$\lambda \ge 1 > \lambda - \min_k \max_{i \ne k} a_{ik}$$ while a sufficient condition is $$\lambda \ge 1 > \lambda - \min_k \min_{i \ne k} a_{ik}$$

share|improve this answer
    
Thanks -- this is very helpful. I especially like the elegant idea of working with an A that has all row sums equal. I wonder if it's possible to use this idea to think about similar conditions involving powers of A and B, where the conditions might be less demanding than the sufficient condition (i.e. they could bite for sparse matrices... –  Ben Golub Oct 23 '12 at 22:41
add comment

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.