Welcome to MathOverflow

Question

2

4

Here's my question --

Let $A$ be an $n \times n$ real matrix, and suppose that the spectral radius $\rho(A)$ is less than one (spectral radius = max eigenvalue). Let's choose some $1 \leq i \leq N$ and look at $A_{N,i}$. Namely, let's replace $A_{N,i}$ with some new value, $a$, to give us a new matrix $\hat A$. I want to characterize the set $\lbrace a : \rho(\hat A) < 1 \rbrace$. It pretty clear that this set is of the form $[0, a_{max})$, but I want to be able to compute $a_{max}$ analytically, given $A$ and $i$. (Also clearly $a_{max} \geq A_{N,i}$, since $\rho(A) < 1$ by assumption.)

This seems like it should be a fairly easy exercise but I haven't been able to make any useful progress on it.

Thanks!

-h

flag	asked Dec 14 2009 at 17:39 hal iii 127●3

What's N? ..... – Scott Morrison♦ Dec 14 2009 at 18:07

3

The spectral radius is a continuous function of the matrix (for the roots of a polynomial depend continuously on tits coefficients). By using explicit continuity results (like those given in [Marden, Morris. The Geometry of the Zeros of a Polynomial in a Complex Variable. Mathematical Surveys, No. 3. American Mathematical Society, New York, N. Y., 1949. ix+183 pp.]) you may find explicit expressions for open intervals contained in the set $\{a:\rho(\hat A)<1\}$. – Mariano Suárez-Alvarez Dec 14 2009 at 19:29

1

Scott: I think $N=n$. – Greg Kuperberg Dec 14 2009 at 23:30

3

I am used to thinking of complex vector spaces, but do you mean the maximum of the absolute values of the real eigenvalues? If so, the spectral radius will often be $-\infty$ I guess. In any case, I'll give you an example where the spectral radius is increased by replacing an element by 0, which by scaling shows that 0 will not generally be in the set. Let n=N=2 and let A be the matrix with -1 in the bottom right and 1 elsewhere. Let i=2. Then the spectral radius of A is $\sqrt2$, but replacing $A_{22}$ by 0 yields a matrix whose spectral radius is $(1+\sqrt5)/2$. – Jonas Meyer Dec 15 2009 at 5:40

1

You can edit your answer. Please correct it and state all of your relevant hypotheses. For instance, are you assuming as in mathoverflow.net/questions/10885 that the matrix is a pseudometric? And the elements that are supposed to replace an entry, are they also assumed to be positive? And did you mean to be replacing entries by a, or perturbing them by a as the answers below assumed? – Jonas Meyer Jan 6 2010 at 5:38

show 4 more comments

fedja · Answer 1 · 2010-02-23 22:41:52Z UTC

Under your assumptions (all matrix elements positive), the spectral radius is the same as the largest positive eigenvalue, so you just need to figure out for which $a$ the determinant of $\widehat A-I$ is zero, which is a linear equation in $a$.

David Lehavi · Answer 2 · 2009-12-15 13:34:05Z UTC

Compute the SVD of A: A = U D V.

Now solve E_{ij} = U B V; getting B= U^* E_{ij} V^*.

Assuming that you mean that your a is the perturbation, and not the value at ij, You are trying to find the the spectral radius of D + a B. This might be a little easier (from a numerical point of) then your original problem if your B is simple enough.

David Bar Moshe · Answer 3 · 2009-12-15 18:37:44Z UTC

I'll assume that A is positive. I'll also assume as in the previous answer that a is the perturbation. The eigenvalues of high powers of a matrix are dominated by its largest eigenvalue. Thus, we obtain the following close expression of the ratio of the larget eigenvalue of the perturbed and nonperturbed matrices:

ratio = lim_n-->inf (trace((A + a *E_Ni)^n/trace(A^n))^(1/n))

In general, this expression converges very fast.

spectral radius of a matrix as one element changes

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

3 Answers

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.