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I have a large sparse nonnegative asymmetric matrix $A$. Since the matrix $A$ is nonnegative, its spectral radius $\rho(A)$ is an eigenvalue of it. But $A$ may have other eigenvalues being the same modulus as $\rho(A)$.

Can the power method give $\rho(A)$?

Eli4ph
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  • Do you need to compute the spectral radius in some specific example or do you need a rigorous proof? The set of matrices with multiple eigenvalues of the same modulus has density zero among all matrices. So for a practical computation you can just assume that there is a unique simple eigenvalue of maximal modulus. – quarague Aug 16 '19 at 12:43
  • @quarague The set of sparse matrices also has measure zero, but the OP is dealing with one. – user1551 Aug 16 '19 at 13:20
  • @user1551 True, but the point I was trying to make still stands. If all you need is the spectral radius of a given matrix, just assuming there will be a unique simple eigenvalue of maximal modulus will work out fine unless you have a very specific reason why your matrix should not satisfy that. – quarague Aug 16 '19 at 13:23

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No. This has nothing to do with the moduli of other eigenvalues. For instance, when $A$ is not primitive, even if the moduli of other eigenvalues of $A$ are strictly smaller than $\rho(A)$, the power method still doesn't always converge. For an illustrative example, consider $$ A=\pmatrix{0&2\\ 1&0},\ x_0=\pmatrix{1\\ 1}. $$ Using the power method, the iterates will oscillate between $x_0$ and $x_1=(\frac23,\frac13)^T$ and none of them is an eigenvector of $A$.

user1551
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  • Wow. What an example. Then is there any way, as simple as possible, to get the spectral radius? – Eli4ph Aug 16 '19 at 13:37
  • @Eli4ph I don't know any such method. You may do a web search for "power method for non-primitive matrices" or the like to see if there are viable solutions. The authors of this paper for instance have proposed an algorithm, which they claim is more robust. – user1551 Aug 16 '19 at 13:50