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Any submultiplicative norm (for example the row-sum-norm) is an upper bound for the spectralradius of a matrix A. But is there a way to get a suitable LOWER bound for the spectralradius ?

Motivation : I would like to have a lower bound for the eigenvalues of a matrix with positive integer entries. In particular, is there a nxn-matrix A with positive integer entries and the integer eigenvalues 1..n ? The answer for n>1 seems to be no. Even more interesting would be the following : If a nxn-matrix A has positive integer entries and distinct positive integer eigenvalues (such as [ [4,1,1] , [1,2,1] , [1,1,2] ]) what is the least possible value for the largest eigenvalue (in the example it is 5 because the eigenvalues are 1,2 and 5).

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  • $\begingroup$ An $n\times n$ matrix with positive integer entries and eigenvalues $1,\ldots ,n$ would be $\mbox{diag}(1,\ldots,n)$. You can add any strictly upper triangular matrix with positive integer entries to this and still maintain the eigenvalues. $\endgroup$ – Jonas Dahlbæk Jul 8 '14 at 9:38
  • $\begingroup$ The entries should be positive, not only non-negative. $\endgroup$ – Peter Jul 8 '14 at 9:39
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For a matrix $A \in \mathbb{R}^{n \times n}$ with strictly positive entries (actually you just need it to be irreducible), you can apply the Perron-Frobenius Theorem which asserts that the eigenvalue $\lambda$ with largest magnitude is positive, simple and the associated eigenvector has strictly positive entries. In particular, with these special settings you may apply the Collatz-Wielandt Theory which tells you that for every $x>0$ (i.e. each component of $x$ is strictly positive) with $\|x\|=1$, $$\min_{i=1,\ldots,n}\frac{(Ax)_{i}}{x_i} \leq \lambda \leq \max_{i=1,\ldots,n}\frac{(Ax)_{i}}{x_i} $$ This is a powerful lower/upper bound generator for the largest eigenvalue $\lambda$ of $A$ (obviously equality holds if and only if $Ax = \lambda x$).

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    $\begingroup$ So, there is indeed no nxn-matrix A (n>1) with positive integer entries and the eigenvalues 1,2,...,n. Thanks a lot! $\endgroup$ – Peter Jul 8 '14 at 10:01
  • $\begingroup$ I just noticed that I was too fast with my conclusion, nevertheless your answer is very helpful! $\endgroup$ – Peter Jul 8 '14 at 10:37

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