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From the Perron-Frobenius theorem, it might be concluded that the spectral radius is the largest eigenvalue for positive matrices, ie, matrices with strictly positive entries. In other words, the largest magnitude eigenvalue of a symmetric, real, positive matrix, is a positive eigenvalue.

However, I'm interested if the theorem is applicable to an arbitrary real symmetric matrix $R$ with no zero entries. Namely, the particular interest is whether the [power iteration] method2, might be used to extract the eigenvector and the associated largest positive eigenvalue of an input matrix.

Obviously, I'm concerned with the possibility of largest magnitude eigenvalue being negative. Is the Power Iteration inappropriate in this case? If not in general, does stating that the eigenvector of all ones, $1_n$, corresponding to the eigenvalue of $0$ spans the null-space of $R$?

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    $\begingroup$ I'm having trouble understanding your question. $\left(\begin{matrix} 1 &0\\0&-2\end{matrix}\right)$ is real, symmetric and the largest positive eigenvalue is less than the spectral radius. There is no null-space in this example. $\endgroup$ – Deathbreath Apr 5 '12 at 12:15
  • $\begingroup$ I've made an edit with a restriction on non-zero entries. In general, I guess, the suggestion would be to shift $R$ across the diagonal by the spectral radius, and then invoke Power Iteration. This way, the spectrum is shifted to the positive side of $0$. Could I get the eigenvector corresponding to the negative dominant eigenvalue in case I do not use the shift? $\endgroup$ – usero Apr 5 '12 at 14:34
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    $\begingroup$ A unitary transform of my matrix will give a symmetric, real, everywhere non-zero matrix with the same eigenvalues. You could shift down by the spectral radius and multiply by -1. $\endgroup$ – Deathbreath Apr 5 '12 at 16:22

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