Is there any ways to compute the eigen vector without computing explicitly the associated eigenvalue?
Actually, I'd like to compute the largest eigenvalue of a positive matrix from its eigen vector, so I have to know its eigenvector first.
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2$\begingroup$ Homework? mathoverflow.net/faq en.wikipedia.org/wiki/Power_iteration $\endgroup$– GoldsternCommented Sep 23, 2012 at 8:43
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5$\begingroup$ Obviously no. If you manage to compute an eigenvector $\mathbf{v}$ of a matrix $A$ then to automatically get the associated eigenvalue $\lambda$ by the formula $A\mathbf{v}=\lambda\mathbf{v}$. $\endgroup$– Fernando MuroCommented Sep 23, 2012 at 9:53
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1$\begingroup$ @Fernando: the comment is a bit tautological, since this is exactly what the OP is proposing to do. $\endgroup$– Igor RivinCommented Sep 23, 2012 at 18:45
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$\begingroup$ @Igor: you're right, it seems I didn't pay attention to the question's second paragraph. Anyway, everything looks tautological in this post. $\endgroup$– Fernando MuroCommented Sep 28, 2012 at 8:22
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2 Answers
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If you pick a random vector $v$ and look at $v_n=A^n v/\| A^n v\|,$ that will converge to the dominant eigenvector.
answered Sep 23, 2012 at 18:46
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$\begingroup$ Small correction: pick a positive vector. If you just pick at random you might get very unlucky and pick one that is orthogonal to the eigenvector. $\endgroup$ Commented Sep 24, 2012 at 10:20
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$\begingroup$ Good point (in the OP's stochastic context...) $\endgroup$ Commented Sep 25, 2012 at 4:09
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What do you know about the matrix?
If we know that the rows all have the same sum (but not what that sum is) then we would essentially find it by multiplying by the corresponding eigenvector, $\mathbb{j}$, the all $1$'s vector. This will be the largest eigenvalue provided that the entries are non-negative.
One way this could happen (but not the only one) is if the rows are identical or merely each is a permutation of the first.
If certain rows are equal then we know that $0$ is a eigenvalue although we never "computed" it. Then we do know an eigenvector.
In a circulant matrix we know all the eigenvectors (not just $\mathbb{j}$) and we essentially use them to compute the corresponding eigenvalues.
answered Sep 23, 2012 at 9:14
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$\begingroup$ Aaron, Actually I'd like to compare the spectral radius of 2 matrices, say A1 and A2. The information about the matrices are: - They are doubly sub-stochastic matrix, with one row sum = a<1 ; their determinant are the same; their traces are the same. we know that they both have spectral radius between a and 1, but I'd like to compare them without computing them explicitly. E.g: A1 = [1/4 1/4 1/4 0; 1/4 3/4 0 0; 1/4 0 1/2 1/4; 0 0 1/4 3/4] A2 = [2/4 1/4 0 0; 1/4 1/4 1/4 1/4; 0 1/4 3/4 0; 0 1/4 0 3/4] We get, rho(A1) < rho(A2), but can we infer w/o computing it? – hayu 0 secs ago $\endgroup$– hayuCommented Sep 23, 2012 at 13:57