can eigenvector be found without computing the eigenvalue - MathOverflow [closed] most recent 30 from http://mathoverflow.net 2013-05-18T20:50:00Z http://mathoverflow.net/feeds/question/107891 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/107891/can-eigenvector-be-found-without-computing-the-eigenvalue can eigenvector be found without computing the eigenvalue hayu 2012-09-23T08:05:34Z 2012-09-23T18:46:54Z <p>Is there any ways to compute the eigen vector without computing explicitly the associated eigenvalue?</p> <p>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.</p> http://mathoverflow.net/questions/107891/can-eigenvector-be-found-without-computing-the-eigenvalue/107893#107893 Answer by Aaron Meyerowitz for can eigenvector be found without computing the eigenvalue Aaron Meyerowitz 2012-09-23T09:14:21Z 2012-09-23T09:14:21Z <p>What do you know about the matrix?</p> <p>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.</p> <p>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.</p> <p>If certain rows are equal then we know that \$0\$ is a eigenvalue although we never "computed" it. Then we do know an eigenvector.</p> <p>In a <a href="http://en.wikipedia.org/wiki/Circulant_matrix#Eigenvectors_and_eigenvalues" rel="nofollow">circulant matrix</a> we know all the eigenvectors (not just \$\mathbb{j}\$) and we essentially use them to compute the corresponding eigenvalues.</p> http://mathoverflow.net/questions/107891/can-eigenvector-be-found-without-computing-the-eigenvalue/107918#107918 Answer by Igor Rivin for can eigenvector be found without computing the eigenvalue Igor Rivin 2012-09-23T18:46:54Z 2012-09-23T18:46:54Z <p>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.</p>