# Given matrix's eigenvectors find the corresponding eigenvalues [closed]

I have found the eigenvectors from a matrix. Now I need the corresponding eigenvalues. How can I get the eigenvalue corresponding to each eigenvector in a "efficient" way for a computer?

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## closed as not a real question by Andrés E. Caicedo, martini, TMM, Thomas, J. M.Nov 25 '12 at 11:43

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## 1 Answer

Well, of course, if $x$ is an eigenvector of $A$ we have $$Ax = \lambda x$$ where $\lambda$ is the eigenvalue. Thus, you can take any non-zero component of $Ax$ and divide it by the corresponding component of the vector $x$ and you will get $\lambda$. Say the $k$:th component of $Ax$ is non-zero and the rows of $A$ are $a_1^T, a_2^T, \dots, a_n^T$, so the $k$:th component of $Ax$ will be $a_k^Tx$ and we get $$\lambda = \frac{a_k^T x}{x_k}.$$

So an efficient way to calculate the eigenvalue $\lambda$, given the eigenvector $x$, is to calculate $a_k^T x$ for $k = 1, 2, \dots, n$ until you get a non-zero value, and then get $\lambda$ from the formula above. If no $a_k^T x$ is non-zero, we get $\lambda = 0$.

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Ok I think I got it. Thanks! – Sheol Nov 22 '12 at 22:18
Of course, with floating-point calculations you need to be careful: something that is non-zero but very small may be an artefact of roundoff error, and if you use that your answer will be way off. It may be best to use the $k$ for which $|x_k|$ is largest. – Robert Israel Nov 22 '12 at 22:34
Yes you are right, I'm going to use the largest. – Sheol Nov 22 '12 at 23:17