I have a quick question about eigenvalues. If I'm given one eigenvector of a 3 by 3 matrix, I can easily calculate its corresponding eigenvalue.

I know I can determine the other two eigenvalues by factorising its characteristic polynomial. My question: Is there a way to work out the other two eigenvalues without factorising the characteristic polynomial? In other words, can I use the fact that the $det(P) = \lambda_1 \lambda_2 \lambda_3$ to work out the other two. For example if $\lambda_1 = 2$ and determinant is 20, $\lambda_2 \lambda_3 = 10$. This means the other two eigenvalues can either be 2 and 5 or 10 and 1.

Any useful insight would be much appreciated.

  • 1
    $\begingroup$ What did in the example is incorrect, there's no reason why eigenvalues should be natural numbers. But you can use the fact that $\text{tr}(P)$ is the sum of the eigenvalues. $\endgroup$ – Git Gud Mar 22 '15 at 21:01
  • $\begingroup$ Really? I suppose eigenvalues can be complex but assuming they are real in this case? Thanks for your comment. $\endgroup$ – John_dydx Mar 22 '15 at 21:32
  • $\begingroup$ Consider $P=\begin{bmatrix} 2 & 0 & 0\\ 0 & \pi & 0\\ 0 & 0 & \frac {10}{\pi}\end{bmatrix}.$ $\endgroup$ – Git Gud Mar 22 '15 at 21:34
  • $\begingroup$ So this matrix agrees with $det(P) = product of eigenvalues$. $\endgroup$ – John_dydx Mar 22 '15 at 21:40

Use $tr(P)$ matrix is equivalent to the sum of the eigenvalues.

Also keep in mind eigenvalues can be negative and/or complex values. Do not restrict the domain to positive integers.


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