I'm trying to solve for the eigenvalues of a matrix $$ A = \begin{pmatrix} 1 & 2 & 0 \\ -1 & -1 & 1 \\ 0 & 1 & 1 \\ \end{pmatrix} $$

I tried expanding $\det(A - \lambda I)$, but that didn't help. I got $-\lambda^3 - \lambda^2 + 2 = 0$, or $(1 - \lambda)(\lambda^2 + 2\lambda + 2)=0$. Either way, I can't factor for $\lambda$. Any ideas?

  • 1
    $\begingroup$ Hint: Quadratic Formula $\endgroup$ – Tim Raczkowski Jun 16 '15 at 18:57
  • $\begingroup$ Oh, sorry. When using that formula, I get (-2 + (-4)^(1/2))/2 or (-2 - (-4)^(1/2))/2... I don't think I can solve for a negative root. $\endgroup$ – ConJoJohn Jun 16 '15 at 19:02
  • $\begingroup$ I find as characteristic polynomial : $$p(\lambda) = \lambda^2 (1-\lambda)$$ $\endgroup$ – Joelafrite Jun 16 '15 at 19:08
  • $\begingroup$ Yeah, just got that too. I messed up. Whoops! $\endgroup$ – ConJoJohn Jun 16 '15 at 19:14

Your calculation of the characteristic polynomial is incorrect. If you redo the calculation, your problem should be resolved.

However, there are other real matrices for which the polynomial is what you've written. In general, The eigenvalues are always given by the roots of the characteristic polynomial.

The polynomial here has only one real root, 1. You can interpret it in one of two ways, depending on the context:

  1. The matrix has a single eigenvalue, 1.

  2. The matrix has, in addition, two nonreal eigenvalues, which you can find by solving the quadratic term. The eigenvectors will, of course, also be complex.

Complex roots of real matrices are very common, for example, with rotation matrices, such as $\{\{0,1\},\{-1,0\}\}$. There is no real vector which is multiplied by a scalar when this matrix is applied.

  • $\begingroup$ I suppose, given this class requires no knowledge of complex numbers, that I should just go with the single value. Thanks. $\endgroup$ – ConJoJohn Jun 16 '15 at 19:03
  • $\begingroup$ @CoJo: Actually, I've now checked your calculation of the polynomial, and it's incorrect. Try again - the correct polynomial has only real roots. $\endgroup$ – Meni Rosenfeld Jun 16 '15 at 19:06
  • $\begingroup$ Oh, wow. That's embarrassing... I'll look for that. Oh, I see. Screwed up the signs. $\endgroup$ – ConJoJohn Jun 16 '15 at 19:08

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