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For the following matrix, compute

  1. its characteristic polynomial
  2. its eigenvalues

$$A = \begin{bmatrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 2 & -5 & 4\end{bmatrix}$$

So I think I know to find the characteristic polynomial, I have to compute det(A-$\lambda$I) = 0. Which gives...

$$A-\lambda I = \begin{bmatrix}-\lambda & 1 & 0 \\ 0 & -\lambda & 1 \\ 2 & -5 & 4-\lambda\end{bmatrix}$$

$$-\lambda\begin{vmatrix}-\lambda & 1 \\ -5 & 4-\lambda\end{vmatrix} - 1 \begin{vmatrix}0 & 1 \\2 & 4-\lambda\end{vmatrix} $$

After calculating the determinant I get, $-\lambda(-4\lambda + \lambda^{2}+5) +2 = 0 $

What do I do next? I'm completely stuck.

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  • $\begingroup$ The sum of the entries on each row is always $1$, so $1$ is an eigenvalue. The now use the trace and the determinant to find the other two. $\endgroup$
    – Git Gud
    Nov 28, 2014 at 0:01
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    $\begingroup$ You can simplify your cubic, and hope for obvious roots. You will find quickly that $\lambda=1$ (and $\lambda=2$) work. If you divide the cubic polynomial by $\lambda-1$ or $\lambda-2$, you will get a quadratic, where root finding is easy. We got "lucky." For a general cubic finding the roots can be messy. Note that there is a double root. $\endgroup$ Nov 28, 2014 at 0:05
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    $\begingroup$ en.wikipedia.org/wiki/Companion_matrix $\endgroup$
    – Will Jagy
    Nov 28, 2014 at 1:57
  • $\begingroup$ What to do next: 1) find the roots of the characteristic polynomial (plot it), 2) solve the (undetermined) linear systems $A-\lambda I$ obtained when plugging the root values. $\endgroup$
    – user65203
    Nov 28, 2014 at 11:08

2 Answers 2

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The sum of the entries on each row is always $1$, so $\left(1,\begin{bmatrix} 1\\1\\1\end{bmatrix}\right)$ is an eigenpair.

Since $\det(A)=2$ (easily seen by Laplace expansion on the first row) and $\text{tr}(A)=4$, the other 'two' eigenvalues follow easily.


Actually answering your question, since $$-\lambda(-4\lambda + \lambda^{2}+5) +2 = 0\iff \lambda^3-4\lambda ^2+5\lambda -2=0$$

(I'd rather work with monic polynomials, this equivalence is unnecessary), the rational root theorem says the rational roots, if any, are in $\left\{-2-1,1,2\right\}$, so you can try these and hope for the best.

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Visual inspection tells us that matrix $\rm A$ is a companion matrix and that $1$ is an eigenvalue of $\rm A$. Hence, the characteristic polynomial of $\rm A$ is

$$q (s) := s^3 - 4 s^2 + 5 s - 2 = (s - 1) (s^2 - 3 s + 2) = \color{blue}{(s-1)^2 (s-2)}$$

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