I’m trying to diagonalize the following matrix:
$$ A = \left[ \begin{matrix} 1&3\\ 4&2 \end{matrix} \right] $$

I’ve found the eigenvalues to be:
$$ \lambda = 2$$ $$ \lambda = 1$$

For $\text{$\lambda=2$}$, the eigenvector I have is
$$\left[ \begin{matrix} 3\\ 1 \end{matrix} \right]$$

But for $\text{$\lambda=1$}$, I am getting an eigenvector of:
$$\left[ \begin{matrix} 0\\ 0 \end{matrix} \right]$$

Are my eigenvalues wrong or does this just mean that I can’t diagonalize the matrix $A$?

  • $\begingroup$ The eigenvalues are $5$ and $-2$. Check your work again. In particular, have you got the characteristic polynomial wrong? It should be $(x-1)(x-2) - 3 \times 4 = 0$. $\endgroup$ – астон вілла олоф мэллбэрг Apr 21 '18 at 23:54
  • $\begingroup$ @астонвіллаолофмэллбэрг nice catch! I forgot to include the -12 in the characteristic equation. Thanks! $\endgroup$ – adamcasey Apr 22 '18 at 0:03
  • $\begingroup$ You are welcome. I hope you managed to find the eigenvectors as well. $\endgroup$ – астон вілла олоф мэллбэрг Apr 22 '18 at 0:42

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