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For the matrix

$$ \begin{pmatrix} 1 & 2 & 3 \\ 0 & 3 & 4 \\ 0 & 0 & 5 \\ \end{pmatrix} $$

I know that $5, 2+\sqrt3, 2-\sqrt3$ are eigenvalues. I am trying to find an eigenvector for $2+\sqrt3$ using $(A-\lambda I)V=0$. But this gives me: $$ \begin{pmatrix} -1-\sqrt3 & 2 & 3 \\ 0 & 1-\sqrt3 & 4 \\ 0 & 0 & 3-\sqrt3 \\ \end{pmatrix}\begin{pmatrix} x \\ y \\ z \\ \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \\ 0 \\ \end{pmatrix} $$

Which implies $x=y=z=0$. But this isnt possible as an eigenvector cannot be a $0$ vector.

What am I doing wrong?

NOTE: Thank you all, I see it now.

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    $\begingroup$ Your eigenvalues are incorrect. Why do you think $2\pm \sqrt3$ are eigenvalues? $\endgroup$ – Misha Lavrov Apr 26 at 21:22
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    $\begingroup$ Since the matrix is triangular, the eigenvalues are $1,3$ and $5$. $\endgroup$ – Bernard Apr 26 at 21:23
  • $\begingroup$ My mistake. I directly tried using the characteristic equation without noticing the roots directly and must've incorrectly factorised something. $\endgroup$ – Mohamad Moustafa Apr 26 at 21:26
  • $\begingroup$ $(1 - \lambda)(3 - \lambda)(5 - \lambda) = 0$. $\endgroup$ – David G. Stork Apr 26 at 21:36
  • $\begingroup$ I multiplied them, then when going back must've made some mistake $\endgroup$ – Mohamad Moustafa Apr 26 at 21:37
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Actually, since this is a triangular matrix, its eigenvalues are the entries of the main diagonal: $1$, $3$, and $5$.

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Indeed, as Jose has rightly pointed out, the eigenvalues are the diagonal entries of a triangular matrix:

$$ \det(A - \lambda I) = 0$$ $$ \begin{vmatrix} 1-\lambda & 2 & 3 \\ 0 & 3-\lambda & 4 \\ 0 & 0 & 5- \lambda \end{vmatrix} = 0$$

$$ (1-\lambda) \begin{vmatrix} 3-\lambda & 4 \\ 0 & 5-\lambda \end{vmatrix} - 2\begin{vmatrix} 0 & 4 \\ 0 & 5-\lambda \end{vmatrix} + 3\begin{vmatrix} 0 & 3 - \lambda \\ 0 & 0 \end{vmatrix} = 0 $$

$$ (1 - \lambda)(3 - \lambda)(5 - \lambda) = 0$$

Its one reason I like linear algebra-the very neat and mesmerising results.

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  • $\begingroup$ You forgot parentheses around $1-\lambda$ when you expand the $3\times3$ determinant. $\endgroup$ – YiFan Apr 26 at 23:00
  • $\begingroup$ Thanks for that! $\endgroup$ – John_dydx Apr 26 at 23:27

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