# Cant find an eigenvector for an eigenvalue

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.

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

Actually, since this is a triangular matrix, its eigenvalues are the entries of the main diagonal: $$1$$, $$3$$, and $$5$$.

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.

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