<s>[INST] 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.
 [/INST] 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.
</s>