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