Question: I have this matrix

$$ \begin{pmatrix}
 4 & -9 & 0 \\
 -9 & 36 & 0 \\
 0 & 0 & 40 \\
 \end{pmatrix} $$

Solving, i can easily find that eigenvalues are $\lambda=40, 20 + \sqrt{337}, 20 +- \sqrt{337} $

I cannot understand why one of the eigenvectors is 

$$ \begin{pmatrix}
 0\\
 0 \\
 1 \\
 \end{pmatrix} $$

I thought eigenvectors are solved by just plugging in the eigenvalues into my characteristic equation and then solving from there.

I did not have one of the best linear algebra teachers, so i did not learn much from him.

Thank you for whomever helps.

Answer: 
> 
> I thought eigenvectors are solved by just plugging in the eigenvalues into my characteristic equation and then solving from there.
> 
> 
> 

The eigenvalues are the roots of your characteristic equation. An eigenvector $x$ is a vector that satisfies the relation
$ Ax = \lambda x$ for some eigenvalue $\lambda$. So for your eigenvalue $\lambda=40$ you get

$ \begin{pmatrix}
 4 & -9 & 0 \\
 -9 & 36 & 0 \\
 0 & 0 & 40 \\
 \end{pmatrix} \pmatrix{0 \\ 0\\ 1}=40\pmatrix{0 \\ 0\\ 1}$
Finding an eigenvalue:

$$\begin{array}{|ccc|}
4-\lambda &-9&0\\
-9&36-\lambda&0\\
0&0&40-\lambda
\end{array}=0$$
Suppose that the eigenvalues are indeed $\lambda=40,20+\sqrt{337},20-\sqrt{337}$.

Finding the eigenvector of $\lambda=40$:

$$ \left(\begin{array}{ccc|c}
-36&-9&0&0\\
-9&-4&0&0\\
0&0&0&0
\end{array}\right)\Rightarrow \begin{pmatrix}
0\\
0\\
u
\end{pmatrix}:u\in\mathbb{R}$$