# Eigenvectors of $\tiny\begin{bmatrix} 4 & 0 \\ 1 & 2 \end{bmatrix}$

Find the eigenvectors of $\begin{bmatrix} 4 & 0 \\ 1 & 2 \end{bmatrix}$.

I did the following:

$\det(A-I\gamma)=0 \Rightarrow \begin{cases}\gamma=4 \\ \gamma=2 \end{cases}$

For $\gamma=4$:

$\begin{bmatrix} 0 & 0 \\ 1 & -2 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \end{bmatrix}$

$x_1=2x_2 \Rightarrow x=\begin{bmatrix} 2 \\ 1\end{bmatrix}$

Then I proceed to $\gamma=2$ to find the second eigenvector:

$\begin{bmatrix} 2 & 0 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \end{bmatrix}$

The system ends up as:

$\begin{cases} 2x_1+0x_2=0 \\ x_1+0x_2=0 \end{cases} \Rightarrow x_1=0$

I don't have a value for $x_2$ so, whats my second eigenvector?

• Your eigenvector is whatever you like satisfying $x_1=0$. For example, $(0,1)$. – vadim123 Jul 9 '15 at 17:01
• I often multiply the the result $v_n$ by $t$, because it gives me confidence in problems which can be interpreted geometrically. – Eemil Wallin Jul 9 '15 at 17:29

"You don't have a value" means that $x_2$ is arbitrary. In this case, the eigenspace contains the vectors $(0,x_2)$ for arbitrary $x_2$. Thus, it is generated by $(0,1)$.
Just like what happened in the first one, there, $x_2$ was also arbitrary. If you had values for both quantities, you wouldn't have had a generating vector.