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?

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


I do agree with vadim123 and BolzWeir.

Another thing that was helpful when I did similar calculations of these kind (especially for exams):

For 2x2 matrices, the general formula helps a lot checking your exercises and avoiding making mistakes in calculation:

General Formula calculated on wolfram alpha


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.