How to find the eigenvectors for this matrix?

Question

So I have already worked out the eigenvalues for the following matrix,

\begin{bmatrix}-3 & 1\\-1 & 1\end{bmatrix}

which happen to be $\lambda_1=-1+\sqrt{3}$ and $\lambda_2=-1-\sqrt{3} $.

However, I am having trouble finding the eigenvectors. This is what I have so far,

From $\lambda_1=1+\sqrt{3}$,

\begin{equation}\begin{bmatrix}-2-\sqrt{3} & 1\\-1 & 2-\sqrt{3}\end{bmatrix} \begin{bmatrix}x_1\\x_2\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}\end{equation}

So using the second equation, \begin{equation} -x_1+(2-\sqrt{3})x_2=0\end{equation} \begin{equation}(2-\sqrt{3})x_2=x_1\end{equation}

Eigenvector is \begin{equation}\begin{bmatrix}(2-\sqrt{3})x_2\\x_2\end{bmatrix}\end{equation}

I don't know how to find the value for $x_2$?

If someone could guide me on finding this value, it would be greatly appreciated!

Answer 1

You cannot find the value of $x_2$ because every value is OK. That is, no matter what $x_2$ is equal to, it is always true that $$\begin{bmatrix}-3&1\\-1&1\end{bmatrix}\cdot\begin{bmatrix}(2-\sqrt{3})x_2\\x_2\end{bmatrix}=\lambda_1\cdot\begin{bmatrix}(2-\sqrt{3})x_2\\x_2\end{bmatrix}$$

This is true in general. If $\lambda$ is an eigenvalue of $A$, and $x$ is an eigenvector, and $\alpha$ is any number, then you have, for $y=\alpha x$, $$Ay=A(\alpha x)=\alpha Ax=\alpha\lambda x=\lambda\alpha x=\lambda y$$

Answer 2

What's been said in both @5xum 's answer and @MisterRiemann 's comment is right. I would like to add the geometric intuition that, when we talk about eigenvectors/eigenvalues, we are really exploring the notion of "invariant spaces". In your case, there are two straight lines that "don't change" after applying the transformation (a matrix is nothing but a linaer application)

What I mean is that, if we call your matrix $A$ and that straight line $R$, then for every $v \in R, Av \in R$ An eigenvector is nothing but a vector that defines that straight line (it could be the case, for double eigenvalues for example, that your eigenvectors define a plane (or a higher dimensional space) instead!) As you already know, there are infinitely many vectors that define the direction of a straight line (just by taking multiples)

Answer 3

For a given Eigen -value there exist infinitely many Eigen-vectors.

So, we can set any non -zero value(as it leads to the trivial solution) to $x_2$ and find an Eigen-vector.

If we set $x_2 = 1$, the corresponding Eigen-vector is

$\begin{bmatrix} 2-\sqrt3 \\1\end{bmatrix}$

Similarly if we set $x_2 = 2$, we get

$\begin{bmatrix} 2(2-\sqrt3) \\2\end{bmatrix} = 2\cdot \begin{bmatrix} 2-\sqrt3 \\1\end{bmatrix}$

Similarly we can obtain for any $c$

$ \begin{bmatrix}c( 2-\sqrt3) \\c\end{bmatrix} = c\cdot \begin{bmatrix} 2-\sqrt3 \\1\end{bmatrix}$

which are all allowable.