My goal is to find the eigenvectors of the matrix $A= \begin{bmatrix} -1 & 0 \\ 2 & 1 \end{bmatrix}$.
I have found the eigenvalues $\lambda_1=-1$ and $\lambda_2=1$. Now, I’m trying to use the formula $(A-\lambda I_2)X= \vec0$ to find the eigenvectors.
For $\lambda_1=-1$, I found the following :
$$\begin{bmatrix} -1-(-1) & 0 \\ 2 & 1-(-1) \end{bmatrix} \times\begin{bmatrix} x \\ y \end{bmatrix}= \begin{bmatrix} 0 \\ 0 \end{bmatrix}$$
$$\begin{bmatrix} 0 & 0 \\ 2 & 2 \end{bmatrix} \times\begin{bmatrix} x \\ y \end{bmatrix}= \begin{bmatrix} 0 \\ 0 \end{bmatrix}\Leftrightarrow \begin{cases} 0x + 0y = 0 \\ 2x + 2y =0 \end{cases}$$
For $\lambda_2=1$, I found the following :
$$\begin{bmatrix} -1-(1) & 0 \\ 2 & 1-(1) \end{bmatrix} \times\begin{bmatrix} x \\ y \end{bmatrix}= \begin{bmatrix} 0 \\ 0 \end{bmatrix}$$
$$\begin{bmatrix} -2 & 0 \\ 2 & 0 \end{bmatrix} \times\begin{bmatrix} x \\ y \end{bmatrix}= \begin{bmatrix} 0 \\ 0 \end{bmatrix}\Leftrightarrow \begin{cases} -2x + 0y = 0 \\ 2x + 0y =0 \end{cases}$$
So, I find myself with 2 systems of equations with infinite solutions. What should I do in this situation to find the eigenvectors ?
Best Answer
Yes, your equations will infinitely many solutions. This is because eigenvectors span an eigenspace; if $a$ is an eigenvectors, then so are $2a,3a,4a,\cdots$.
You just need to pick an easy example for your eigenvectors, and then all multiples of them will still be eigenvectors. For the first system, you can extract the equation $y=-x$, which is spanned by the vector $(1,-1)^T$. For the second, $x=0$ and $y$ can be anything, so this is spanned by, say, the unit $y$ vector: $(0,1)^T$. You can go further and say that this line ($x=0$) is an invariant line of the matrix.
These are two possible choices of representative eigenvector; any choice will do!
An exercise/thought experiment for you would be to show that the set of eigenvectors, for a given eigenvalue of a matrix, really is a vector subspace (an eigenspace).