# Find an eigenvector

I am trying to find an eigenvector of the following matrix, where I get an eigenvector zero. But, to my knowledge, an eigenvector cannot be zero. $$\pmatrix{0&1\\0&i-1}\pmatrix{x\\y} = \pmatrix{0\\0}.$$

• Is this the equation $(A-\lambda I)\mathbf x=\mathbf 0$? Nov 26, 2019 at 10:18

It's not zero. The two equations corresponding to the linear system gives $$y=0$$, $$\forall x \in \mathbb{R}$$. That said, any vector of the form $$[a,0]^T$$ works. Assuming you want normalized eigenvectors, then you can choose $$a = 1$$.
You are looking for an eigenvector $$(x,y)$$ associated to the eigenvalue $$\lambda=0$$.
$$\begin{pmatrix}0&1\\0&i-1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} = 0\begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}0\\0\end{pmatrix}$$
On the other hand, $$\begin{pmatrix}0&1\\0&i-1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}y\\(i-1)y\end{pmatrix}$$
This implies $$y=0$$. So, an eigenvector is given by $$\alpha\begin{pmatrix}1\\0\end{pmatrix}$$ with $$\alpha\in\mathbb{R}^*$$.