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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. $$ \pmatrix{0&1\\0&i-1}\pmatrix{x\\y} = \pmatrix{0\\0}.$$

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

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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 $a = 1$.

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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 $\alpha\in\mathbb{R}^*$.

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