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I'm trying to solve an exercise which specify the subspace $Im(T)$ (the image of a linear transformation $T$) and a vector $v$ which is an eigenvector of the transformation matrix (in the standard basis) but $v\notin Im(T) $.

How is that possible?

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    it could be an eigenvector of eigenvalue 0? does it make sense? – 2017-02-24
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    Yes: only eigenvectors of eigenvalue $0$ may not belong to $Im(T)$. – 2017-02-24

1 Answers 1

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Let $v \neq 0$ be an eigenvector of eigenvalue $\lambda$. If $\lambda \neq 0$, then $$v=T (v/\lambda) \in \mathrm{Im}(T)$$

so you need an example with $ \lambda =0$.

Now, if you consider any direct sum $V= W \oplus U$, the projection $T$ of $V$ over $W$ defined by $$T(w+u)=w$$ any nonzero vector $u \in U$ is mapped to $0$ (hence it is an eigenvector), and obviously does not belong to the image of $T$.