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Question: $V$ is a vector space over $F$, Let $ T\colon V \to V$

$U=\{S \in \operatorname{Hom}(V,V): S \circ T=0\}$. $\dim V=2$. If $\dim\operatorname{Im} T=1$ what is $\dim U$?

Thoughts: Tried to translate it somehow to a product of matrices, but didn't really know how to use the fact about the rank (how to build the matrices in a way I'll be able to figure the dimension of $U$.

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Hint: Note that $U$ can be identified with the space of linear functions from the $1$-dimensional space $V/\operatorname{Im} T$ to $V$.

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Thanks. But why can't S get vectors that are not in ImT as well? – user1685224 Jul 6 '13 at 13:50

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