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Let $ f, g: V \to W . $ Proof, that the set of vectors from V where linear transformation f and g are equal, forms a vector subspace.

Well, f and g are equal if both of those linear transormations are equal to zero, so that set is in kernel which is the subspace itself by definition.

Linear transformations must satisfy: $ f(u+v) = f(u) + f(v)$ and $f(ru) = rf(u) $

where u,v are vectors from V and r is scalar.

Vector subspace satisfies: $ {u}\in V, v\in V $, then $ {{(u + v)}\in} V$ and for $ r\in R$, $ ru \in V $

Not sure how to prove it.

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  • $\begingroup$ Are you asking why $\ker(f-g)$ is a subspace of $V$ ? If so, the answer is yes : the kernel of any linear map $\varphi:X\to Y$ is a subspace of $X$. $\endgroup$ – Adren Jan 28 '17 at 15:54
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Consider the linear transformation $f-g$. The vectors $v\in V$ map to the same $w\in W$ for both $f$ and $g$ iff $(f-g)(v)=0$. Thus, the set of vectors you want to show is a subspace is in fact the kernel of $f-g$, hence a subspace.

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