# Vector subspace of two linear transformations

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.

• 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 $X$. – Adren Jan 28 '17 at 15:54

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.