HOW TO SHOW $S:L(V,W) \rightarrow W, T \rightarrow T(v)$ is linear if it's a finite-dimensional vector space over $F$

Question

Let $V,W$ be finite-dimensional vector spaces over a field $F$, and $L(V,W)$ the vector space of all linear maps $T:V\rightarrow W.$

a) Show that for every given $v \in V,$ the map

$S:L(V,W) \rightarrow W, T\rightarrow T(v)$ is linear.

b) Find $\ker(S)$. You may use the fact that $\dim(L(V,W))=\dim(V)\dim(W).$

To prove a function is linear I have to show $T(v + v') = T(v) + T(v')$ and $T(\alpha v) = \alpha T(v)$. But its just the Definition,how do I apply in this question if I just know its a finite dimensional, and how do I know the $\ker(S)$?

Answer 1

By definition of addition in $L(V,W)$, $(T+T')(v)=T(v)+T'(v)$ and likewise (\alpha T)(v)=\alpha T(v)$..

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If $v=0$ then $S=0$, so $\ker S=L(V,W)$. In all other cases, $S$ is clearly onto (why?), hence $\dim\ker S=\dim L(V,W)-\dim W.

Answer 2

i)For infinity linear map $f : V \rightarrow W$, we have $$f(\sum_{i=1}^na_iv_i=a_i\sum_{i=1}^nv_i)$$ ii)If $\{v_1, . . . , v_k\}$ is a basis for a finite dimensional vector space, then every element of $V$ is a linear combination of these $v_i$ , say $v =\sum_{i=1}^na_iv_i$. Therefore from (i) we get $$f(v)=\sum_{i=1}^na_if(v_i)).$$ Thus, it follows that $f$ is completely determined by its value on a basis of $V$. This simply means that if $f$ and $g$ are two linear maps such that $f(v_i) = g(v_i)$ for all elements of a basis for $V$, then $f = g$. We are free to choose the value of $f$ on the elements of a given basis, as far as they all belong to the same vector space $W$. Thus if $V$ has a basis consisting of $k$ elements as above then for each ordered $n$ tuple of elements of $W$ we obtain unique linear transformation $f : V \rightarrow W$ and vise versa.