Prove the linear transformation that takes all linear maps $T: V\to W$ to their respective matrix representations is an isomorphism.

Question

Let $V$, $W$ be finite dimensional vector spaces. Prove the linear transformation that takes all linear maps $T: V\to W$ to their respective matrix representations is an isomorphism. Thanks in advance!

I know it has to be onto and one-to-one, but I don't know how to explicitly show it is both injective and surjective.

Answer 1

Showing that this map is injective means showing that if two transformations have the same matrix representation, they must be the same.

Showing that this map is surjective means showing that any matrix has an associated matrix representation.

You can do both in one move as follows: if you have a matrix, show that there is one and only one linear transformation which that matrix represents. Namely, show (probably using the definition of a "matrix representation") the matrix $A$ can only represent the transformation given by $$ T(x) = Ax $$

Answer 2

If you have a basis $B_1 :v_1, v_2, ... v_n$ of $V$ and basis $B_2 : w_1, w_2, ... w_m$ of $W$, a linear transformation $f : V \rightarrow W$ is determined entirely by the values of $f(v_1), f(v_2), ... f(v_n)$ in $W$. So if $g$ and $h$ are different linear maps from $V$ to $W$, then $g(v_i) \neq h(v_i)$, for some $1 \leq i \leq n$. Therefore the coordinates of $g(v_i)$ and $h(v_i)$ in $B_2$ are different, hence the $i^{th}$ column in the corresponding matrices are different. That shows that the map is one-to-one. The 'onto' part is even easier, because for every $a_1, a_2, ... a_n$ vectors of $W$ there is linear map $t$, such that $t(v_i) = a_i$ and every matrix $m$ x $n$ defines such $n$ vectors.