[Math] Show that the dimension of the set of all linear transformations from $V$ to $W$ has dimension equal to $mn$.

linear algebralinear-transformations

Let $V, W$ be vector spaces. The set $L(V,W)$ is the set of all linear transformations $T: V \to W$, with the operations:

$(T_1 + T_2)(v) = T_1(v)+T_2(v)$

$(\alpha T)(v) = \alpha T(v)$

for $v\in V$.

Prove that the dimension of the set $L(V,W)$ is $\mathrm{dim}(L(V,W)) = \mathrm{dim}(V) \cdot \mathrm{dim}(W)$

I know I have to (or can) use the bases of the vector spaces to prove this but I cannot come up with a solution.

Best Answer

A longer hint:

Pick a basis $\{e_i\}$ of V and a basis $\{f_j\}$ of W. Consider the set of linear transformations $S=\{T_{mn}\}$ where we define $T_{mn}$ as $$T_{mn}(\sum a_ie_i) =a_mf_n$$ In other words, $T_{mn}$ maps $e_m$ to $f_n$ and maps the other basis vectors of V to 0 (in W), and then we extend this over all of V while keeping $T_{mn}$ linear.

Can you show that the transformations in S are independent i.e. you cannot make one of them by adding together multiples of the others ?

Can you see how to add together multiples of transformations in S to create a transformation that maps $e_m$ in V to any given vector w in W ?

Now note that a linear transformation in T is defined by its action on each of the basis vectors in V. Can you see how to add together multiples of transformations in S to create any given linear transformation in T ?

Once you have shown that the transformations in S are independent and that they cover all of T, then you know that S is a basis of T. Count the size of S to find the dimension of T

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