Prove that the map that takes $\mathcal{L}(V,W)$ to $\mathcal{L}(W',V')$ is an isomorphism of $\mathcal{L}(V,W)$ onto $\mathcal{L}(W',V')$

Question

I was reading Axler's Linear Algebra Done Right, and the following question appears as exercise $16$ of chapter $3$, section F:

Suppose V and W are finite-dimensional.

Prove that the map that takes $\mathcal{L}(V,W)$ to $\mathcal{L}(W',V')$ is an isomorphism of $\mathcal{L}(V,W)$ onto $\mathcal{L}(W',V')$.

My attempt to prove that goes as follows:

Let $\Gamma : \mathcal{L}(V,W) \rightarrow \mathcal{L}(W',V')$ defined by: $$\Gamma (T) = T'$$

Now since V and W are finite-dimensional we conclude that $\mathcal{L}(V,W)$ also finite-dimensional, therefore there is a basis: $$T_1, \cdots, T_p \quad : p = dim \space \mathcal{L}(V, W)$$ Now for any $T \in \mathcal{L}(V, W)$ we can write: $$T = a_1T_1 + \cdots + a_pT_p$$ $$\implies \Gamma(T) = a_1\Gamma(T_1) + \cdots + a_p\Gamma(T_p)$$ $$\implies T' = a_1T_1' + \cdots + a_pT_p'$$

Since $dim \space \mathcal{L}(V, W) = dim \space \mathcal{L}(W', V')$, then the list: $$T_1', \cdots, T_p'$$ spans $\mathcal{L}(W', V')$ (I assume that because each $T'$ is related to one $T$ because $T'(\phi) = \phi T$), and has the right length so it's a basis for $\mathcal{L}(W', V')$, since $\Gamma$ maps basis to basis it's an isomorphism.

Is that correct?

Answer 1

Okay, given $T \in \mathcal L(V,W)$, $T'$ can be written as a linear combination of $T_1',\dots,T_p'$; but that just prove that $T_1',\dots,T_p'$ spans $\operatorname{im} \Gamma$! That is, how do you know that any element of $\mathcal L(W',V')$ can be written as a linear combination of $T_1',\dots,T_p'$?

Hint: Choose ordered bases $\alpha$ and $\beta$ for $V$ and $W$, respectively. Then, if $\alpha'$ and $\beta'$ are they respective dual bases, consider the isomorphisms $$\mathcal M : \mathcal L(V,W) \to \mathbf F^{m,n} \quad \& \quad \mathcal M' : \mathcal L(W',V') \to \mathbf F^{n,m}$$ (here $n := \dim V$ and $m := \dim W$) given by $$\mathcal M(T) := \mathcal M(T,\alpha,\beta) \quad \& \quad \mathcal M'(S) := \mathcal M(S,\beta',\alpha').$$ Finally, prove that $\Gamma = (\mathcal M')^{-1} \circ ()^{\rm t} \circ \mathcal M$, where $()^{\rm t} : \mathbf F^{m,n} \to \mathbf F^{n,m}$ sends a matrix to its transpose.