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

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?

• Are there some additional restrictions on $\Gamma$? How do we know $\Gamma(T_1) \neq \Gamma (T_2)$? Commented Apr 15, 2021 at 19:01

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.
• Isn't every element $T' \ in \mathcal{L}(W', V')$ defined by $T'\phi = \phi T$? So I assumed in my proof that this relation is the reason for that. Commented Apr 16, 2021 at 14:27
• @KaisHasan There is no reason to assume that every element of $\mathcal L(W',V')$ is $T'$ for some $T \in \mathcal L(V,W)$, in other words, there is no reason to assume that $\Gamma$ is surjective. Commented Apr 16, 2021 at 19:01
• Can you explain that in more detail? why is $\Gamma$ not surjective if $T'$ defined in terms of $T$? How can $T'$ be such that no $T$ satisfying $\Gamma T = T'$ exist? Commented Apr 17, 2021 at 18:47
• What I mean is that, given $S \in \mathcal L(W',V')$, can you find $T \in \mathcal L(V,W)$ such that $S = T'$? Commented Apr 17, 2021 at 19:12