# dual of linear operator space

Let $U$ and $V$ be finite dimensional vector spaces.

Let $L(U,V)$ be the space of linear maps $U\rightarrow V$.

Let $A \in L(U,V)^*$, i.e., the dual space of $L(U,V)$.

Is it true that $A : U^* \rightarrow V^*$?

Edit: (from the comments)

$L(U,V)^*$ is the dual space of $L(U,V)$.

I want to show that $L(U,V)^*$ is isomorphic to $L(U^*,V^*)$. I see that for each $A\in L(U^*,V^*)$ there is a map $l_A\in L(U,V)^*$ defined by $l_A(B)=\text{trace}(A^*B)$, where $A^*$ is the adjoint of $A$.

Can we show that for each $l\in L(U,V)^*$ there is an $A$ such that $l=l_A$?

• How do you define $L(U, V)^*$? – EPS May 14 '14 at 21:02
• $L(U,V)^*$ is the dual space of $L(U,V)$ – user150113 May 14 '14 at 23:39
• If by dual space of $L(U, V)$ you mean the set of (say) real linear functions from $L(U, V)$ to $\mathbb{R}$. Then of course $A\not \in L(U, V)^*$. As long as you know what the notation means answering this question should be fairly easy. – EPS May 15 '14 at 0:05
• Yes, I meant $L(U,V)^*$ is the set of linear functions from $L(U,V)$ to $\mathbb{R}$. I am wondering if $A \in L(U,V)^*$, what are the domain and codomain of $A$? – user150113 May 15 '14 at 1:12
• The domain of $A$ is $L(U, V)$ and the codomain is $\mathbb{R}$ according to your definition! That is, $A$ eats some function $f\colon U\to V$ and spits out a number $A(f)$ in $\mathbb{R}$. – EPS May 15 '14 at 1:22

## 1 Answer

First note that $L(U, V)^*$ is isomorphic as a vector space to $L(U^*, V^*)$, because of obvious reasons: they have the same dimension.

The answer to your last question is affirmative. The linear map that you have defined between the two spaces is onto because of the Riesz Representation Theorem. You have defined an inner product on $L(U^*, V^*)$ by setting $\langle A, B\rangle=\text{trace}(A^*B)$.