$L(V)$ isomorphism $L(V*)$.

Question

Let be $V$ a vector space with finite dimension over F. Show that $\mathscr{L}(V^{*})\cong \mathscr{L}(V)$.

My attemp was try to show that $\mathscr{L}(V^{*})$ is isomorphic with the space $\matrix{M^{n×n}(F)}$, but i don't know how to start. I know that $Dim(V)$ = $Dim(V^{*})$, but it is sufficient to use that?

Sorry for my english

Answer 1

From the linear algebra book we learn that if $V$ is a finite $n$-dimensional vector space over $F$, then $\mathscr L(V)$ is isomorphic to $M_n(F)$. Since both $V$ and $V^*$ have the same dimension, say $n$, we have $\mathscr L(V)\cong M_n(F) \cong \mathscr L(V^*)$.

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EDIT: As a reply to the comment of @Rodrigo Palacios

The isomorphic relation is transitive. Actually it is an equivalent relation. Suppose $\phi:U\cong V$, $\psi:V \cong W$ are the bijections preserving addition and scalar multiplication. Then we can examine that $\phi\circ\psi:U\to W$ is also a bijection preserving the two operations. This implies $U\cong W$.