linear endomorphism between $V$ and dual of $V$ Let $V$ be a finite dimension vector space over field $K$. $V^*=\{l:V\to K\}$.
Prove $\operatorname{End}(V)$ linear isomorphic to $\operatorname{End}(V^*)$.
My attempt:Since for finite dimension vector space $\dim V^*=\dim V$
so they are linearly isomorphic by $\psi:V\to V^*$.
So given element  $T\in \operatorname{End}(V)$ we can find $\hat{T} = \psi T\psi^{-1}$ is easy to check it's a linear endomorphism.
And the map is onto since for any $\hat{T}$ we can construct $T=\psi^{-1}\hat{T}\psi \in \operatorname{End}(V)$. It's injective since $\hat{T} = 0$ implies $T = 0$ is the zero map, so it has trivial kernel.
Finally we need to show $\phi:\operatorname{End}(V) \to \operatorname{End}(V^*)$ is also linear. i.e. $\phi(T+S) = \phi(T)+\phi(S)$ by definition of $\hat{T}$ it holds.
Is my proof correct?

Your proof is correct. However, there is another vector space isomorphism between $\operatorname{End}(V)$ and $\operatorname{End}(V^*)$ that doesn't require an isomorphism $V \rightarrow V^*$. Namely, map $A \in \operatorname{End}(V)$ to $A^* \in \operatorname{End}(V^*)$ by defining $(A^*\phi)(x) = \phi(Ax)$. Here, $ x\in V$ and $\phi \in V^*$.
