Exercise 7, Section 3.3 of Hoffman’s Linear Algebra 
Let $V$ and $W$ be vector spaces over the field $F$ and let $U$ be an isomorphism of $V$ onto $W$. Prove that $T \to UTU^{−1}$ is an isomorphism of $L(V, V)$ onto $L(W, W)$.

My attempt: $f:L(V,V)\to L(W,W)$ is defined by $f(T)=U\circ T\circ U^{-1}$. Let $T,L\in L(V,V)$. Then $f(T+L)$ $=U\circ (T+L)\circ U^{-1}$. We need to show $U\circ (T+L)\circ U^{-1}$ $= U\circ T\circ U^{-1} + U\circ L\circ U^{-1}$. Let $y\in W$. So $U\circ (T+L)\circ U^{-1}(y)$ $= U[ (T+L)( U^{-1}(y))]$ $=U[T(U^{-1}(y))+L(U^{-1}(y))]$. Since $U$ is a linear map, we have $U[T(U^{-1}(y))+L(U^{-1}(y))]$ $=U(T(U^{-1}(y)))+U(L(U^{-1}(y)))$ $=U\circ T\circ U^{-1}(y) +U\circ L\circ U^{-1}(y)$ $= (U\circ T\circ U^{-1} + U\circ L\circ U^{-1})(y)$. Hence $U\circ (T+L)\circ U^{-1}$ $= U\circ T\circ U^{-1} + U\circ L\circ U^{-1}$ $=f(T+L)$ $=f(T)+f(L)$.
Let $c\in F$ and $T\in L(V,V)$. Then $f(c\cdot T)=U\circ (c\cdot T)\circ U^{-1}$. We need to show $U\circ (c\cdot T)\circ U^{-1}$ $=c\cdot (U\circ T\circ U^{-1})$. Let $y\in W$. So $U\circ (c\cdot T)\circ U^{-1}(y)$ $=U[(c\cdot T)(U^{-1}(y))]$ $=U(c\cdot T(U^{-1}(x)))$. Since $U$ is linear map, we have $U(c\cdot T(U^{-1}(x))$ $=c\cdot U(T(U^{-1}(y)))$ $=c\cdot (U\circ T\circ U^{-1})(y)$. Hence $U\circ (cT)\circ U^{-1}$ $=c(U\circ T\circ U^{-1})$ $=f(cT)$ $=c\cdot f(T)$. I didn’t make distinction for addition and scalar multiplication for different vector spaces, because there are total $4$ vector space: $V$, $W$, $L(V,W)$ and $L(W,W)$.
There are two ways to show $f$ is bijective, $(1)$ $f$ is injective and surjective, $(2)$ $f$ is invertible, i.e. $\exists g:L(W,W)\to L(V,V)$ such that $g\circ f=\text{id}_{L(V,V)}$ and $f\circ g=\text{id}_{L(W,W)}$.
Approach(1): If $f(T)=f(L)$, for some $T,L\in L(V,V)$. Then $U\circ T\circ U^{-1}$ $=U\circ L\circ U^{-1}$. We need to show $T=L$, i.e. $T(x)=L(x)$, $\forall x\in V$. Let $x\in V$. Since $U:V\to W$ is invertible, we have $\exists !$$y\in W$ such that $U^{-1}(y)=x$. So $U(T(U^{-1}(y)))$ $=U(L(U^{-1}(y)))$ $=U(T(x))$ $=U(L(x))$. Since $U$ is injective, we have $T(x)=L(x)$. Hence $T=L$. Another way to show $f$ is injective is $N_{L(V,V)}$ $=\{T\in L(V,V)| f(T)=0\}$ $=\{0\}$. Again I didn’t make any distinction for zero vector for different vector spaces. If $f(T)=0$, for some $T\in L(V,V)$. Then $U\circ T\circ U^{-1}=0$. We need to show $T=0$, i.e. $T(x)=0$, $\forall x\in V$. So $U(T(U^{-1}(W)))=\{0\}$. Since $U:V\to W$ is invertible, we have $U^{-1}(W)=V$. So $U(T(U^{-1}(W)))$ $=U(T(V))$ $=\{0\}$. Since $U$ is injective, i.e. $N_U=\{0\}$, we have $T(V)=\{0\}$. Thus $T=0$. Hence $T$ is injective.
Let $Z\in L(W,W)$. Take $T=U^{-1}\circ Z\circ U \in L(V,V)$. Then $f(T)=U\circ (U^{-1}\circ Z\circ U)\circ U^{-1}$. It’s easy to check, composition is associative. so $U\circ (U^{-1}\circ Z\circ U)\circ U^{-1}$ $= (U\circ U^{-1})\circ Z\circ (U\circ U^{-1})$ $=\text{id}_W \circ Z\circ \text{id}_W$ $= \text{id}_W\circ Z$ $=Z$. Thus $\exists T\in L(V,V)$ such that $f(T)=Z$. Hence $f$ is surjective. So $f$ is bijective.
Approach(2): We show $f$ is invertible, i.e. $\exists g:L(W,W)\to L(V,V)$ such that $f\circ g=\text{id}_{L(W,W)}$ and $g\circ f=\text{id}_{L(V,V)}$. Define $g:L(W,W)\to L(V,V)$ such that $g(Z)=U^{-1}\circ Z\circ U$. It’s easy to check $U^{-1}\circ Z\circ U:V\to V$. By theorem 6 section 3.2, $U^{-1}\circ Z\circ U$ is a linear map. So $g$ is a well defined map. Let $Z\in L(W,W)$. Then $f\circ g(Z)$ $=f(g(Z))$ $=f(U^{-1}\circ Z\circ U)$ $=Z$. Hence $f\circ g=\text{id}_{L(W,W)}$. Let $T\in L(V,V)$. Then $g\circ f(T)$ $=g(f(T))$ $=g(U\circ T\circ U^{-1})$ $=T$. Hence $g\circ f=\text{id}_{L(V,V)}$. So $f$ is invertible, i.e. bijective. Is my proof correct?
Question: In approach(2), it seems like we didn’t use full strength of $V\cong W$, mainly $U$ is bijective. Similarly in proof (subjective part) of approach(1), it seems like we didn’t use $U$ is bijective. In proof (injective part) of approach(1), we explicitly use $U$ is bijective. Before writing proof, I was thinking, I have to use injective condition of $U$ to show $f$ is injective and surjective condition of $U$ to show $f$ is surjective. But that is not the case, I think.
 A: $V, W$ : two finite dimensional vector spaces over $F$ .
$T\in \mathcal{L}(V) $ and $U\in\mathcal{L}(W) $
Given $U\in \mathcal{L}(V, W) $ is an isomorphism i.e an invertible linear map. Then it follows that $V\cong W$
Hence $\mathcal{L}(V) $ and $\mathcal{L}(W)$ are same dimensional vector spaces over $F$.
Let us consider $\varphi:\mathcal{L}(V)\to \mathcal{L}(W)$ defined by$$ \varphi(T )=UTU^{−1}$$
Then
$\varphi$ is a linear map.
$\begin{align}\ker \varphi &=\{T\in\mathcal{L}(V): \varphi(T)=\textbf{0}\}\\&=\{   T\in\mathcal{L}(V) : UTU^{-1}=\textbf{0} \}\\&= \{\textbf{0}\} \end{align}$
A linear map between two same finite dimensional linear spaces is injective iff surjective iff invertible.
Hence $\varphi$ is an isomorphism between
$\mathcal{L}(V) $ and $\mathcal{L}(W)$

Note:

*

*Suppose  $V, W$ are not finite dimensional .

We use finite dimensionality of $V, W$ to prove $\varphi$ is injective iff surjective iff bijective.
In general we need to show $\phi$ is injective as well as surjective.
Let us take $S\in \mathcal{L}(W) $ .
Then we need to find a $T\in \mathcal{L}(V) $ such that $\varphi(T) =S$
i.e $UTU^{-1}=S$ implies $T=U^{-1}SU$
$U^{-1}SU$ is the required preimage as $\varphi(U^{-1}SU) = UU^{-1}SUU^{-1}=S$


*First half of the proof is enough to conclude $\varphi$ is an isomorphism even when vector spaces are infinite dimensional.Because the proof injective iff surjective depends on Rank-Nullity theorem and we can also restore Rank-Nullity theorem for infinite dimensional vector spaces upto cardinality.(interesting!)

