Help Showing that the Adjoint Operator $T^*$ is Surjective if and only if $T$ is Injective

Question

Let $T\in L(V,W)$,where $L(V,W)$ denotes a linear map from a vector space $V$ to vector space $W$. I want to prove that $T$ is injective iff $T^*$ is surjective, where $T^*$ is the adjoint of $T$. I start with the definition of adjoint: $\langle w,Tv \rangle= \langle T^*w,v \rangle$ for all $w \in W $, $v\in V$. What should I do next? Take $v=0$?

Answer 1

Let us note for the sake of clarity that given $T:V\to W$, $T^*:W\to V$, we have $\text{im}(T^*)\subset V$ and $\text{ker}(T)\subset V$. Now, $$ \text{im}(T^*)^\perp=\{x\in V: \langle x,T^*y\rangle=0,\forall y\in W\}=\{x\in V:\langle Tx,y\rangle=0, \forall y\in W\}$$ $$ =\{x\in V: Tx=0\}=\text{ker}(T).$$ Therefore, we have that $$ \text{im}(T^*)^\perp=\text{ker}(T)=\{0\}.$$ This implies that
$$ \text{im}(T^*)=(\text{im}(T^*)^\perp)^\perp=\{0\}^\perp=V.$$ So, we observe that $T^*$ is surjective. Suppose, conversely, that $T^*$ is surjective. Then, we see that $\text{im}(T^*)=V$. From before, we know that $\text{im}(T^*)^\perp=\text{ker}(T)$. This implies that $$ V=(\text{im}(T^*)^\perp)^\perp=\text{ker}(T)^\perp.$$ Thus, we see that $\{0\}=\text{ker}(T)$, so that $T$ is injective. Do be sure that you can justify each of these steps.