Help Showing that the Adjoint Operator $T^*$ is Surjective if and only if $T$ is Injective 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 $v=0$?
 A: Suppose $T^*$ exists and is surjective. 
Let $\mathbf{v} \in \ker(T)$. Since $T^*$ is surjective, there exists $\mathbf{w} \in W$ such that $T^*(\mathbf{w}) = \mathbf{v}$. Then
$$\langle \mathbf{v}, \mathbf{v} \rangle = 
\langle \mathbf{v}, T^*(\mathbf{w}) \rangle = 
\langle T(\mathbf{v}), \mathbf{w} \rangle = 
\langle \mathbf{0}_W, \mathbf{w} \rangle = 0$$
This only happens when $\mathbf{v} = \mathbf{0}_V$. Hence, $\ker(T) = \{\mathbf{0}_V\}$ and $T$ is injective.
The converse is not true in general, here is a counter example. Let $V = W = l^2$, the space of square-summable sequences. Define a linear operator $T$ such that
$$T((a_n)) = (a_n - a_{n+1}),$$ for all $(a_n) \in V$. (Identity minus left-shift)


*

*$T$ is injective. (The only constant, square-summable sequence is the zero-sequence.)

*$T^*$ exists (Identity minus right-shift)

*$T^*$ is not surjective. (The sequence $(1,0,0,\ldots) \in V$ is not in $\mathcal{R}(T^*)$)

A: 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.
