Help Showing that the Adjoint Operator $T^*$ is Surjective if and only if $T$ is Injective
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 $\mathcal{R}(T^*)$)