Conjecture: A linear transformation is onto if and only if its adjoint is 1-1 Suppose $V$ and $W$ are inner product spaces and $T:V\rightarrow W$ is a linear mapping between them. Is $T$ onto if and only if $T^*$ is 1-1? 
This is either a basic fact or a total fantasy. I can find support for the "if" but not the "only if", and I am curious about showing that $T$ cannot be onto if $T^*$ has nontrivial kernel. 
A proof for the matrix case is good enough for my purposes. 
 A: One has the following result for normed spaces.

If $T:X\to Y$ is a bounded linear operator between normed spaces, then $T^*$ is injective iff the range of $T$ is dense in $Y$. (Recall that $T^*:Y^*\to X^*$ is defined by $T^*(f)=f\circ T$.)

Proof:
For the forward direction, suppose by contraposition that $y\in Y\setminus\overline{TX}$.
Use the Hahn-Banach theorem to find $f\in Y^*$ such that $f(y)\ne0$ and $f|_{\overline{TX}}=0$.
Then $f$ is a nonzero element of the kernel of $T^*$ because $(T^*f)(x)=f(Tx)=0$ for every $x\in X$. Consequently $T^*$ is not injective.
Conversely suppose that $TX$ is dense in $Y$ and that $f,g\in Y^*$ such that $T^*f=T^*g$.
Given $y\in Y$, take a sequence $(x_n)$ in $X$ such that $Tx_n\to y$.
Then, by continuity we obtain
$$
f(y) = \lim f(Tx_n) = \lim (T^*f)(x_n) = \lim (T^*g)(x_n) = \lim g(Tx_n) = g(y).
$$
Therefore $f=g$, so $T^*$ is injective. $\square$
Thus if there is some way to guarantee that the range of $T$ is closed (say $Y$ is finite dimensional), then we could say that $T$ is onto iff $T^*$ is injective. Since this formulation is a little more general than what you are asking, one might wonder if things can be improved if we have an inner product. The answer turns out to be no.
Consider $T:\ell_2\to\ell_2$ defined by $(Tx)(n)=x(n)/\sqrt{n}$. This is an injective bounded linear operator. It is not onto because no element of $\ell_2$ maps to the element $y\in\ell_2$ given by $y(n)=1/n$. Indeed, if we had $Tx=y$, then we would have $x(n)=1/\sqrt{n}$ - which is not square summable.
However its adjoint is injective.
To see this, note that $T^*=T$ because
\begin{align*}
\langle Tx,y\rangle
&= \sum_{n=1}^\infty (Tx)(n)\overline{y(n)}
= \sum_{n=1}^\infty \frac{x(n)}{\sqrt n}\overline{y(n)}
= \sum_{n=1}^\infty x(n)\overline{\left(\frac{y(n)}{\sqrt n}\right)} \\
&= \sum_{n=1}^\infty x(n)\overline{(Ty)(n)}
= \langle x,Ty \rangle
\end{align*}
holds for every $x,y\in\ell_2$, and $T$ is injective.
A: The relevant equality is 
$$\tag{1}
\ker T^*=(\text{ran}\,T)^\perp
$$
Proof. 
Let $v\in\ker T^*$. Then $$ \langle v,Tw\rangle=\langle T^*v,w\rangle=0,$$which shows that $\ker T^*\subset (\text{ran}\,T)^\perp$. Conversely, if $v\in (\text{ran}\, T)^\perp$, then for any $w$ we have 
$$
\langle T^*v,w\rangle=\langle v,Tw\rangle=0. 
$$
As $w$ is arbitrary, we conclude that $T^*v=0$, so $(\text{ran}\,T)^\perp\subset \ker T^*$, establishing $(1)$. 
If we now consider the orthogonal version of $(1)$, we get
$$\tag{2}
(\ker T^*)^\perp=\overline{\text{ran}\,T}.
$$
This shows that, in the finite-dimensional case, indeed $T^*$ is 1-1 if and only if $T$ is onto. In the general case you can only say that $T$ has dense range. 
