Linear Maps, Null Space Suppose $S$ and $T$ are linear maps from $V$ to $\Bbb F$ that have the same null space. Show that there exists a constant $c\in\Bbb F$ such that $S = cT.$
 A: Here is a "simple solution" which just uses the definition of null space.
Case 1) If $T=0$, then $S=0$ (because $S$ and $T$ have the same null space) and thus $S=cT$ for any $c\in F$.
Case 2) If $T\neq 0$, then there exists $x_0\in V$ such that $T(x_0)\neq 0$. Furthermore, for all $x\in V$,
$$\begin{align*}
0&=T(x_0)T(x)-T(x_0)T(x)\\
&=T(T(x_0)x-T(x)x_0)&&\text{(linearity of $T$)}\\
&=S(T(x_0)x-T(x)x_0)&&\text{($S$ and $T$ have the same null space)}\\
&=T(x_0)S(x)-T(x)S(x_0)&&\text{(linearity of $S$)}.
\end{align*}$$
It follows that
$$S(x)=\frac{S(x_0)}{T(x_0)}T(x)\quad\forall \ x\in V$$
and thus $S=cT$ where $c=S(x_0)/T(x_0)$.
A: Since $S,T$ are linear functionals, we know that their null spaces have codimension $1$. This means that they are only nonzero on the orthogonal complement of their null spaces and that the orthogonal complement has dimension $1$.
Denote $U$ to be the orthogonal complement of the null space in $V$. Let $x$ be a basis element for $U$. Then $Sx = a$ and $Tx = b$ for some $a,b\in \Bbb F$ nonzero. Or equivalently, $Tx = \frac{b}{a}Sx$.
Do you see how to proceed?
If $V$ is not necessarily a Hilbert space, then you can adjust the terminology so that $U = V/\text{null}(S)$ and proceed from there.
A: Assuming $V$ is finite dimensional with dimension $n$, and that $S$ and $T$ are non-zero:
Let $\{v_1,v_2,\dots,v_{n-1}\}$ be a basis for the common nullspace of $S$ and $T$ (how do we know the dimension of this subspace?).  Extend this basis by one element so that $\{v_1,v_2,\dots,v_{n-1},v_n\}$ is a basis of $V$.  Any vector $v \in V$ can be written in the form $v = \sum_{i=1}^n a_i v_i$.   We note that
$$
S\left(\sum_{i=1}^n a_i v_i \right) = \sum_{i=1}^n a_i S(v_i) = a_n S(v_n)
$$
Similarly, $T(v) = a_n T(v_n)$.  
Conclude that $S(v) = \frac{S(v_n)}{T(v_n)}T(v)$ for all $v \in V$.
The proof is similar for the infinite dimensional case, but we no longer use "$n$" in our indexing, and rely on the notion of codimension rather than the rank-nullity theorem, hence Cameron's more generalized answer.
