proof that $\rho,\psi \in V^{*}$ s.t. $\rho(v)\psi(v) = 0, \forall v \in V,$ implies $\rho = 0$ or $\psi = 0$ I'm in the middle of revising for an exam, an this question came up on an old midterm that the professor gave us as an example of what to expect.
My attempt to show that; given $\rho,\psi \in V^{*} \textrm{ s.t. } \rho(\textbf{v})\psi(\textbf{v}) = 0, \forall \textbf{v} \in V\Rightarrow \rho = 0 \textrm{ or } \psi = 0.$ is as follows:
First, note that it is logically equivalent to prove that given
$\rho,\psi \in V^{*} \textrm{ s.t. } \rho(\textbf{v})\psi(\textbf{v}) = 0, \forall \textbf{v} \in V$ and $\rho(\textbf{v}) \neq 0 \Rightarrow \psi = 0$.
So with this in mind we let $\textbf{x} \in V$, and observe that $\rho(\textbf{x}),\psi(\textbf{x}) \in F$, where $F$ is a field, since $\rho: V \rightarrow F$ and $\psi: V \rightarrow F$.  Thus, since $\rho(\textbf{x}) \neq 0$, we know that $\exists \big(\rho(\textbf{x})\big)^{-1} \in F$ s.t. $\rho(\textbf{x})\big(\rho(\textbf{x})\big)^{-1} = \big(\rho(\textbf{x})\big)^{-1}\rho(\textbf{x}) = 1_{F}.$  Hence: 
\begin{align}
0 &= \big(\rho(\textbf{x})\big)^{-1}0 = \big(\rho(\textbf{x})\big)^{-1}\big(\rho(\textbf{x})\psi(\textbf{x})\big) \\
&= \Big(\big(\rho(\textbf{x})\big)^{-1}\rho(\textbf{x})\Big)\psi(\textbf{x}) \\
&= (1_{F})\psi(\textbf{x}) \\
&= \psi(\textbf{x}).
\end{align}
It follows that $\psi: V \rightarrow \{0\} \hspace{4.5in}\square$ 
I explained my solution to the instructor and they said that it doesn't work, but didn't have time to explain why.  Help?!
 A: The argument you give does nothing: you are showing that if $\rho(x)\ne0$, then $\psi(x)=0$; that's an obvious conclusion from $\rho(x)\psi(x)=0$. Note in particular that your argument doesn't use linearity at all. 
Here is one way of doing it. Let $x,y\in V$. Then, since $\rho(x)\psi(x)=\rho(y)\psi(y)=0$,
$$\tag{1}
0=\rho(x+y)\psi(x+y)=\rho(x)\psi(y)+\rho(y)\psi(x).
$$
Suppose that $\psi\ne0$. This means that there exists some $x_0\in V$ with $\psi(x_0)\ne0$. Note that by hypothesis, $\rho(x_0)=0$. Then, from $(1)$, we get
$$
\rho(y)=-\frac{\rho(x_0)}{\psi(x_0)}\,\psi(y)=0.
$$
As this can be done for arbitrary $y$, we get that $\rho(y)=0$ for all $y\in V$, i.e. $\rho=0$.
A: Your logic is not very clear. The use of $\exists$ and $\forall$ is not precise in what you write. Anyway if $Z=\ker \rho$ and $\rho$ is non-zero then there is $u_0$ so that $\rho(u_0)=1$. Then for every $z\in Z$ the condition on the product means that:
 $$ 0= \psi(u_0+z) \rho(u_0+z) = \psi(u_0+z)(1+0)=\psi(u_0+z)$$
Now use linearity to show that $\psi\equiv 0$.
