Solution to Problem 11.20 Schaum's Outline of Linear Algebra (4th Edition) Problem: Suppose $u,v\in V$ and that $\phi(u)=0$ implies that $\phi(v)=0$ for all $\phi\in V^{*}$, which is the dual of $V$ then show that $v=ku$ for some scalar $k.$
My Attempt:  I tried doing this by contrapositive. Suppose $v\not =ku$ then we can say that for every scalar $k$ there exists a non-zero vector $b$ such that $v=ku+b.$ In this case we get that $\phi(v)=\phi(b).$ Now if we set $\phi(b)=t,$ where $t\not =0$ then $\phi(v)\not =0$ Thus we have found a linear functional $\phi$ such that $\phi(u)=0$ but $\phi(v)\not =0.$
PS: The way I negated the first claim was as follows: If $A:=\forall \phi \in V^{*}(\phi(u)=0\implies \phi(v)=0)$ then $\neg A:=\exists\phi\in V^{*}(\phi(u)=0\land\phi(v)\not =0).$
Is this proof correct?
 A: No, is incorrect, because is false that $\phi(v)=\phi(b)$ for all $\phi$. The hypothesis is if $\phi(u)=0\to \phi(v)=0$ (1)
A hint: if $v\neq ku$, then $\{u,v\}$ is l.i. set and you can extend it to a basis ($V$ has finite dimention?) $B=\{u,v,x_1,...,x_s\}$ and construct a dual basis $B^*=\{\phi_1,...,\phi_{s+2}\}$. 
Now, $\phi_2$ satisfies $\phi_2(u)=0$ but $\phi_2(v)=1$.
Edit: (1) you must construct a $\phi$ such that $\phi(u)=0$.
A: Martin's answer does a fine job of pointing out the flaw in your proof.
Here's a proof that works without assuming that $V$ is finite dimensional (and without constructing a basis).
The proof is trivial in the case where $u = 0$ (this proof still requires care, though).  So, suppose that $u \neq 0$. Consider the map $\Psi: V^* \to \Bbb F^2$ given by
$$
\Psi(\phi) = (\phi(u),\phi(v))
$$
Since $u \neq 0$, there exists a $\phi$ such that $\phi(u) \neq 0$.  So, the image of $\Psi$ has dimension at least $1$.  However, the vector $(0,1)$ is not in the image of $\Psi$, which means that the image of $\Psi$ is of dimension exactly $1$.
That is, there exists a vector $(k_1,k_2) \in \Bbb F^2$ such that for all $\phi \in V^*$, there is an $\alpha \in \Bbb F$ such that
$$
(\phi(u),\phi(v)) = \alpha(k_1,k_2)
$$
In other words: for all $\phi \in V^*$, we have
$$
k_2 \phi(u) = k_1\phi(v)
$$
(it may help to note that this can be rearranged to $\frac{\phi(v)}{\phi(u)} = \frac{k_2}{k_1}$ in the case where $\phi(u) \neq 0$).  We can rearrange the above to state that for all $\phi \in V^*$, we have
$$
\phi(k_2 u - k_1 v) = 0
$$
However, this must mean that $k_2 u - k_1 v = 0$.  That is, the vectors $u$ and $v$ are linearly dependent, which was the desired conclusion.
