A functional is a multiple of another In halmos book there's a question that says: If $y $ and $z$ are linear functionals on the same vector space and $[x, y]=0$*  whenever $[x, z]=0$ then show that there exists a scalar $\alpha$ such that $y=\alpha z$. 
Then he gives the hint. If $[x_0, z]\ne 0$ then write $\alpha =\frac {[x_0, y]}{[x_0, z]}$
Attempt:
So call the space $V$ and let $U$ be the subspace of $V$ such that for all $x \in U, \; [x, z] =0$. By extension $[x, y]=0$ also. 
We consider only the case that $z$ is a non zero functional as otherwise it's trivial. So there exists an $x_0 \in V\backslash U$ such that $[x_0, z]\ne 0$. Now define alpha as in the hint and consider the functional $w:=y-\alpha z$. We have $[x_0, w]=0$ and furthermore if we call $W$ the space spanned by $U\cup \{x_0\}$ then $[x, w]=0$ for all $x\in W$ and hence $y=\alpha z$ if restricted to this subspace. 
Now I'm stuck. If I could prove that $V=W$ I'd have the result but I feel I'm missing something...
*and by that he means $y (x)$ although he moves to calling it a bracket and says it's a bilinear functional.
 A: I think you could use that a kernel has codimension 1.  Specifically we can write any $x$ in the space as a sum $x_z + c x_0$ where $x_z \in \ker z$, $c$ is a scalar depending on $x$, and $x_0$ is some fixed vector.
In this case, since $x_z \in \ker z$ it also also true that $x_z \in \ker y$.  Thus 
$$(x,y) = (x_c, y) + c(x_0, y) = c (x_0, y)$$
Compare this to the value of $(x,z)$ and you should be able to infer your result.
A: For a vector space $\mathscr{U}$ over a field $\mathscr{F}$, for every $l \in \mathscr{U}'$ ($l$ non-trivial), there exists $u \in \mathscr{U}$ such that $l(u)=1$ (this is actually to be proved -in fact, the previous exercise in Halmos book-). Let $w \in \mathrm{ker}\ l$. Then, any vector $v \in \mathscr{U}$ can be written as $ v = l(v)u+w$ since $l(w) = l(v-l(v)u)=0$.
From the above, we see that $\mathscr{U} = \mathscr{F}v + \mathrm{ker}\ l$ so that $\mathrm{ker}\ l$ has codimension 1.
Since $l$ is arbitrary, this holds for any linear functional in $\mathscr{U}$.
Now, the exercise implies that $\mathrm{ker} z \subseteq \mathrm{ker}\ y$. If $\mathrm{ker}\ z \subset \mathrm{ker}\ y$, the codimension of $\mathrm{ker}\ z$ must be greater than that of $\mathrm{ker}\ y$. But this means that $\mathrm{ker}\ y$ has codimension $0$, and thus $y \equiv 0$ (trivial case). For non-trivial $y,z$, it must hold that $\mathrm{ker}\ z$ = $\mathrm{ker}\ y$.
Let $\mathscr{K} = \mathrm{ker}\ z = \mathrm{ker}\ y$, then, as before, $\mathscr{U} = \mathscr{F}v + \mathscr{K}$ for any $v \in \mathscr{U}$. That means that $v = cu + w$ for any $v,u \in \mathscr{U}$, $c \in \mathscr{F}$ and $w \in \mathscr{K}$. Subsequently,
\begin{gather*}
y(v) = y(cu+w) = cy(u)\\
z(v) = z(cu+w) = cz(u)
\end{gather*}
and
\begin{gather*}
\frac{y(v)}{z(v)} = \frac{y(u)}{z(u)}
\end{gather*}
for any $u,v \in \mathscr{U}\setminus\mathscr{K}$. This proves that $y = \alpha z$ with $\alpha$ the scalar as given by Halmo's hint.
