2 linear functionals on a vector space so one can be represented as a multiple of the other. Prove that if $y$ and $z$ are linear functionals (on the same vector space) such that $\left[ x,y\right] = 0$ whenever $\left[ x,z\right] = 0$, then there exists a scalar $\alpha $ such that $y=\alpha z$. 
Now I am aware that linear functionals dispense scalars and the defining property of a linear functional is 
$$y\left( \alpha _{1}x_{1}+\alpha_{2}x_{2}\right) =\alpha _{1}y\left( x_{1}\right) +\alpha _{2}y\left( x_{2}\right)\;.$$
Let $x_{0}$ be some vector in our vector space such that $\left[ x_{0},z\right] \neq 0$; then based on the question $\left[ x_{0},y\right] \neq 0$, hence we have $\alpha_{0} =\dfrac {\left[ x_{0},y\right] } {\left[x_{0},z\right] }$, but how to extend this to all cases?
Any help would be much appreciated.
 A: We have to assume $y$ is not identically $0$ otherwise we just take $\alpha=0$. Let $y_0$ such that $y(y_0)=1$. We can write $x=x-z(x)\frac{y_0}{z(y_0)}+z(x)\frac{y_0}{z(y_0)}$ becausse $z(y_0)\neq 0$. We have $z\left(x-z(x)\frac{y_0}{z(y_0)}\right)=0$ so $y\left(x-z(x)\frac{y_0}{z(y_0)}\right)=0$ and $y(x)=z(x)\frac{y(y_0)}{z(y_0)}=\frac 1{z(y_0)}z(x)$.
A: I'm going to use $f$ and $g$ for the functionals.  Something about using $x$ and $y$ to stand for completely different objects bothers me.
I'm assuming that $f(x) = 0$ whenever $g(x) = 0$, so if we plug $x$ into $g$ and get $0$, we must also get $0$ when plugging the same $x$ into $f$ - but there may be $x$s we can plug into $f$ to get $0$ which don't evaluate to $0$ when plugged into $g$.  The goal is to show that $f(x) = \alpha g(x)$ for some scalar $\alpha$.
First a stupid case:  If $g(x) = 0$ for all $x$, then $f$ and $g$ are both the $0$ functional, so are multiples of each other with any $\alpha$.
So, we may assume there is an $x$ with the property that $g(x)\neq 0$.  The rank-nullity theorem asserts that the kernel $\ker g$ of $g$ is a codimension $1$ hyperplane in the vector space.  Choose a basis $\{w_1,w_n,...\}$ for this kernel and extend this basis to the set $\{w_1,..., x\}$.  I claim that this new set is a basis for the whole vector space.  Since it has the right number of elements, we must only show it's independent.
So, assume $\sum c_i w_i + dx = 0$.  Applying $g$ to both sides gives $\sum c_i g(w_i) + dg(x) = 0$.  Since each $w_i\in\ker g$, we get $dg(x) = 0$.  This implies $d = 0$.  Since the $w_i$ are assumed to be independent, this implies all the $c_i$ are $0$, so we really do have a basis.
Now, define $\alpha = f(x)/g(x)$.  I claim that $f = \alpha g$.
To see this, notice that given any linear combination of $w_i$ and $x$, $\sum c_i w_i + dx$, we have $g(\sum c_i w_i + dx) = dg(x)$ while $f(\sum c_i w_i + dx) = df(x) = d\alpha g(x)$.
So $f(\sum c_i w_i + dx) = \alpha g(\sum c_i w_i + dx)$ and since $\{w_1,...,w_n, x\}$ forms a basis, we're done.
