Linear functionals on a vector space Given $f$ and $g$, linear functionals on the same vector space such that $[x,f]=0$ whenever $[x,g]=0$, where $[x,f]=f(x).$ I am supposed to show that there exists a scalar $\alpha$ such that $f=\alpha g$. What I have done is below. Please see if it is right.
Fix $x$ in V. If $[x,g]\ne0$, we define $\alpha=[x,f]/[x,g].$ Then, $[x,f]=\alpha[x,g].$ i.e $f(x)=\alpha g(x)$, which follows that $f=\alpha g$
 A: As an answer, because otherwise it would run several comments.
Assuming that you can find an $x$ such that $[x,g]\neq 0$, you have defined a scalar $\alpha$ as $\alpha = [x,f]/[x,g]$. This implies that the specific scalar $f(x)$ is equal to the scalar $\alpha g(x)$. From this, we can conclude that $f$ and $\alpha g$ agree at $x$; but this is not enough. In order to show that $f=\alpha g$ as functions, you need to show that $f$ and $\alpha g$ agree on all of $V$. So you would still need to show that for every $y\in V$ you have $f(y)=\alpha g(y)$; that is, that
$$f(y) = \left(\frac{[x,f]}{[x,g]}\right)g(y).$$
You have not done so: you have only shown that this equation holds for $y=x$, not for every $y\in V$.
If you can prove that this is indeed true (that the displayed equation does hold), then you will be almost done, but not quite: you would still need to consider what happens if you cannot find an $x$ with $[x,g]\neq 0$ (that is, what happens if $g(x)=0$ for all $x\in V$?).
Now, going back to what you still need to prove, there are a number of ways of doing this.
One way would be to show that for every $x$ and $y$, if $g(x)\neq 0$ and $g(y)\neq 0$, then 
$$\alpha = \frac{[x,f]}{[x,g]} = \frac{[y,f]}{[y,g]}.$$
One possibility for doing this would be to try to come up with some linear combination of $x$ and $y$ which is sent to $0$ by $g$; then use the hypothesis that everything sent to $0$ by $g$ is sent to $0$ by $f$, and see if you can deduce the above equality.
Another possibility is to think about the dimensions of the kernels of $f$ and $g$, and what that tells you about $x$ and $y$.
