# 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 $f=\alpha g$

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There are a couple of reason why the above is not correct: first, your $\alpha$ depends on $x$: you have not proven that $f=\alpha g$, just that, for that particular $x$ you have $f(x)=\alpha g(x)$. And you don't say what to do if $[x,g]=0$ (it's clear what to do, but you should say it). Really, the first part is the big problem: you would need to show that if $x$ and $y$ are such that $[x,g]\neq 0$ and $[y,g]\neq 0$, then $[x,f]/[x,g] = [y,f]/[y,g]$. Otherwise, your $\alpha$ will change with $x$. –  Arturo Magidin Sep 13 '11 at 3:24
P.S. You can edit your question and modify your attempt; there are a couple of ways to finish off your argument... perhaps the Rank-Nullity Theorem can be of some help? –  Arturo Magidin Sep 13 '11 at 3:41
okay. I'm a bit confused as to what to do next. what exactly am I to show. –  smanoos Sep 13 '11 at 3:42

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 $y$.

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Thank you for your help –  smanoos Sep 13 '11 at 12:46
Can you please give me a hint on the linear combination of x and y? The following is what I assumed and it works only if $g(y)=0$: $u=(\alpha y+x)-x$ –  smanoos Sep 13 '11 at 14:58
@smanoos: Suppose $g(x)=a$, and $g(y)=b$. How much is $g(bx-ay)$? (Note that his makes sense, because $a$ and $b$ are scalars, so $bx-ay$ is a vector). –  Arturo Magidin Sep 13 '11 at 15:54
Thank you so much. I think this one works perfectly. You are a genius. –  smanoos Sep 13 '11 at 23:33