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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.

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What do you mean by $[x, y]$? –  Qiaochu Yuan Feb 14 '12 at 18:27
    
I am sorry i thought that was a standard convention it means y(x). Where y is the functional and x is the vector. obviously it evaluates to a scalar. –  Hardy Feb 14 '12 at 18:31
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It's unfamiliar to me. Usually by $[x, y]$ I mean a Lie bracket. –  Qiaochu Yuan Feb 14 '12 at 18:47
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@Qiaochu: The bracket notation for functionals is common but a bit old-fashioned. The last place I remember seeing it is in Halmos's Finite Dimensional Vector Spaces. –  Pete L. Clark Feb 14 '12 at 18:58
    
This is actually an exercise question from that text. –  Hardy Feb 14 '12 at 19:00

2 Answers 2

up vote 2 down vote accepted

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 $y(x)=z(x)\frac{y(y_0)}{z(y_0)}=\frac 1{z(y_0)}z(x)$.

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i could not follow last line of your proof. Could you please elaborate a bit more. –  Hardy Feb 14 '12 at 19:04
    
Do you agree with the fact that $z\left(x-z(x)\frac{y_0}{z(y_0)}\right)=0$? Then we just use the fact that $z(u)=0$ implies $y(u)=0$ for any vector $u$. –  Davide Giraudo Feb 14 '12 at 19:08
    
actually that is what i was having trouble with why is $z\left(x-z(x)\frac{y_0}{z(y_0)}\right)=0$ true ? –  Hardy Feb 14 '12 at 19:12
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Use linearity to get it. –  Davide Giraudo Feb 14 '12 at 19:15
    
did u mean linear independence ? Sorry i am a newbie i am not sure which linearity u were referring to. Thanks for your help i really hope that u would n't get frustrated atleast yet and tell me this one. It could be a real eye opening trick use case for me. –  Hardy Feb 14 '12 at 19:20

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.

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i have to say i really like your proof style although i am not familiar with the rank-nullity theorem and kernels etc. I 'll try to read up to decrypt your answer. –  Hardy Feb 14 '12 at 19:15
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Given a linear map $f$ between vector spaces, the kernel of $f$ is the subset of all things in the domain which, when plugged into f, give $0$ in the range. One can prove it's a vector subspace. The rank nullity theorem states the dimension of the domain is equal to the dimension of the kernel + the dimension of the image of $f$. –  Jason DeVito Feb 14 '12 at 19:17

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