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I want to prove the following

If $y$ is a non-zero linear functional on a vector space $\mathbb V$, and if $\alpha$ is an arbitrary scalar, does there neccessarily exist a vector $x \in \mathbb V$ such that $[x,y] = \alpha$ ?

EDIT: Removed my attempt at it.

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What you've written is mostly nonsense, I think. To say $y$ is a functional on $V$ is to say $[x,y]$ is defined for $x$ in $V$. "arbitrary $x$" makes no sense - $x$ has to be somewhere, in some set. Also, you introduce $p$ and $q$ and the lambdas with no hint as to what on Earth they mean. You really have to go back to the definitions and think things through. –  Gerry Myerson Mar 20 '12 at 12:05
    
I tried to improve the question. Please see if it is somewhat more sensible, –  Dilawar Mar 20 '12 at 12:09
    
No, you are still writing $[x,y]$ where $x$ need not be in $V$, when the fact that your functional is defined on $V$ means precisely that $x$ has to be in $V$ for $[x,y]$ to be defined. It's like saying let $f$ be defined on the real numbers, now consider $f(x)$ where $x$ may not be a real number. The rest of it isn't much better, I'm afraid. –  Gerry Myerson Mar 20 '12 at 12:14
    
Otherwise just consider the problem statement and ignore my attempt at it. –  Dilawar Mar 20 '12 at 12:15
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The problem with removing your attempt is that it removes valuable information about the state of your knowledge, and that makes it hard for others to know what might and what might not help you. –  Gerry Myerson Mar 20 '12 at 12:46

3 Answers 3

up vote 4 down vote accepted

If $y$ is nonzero, then there is a $v_0 \in V$ such that $y(v_0)=r\neq 0$. And if $\alpha \in \mathbb R$, then $y({\alpha \over r}v_0)={\alpha \over r}y(v_0)=\alpha$. So $y$ is surjective.

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This is helpful...I used this to show that if $f$ is a nonzero linear functional, then the dimension of the image of $f$ is one. –  Philip Benj Marcoby Eragon Feb 24 '13 at 4:41

Maybe it will help if I remind you of some definitions.

$y$ is a functional on $V$ means $y$ is a function with domain $V$ and codomain the reals.

$y$ is a linear functional means if $a$ and $b$ are reals and $u$ and $v$ are in $V$ then $y(au+bv)=ay(u)+by(v)$ (or, in your notation, $[au+bv,y]=a[u,y]+b[v,y]$).

$y$ is non-zero means there is at least one $v$ in $V$ such that $y(v)\ne0$ (in your notation, $[v,y]\ne0$).

Now I hope you can put it all together and solve the question.

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Recall that the dual space $V^{\ast}$ is the vector space of all linear transformations $T$ from $V$ to $\mathbb{R}$.

So what you're asking is given a $T \in V^{\ast}$, and $a \in \mathbb{R}$ does there always exist a vector $v \in V$ such that $T(v) = a$. This is equivalent to asking if there is a solution for every right hand side, or if your map $T$ is surjective. Now if $V$ is finite dimensional, we can apply the the Rank - Nullity Theorem:

If $T : V \longrightarrow \Bbb{R}$ for $V$ finite dimensional then $$\dim V = \dim \ker T + \dim \operatorname{Im} T.$$

Now if the dimension of the image is zero then this means that the dimensional of the kernel is equal to the dimension of $V$. Since $\ker T$ is a subspace of $V$, this means that $V = \ker T$ (exercise). In other words, $T$ is the zero map so we exclude this possibility.

Therefore this means that $1 \leq \dim \operatorname{Im} T \leq \dim \Bbb{R} = 1$ so that by the same reasoning as before and noting that the image of a linear transformation is always a subspace of the codomain that

$$\operatorname{Im} T = \mathbb{R}.$$

In other words, your map is surjective.

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Very nice, Benjamin, but read the unedited version of the question and tell me whether you thnk OP will understand a word of your answer. –  Gerry Myerson Mar 21 '12 at 1:21
    
@GerryMyerson I understand what you mean. The OP was assumed what he was going to prove. However, I assumed that one learns about rank - nullity before linear functionals. I was just trying to show in the case that $V$ is finite dimensional we can do it like this. –  user38268 Mar 21 '12 at 4:29

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