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Let $f$ be a linear map from $V$ to its field set, say $F$. Now if $f(v)=0$ does it necessarily imply that $v=0$. I don't think so. $v=0$ is one solution, but it need not be the only solution because $f$ need not be one-to-one. Is this correct ?

Actually this is used in some part of one proof in a book which is creating confusion. Let $V'$ be the set of all linear maps from $V$ to $F$ and $V''$ be the set of all linear maps from $V'$ to $F$ .

Now, let $g$ be a map from $V$ to $V''$. Take any element $v \in V$. Then $g(v) \in V''$ is a map from $V'$ to $F$. Now lets define this map as $g(v)(f)=f(v)$. Now to prove that $g$ is one to one we need to prove that $g(v)=0$ implies $v=0$. Now $g(v)=0$ implies $g(v)(f)=0 \forall f \in V'$ which implies $f(v)=0$ then it is concluded that $v=0$. This is confusing to me.

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3 Answers 3

up vote 10 down vote accepted

You're correct. For example $f(1,1)=0$ when $f(x,y)=x-y$, $f:\Bbb R^2\to\Bbb R$. In fact, if $f$ is a linear transformation $f:V\to F$, $f$ is one-one if and only if $f(v)=0\implies v=0$. Can you show this? Hint: $f(x)-f(y)=f(x-y)$.

The space of linear transformations (also linear forms) $f:V\to F$ is called the dual space of $V$, and denoted by $V^\ast$. These functions are fed a vector from $V$, and return a scalar in $F$. This dual space is indeed a vector space in its own right, thus, we can talk about its dual, $(V^\ast)^\ast=V^{\ast\ast}$, the space of linear functionals $g:V^*\to F$. These are fed a function, a vector of $V^*$, and return a scalar in $F$. What you're being shown is almost that $V$ and $V^{\ast\ast}$ are isomorphic: the proof shows the map is one-one. But since $V$ and then $V^\ast,V^{\ast\ast}$ are finite dimensional, this finishes it -- this is a standard use of the rank-nullity theorem$^{1}$. The map that is being proposed is $$\phi: V\to V^{\ast\ast}$$ defined as follows: given $v\in V$, $\phi(v)$ is a map $\phi(v):V^\ast \to F$ that sends the functional $\varphi$ to $\varphi(v)$. This is usually written as $$\phi(v)(\varphi)=\varphi(v)$$

This is indeed a linear transformation. Now, suppose that $\phi(v)=0$. This is saying $\phi(v)$ is the null transformation, that is, for any $f$ we pick in $V^\ast$; $\phi(v)(f)=f(v)=0$. This means that $v$ is in the kernel of all linear functionals $f:V^\ast\to F$, which forces $v=0$. To show that $$\bigcap_{\varphi\in V^\ast}\ker \varphi=\{0\}$$ show that for any nonzero $v$, there is a functional that is nonzero at $v$, i.e. there is some $\varphi\in V^\ast$ such that $\varphi(v)\neq 0$. (Hint: Without loss of generality, you can assume the first coordinate of $v$ is nonzero, then look at the projection mapping onto the first coordinate).


$1.$ In general, it is false that $V\simeq V^{\ast\ast}$ for infinite dimensional spaces.

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Plz see the elaborated problem definition. –  sosha Aug 20 '13 at 2:51
    
You are asking to define a define a linear map from $V$ to $F$ such that all non-zero elements in $V$ go to non-zero element in $F$. But if we take $V=\Bbb R^n$ and $F=\Bbb R$ then we can't define such a map (rank-nullity theorem). –  sosha Aug 20 '13 at 5:44
    
@prasenjit I did not say that. I said that given one non zero $v$ there exist a functional $f_v:V\to F$ such that $f_v(v)\neq 0$. –  Pedro Tamaroff Aug 20 '13 at 5:53
    
Note the above is "For all $v\neq 0$ there is a functional $f$ such that $f(v)\neq 0$ , which would be something like $\forall v\neq 0,\exists f(f(v)\neq 0)$; not "There is a functional $f$ such that for all $v$ nonzero, $f(v)\neq 0$, which is something like $\exists f\forall v\neq 0,( f(v)\neq 0)$. –  Pedro Tamaroff Aug 20 '13 at 5:59
    
@prasenjit I'm glad. –  Pedro Tamaroff Aug 20 '13 at 6:14

That is correct! You can go further and prove that if the dimension of $V$ is greater than $1$, then for any linear map $f:V\to F$, there exists some $v\neq0$ such that $f(v)=0$. Hint: Start with any two linearly independent vectors and see what happens when you apply $f$...

Edit: To address the expanded question, one of the steps in the proof shows that for all $f$, we have $f(v)=0$. Under these circumstances, $v$ must indeed be $0$. Do you see the difference?

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Plz see the elaborated problem definition. –  sosha Aug 20 '13 at 2:51
    
understood now. –  sosha Aug 20 '13 at 2:59

Yes, you are right. For example, $f:\mathbb{R}^2\to \mathbb{R}$. Define the map by $f((v_1,v_2))=v_1$. Then any vector $(0,v_2)$ will give you zero. Hope this helps.

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Plz see the elaborated problem definition. –  sosha Aug 20 '13 at 2:52

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