1. ## Dimensional

Let V be an n-dimensional vector space and let $f$ be a nonzero linear form on V, that is a nonzero element of the dual space V* of V. Prove that the kernel of $f$ is (n-1)-dimensional. Conversely, show that every (n-1)- dimensional linear subspace of V is the kernel of a linear form.
Plx help...

2. Originally Posted by mathbeginner
Plx help...
I'll give hints

The first part is trivial if you know the rank-nullity theorem since, if $V$ is an $F$-space then a linear functional $\varphi:V\to F$ can be viewed as a linear transformation with $F$ viewed as a one dimensional vector space over itself.

Otherwise, try to show that if $\varphi(x_0)\ne 0$ then $V=\text{span}\{x_0\}\oplus\ker\varphi$.

For the other part, what if $W$ is an $n-1$ dimensional subspace of $V$ and you considered some complement to it $\text{span}\{x_0\}$. Then, what if you considered the mapping $\varphi:V\to W:w+\alpha x_0\mapsto \alpha$?

3. Originally Posted by Drexel28
I'll give hints

The first part is trivial if you know the rank-nullity theorem since, if $V$ is an $F$-space then a linear functional $\varphi:V\to F$ can be viewed as a linear transformation with $F$ viewed as a one dimensional vector space over itself.

Otherwise, try to show that if $\varphi(x_0)\ne 0$ then $V=\text{span}\{x_0\}\oplus\ker\varphi$.

For the other part, what if $W$ is an $n-1$ dimensional subspace of $V$ and you considered some complement to it $\text{span}\{x_0\}$. Then, what if you considered the mapping $\varphi:V\to W:w+\alpha x_0\mapsto \alpha$?
FOR the first part: as I know V*=L(V,F)
and dimV*=dimV
by using the rank-nullity theorem
I get dim F(v) + dim(Ker F)
I want ask that is it dim F(v) = dim F then I can have dim F =1???

4. Originally Posted by mathbeginner
FOR the first part: as I know V*=L(V,F)
and dimV*=dimV
by using the rank-nullity theorem
I get dim F(v) + dim(Ker F)
I want ask that is it dim F(v) = dim F then I can have dim F =1???
Ok, so from the above I assume you know the Rank-Nullity theorem.

I first ask you to realize that every non-zero linear functional is surjective. The easy, non-dimension argument, comes from the fact that if $\varphi(v_0)\ne0_F$ then for any $\alpha\in F$ we have that $\displaystyle \varphi\left(\frac{\alpha}{\varphi(v_0)}v_0\right) =\frac{\alpha}{\varphi(v_0)}\varphi(v_0)=\alpha$.

So, you would agree that if we take the field operations of $F$ and turn them into vector operations (i.e. the scalars and vectors are elements of $F$) we're left with a vector space, call it $\mathcal{F}$. It's fairly easy to see though that $\{1_F\}$ is linearly independent and $\text{span}\{1_F\}=\mathcal{F}$ and thus $\dim_F \mathcal{F}=1$. Note though that with this in mind we can see that any linear functional $\varphi:V\to F$ is, in fact, also a linear transformation $\varphi:V\to\mathcal{F}$. But, the Rank-Nullity theorem then states that $\dim_F V=\dim_F\ker\varphi+\dim_F\varphi\left(V\right)$. Note though that by the surjectivity of $\varphi$ we have that $\varphi\left(V\right)=\mathcal{F}$ and so $\dim_F\varphi\left(V\right)=1$ and thus plugging this into our nice little formula gives $\dim_F \ker\varphi=\dim_F V-1$.

5. Originally Posted by Drexel28
I'll give hints

The first part is trivial if you know the rank-nullity theorem since, if $V$ is an $F$-space then a linear functional $\varphi:V\to F$ can be viewed as a linear transformation with $F$ viewed as a one dimensional vector space over itself.

Otherwise, try to show that if $\varphi(x_0)\ne 0$ then $V=\text{span}\{x_0\}\oplus\ker\varphi$.

For the other part, what if $W$ is an $n-1$ dimensional subspace of $V$ and you considered some complement to it $\text{span}\{x_0\}$. Then, what if you considered the mapping $\varphi:V\to W:w+\alpha x_0\mapsto \alpha$?
so the second part: if I say W is and n-1 dim subspace of V
then dimV is dimW + span{x0}. But what do you mean by $\varphi:V\to W:w+\alpha x_0\mapsto \alpha$?[/QUOTE] what is $\alpha x_0\$ and why it go to $\alpha\$? What does that mean???
THX

6. Originally Posted by mathbeginner
so the second part: if I say W is and n-1 dim subspace of V
then dimV is dimW + span{x0}. But what do you mean by $\varphi:V\to W:w+\alpha x_0\mapsto \alpha$?
what is $\alpha x_0\$ and why it go to $\alpha\$? What does that mean???
THX[/QUOTE]

Why do I get the niggling feeling that you're trying to milk me for answers?

7. Originally Posted by Drexel28
what is $\alpha x_0\$ and why it go to $\alpha\$? What does that mean???
THX
Why do I get the niggling feeling that you're trying to milk me for answers?[/QUOTE]

nonono..
I am trying to understand what are u talking abt because I reli don't get what is it.

8. Originally Posted by mathbeginner

nonono..
I am trying to understand what are u talking abt because I reli don't get what is it.
We know that if $W$ is a $n-1$ dimensional subspace that we can fix a complement $U$ of it so that $V=W\oplus U$. But, since $n=\dim_F W+\dim_F U$ we know that $\dim_F U=1$ and thus $U=\text{span}\{x_0\}$ for some $x_0\in V$. But, since $V=W\oplus U$ we know that every vector $v\in V$ can be represented uniquely as $w+\alpha x_0$ for some $w\in W$ and some $\alpha\in F$. So consider the function $\varphi$ that takes $w+\alpha x_0$ and maps it to $\alpha$. Show that $\varphi\in\text{Hom}\left(V,F\right)$ (if you aren't familiar with this notation, it just means $V^*$) and $\ker\varphi=W$.