Plx help...Let V be an n-dimensional vector space and let $\displaystyle f$ be a nonzero linear form on V, that is a nonzero element of the dual space V* of V. Prove that the kernel of $\displaystyle f$ is (n-1)-dimensional. Conversely, show that every (n-1)- dimensional linear subspace of V is the kernel of a linear form.
I'll give hintsOriginally Posted by mathbeginner
The first part is trivial if you know the rank-nullity theorem since, if $\displaystyle V$ is an $\displaystyle F$-space then a linear functional $\displaystyle \varphi:V\to F$ can be viewed as a linear transformation with $\displaystyle F$ viewed as a one dimensional vector space over itself.
Otherwise, try to show that if $\displaystyle \varphi(x_0)\ne 0$ then $\displaystyle V=\text{span}\{x_0\}\oplus\ker\varphi$.
For the other part, what if $\displaystyle W$ is an $\displaystyle n-1$ dimensional subspace of $\displaystyle V$ and you considered some complement to it $\displaystyle \text{span}\{x_0\}$. Then, what if you considered the mapping $\displaystyle \varphi:V\to W:w+\alpha x_0\mapsto \alpha$?
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 $\displaystyle \varphi(v_0)\ne0_F$ then for any $\displaystyle \alpha\in F$ we have that $\displaystyle \displaystyle \varphi\left(\frac{\alpha}{\varphi(v_0)}v_0\right) =\frac{\alpha}{\varphi(v_0)}\varphi(v_0)=\alpha$.
So, you would agree that if we take the field operations of $\displaystyle F$ and turn them into vector operations (i.e. the scalars and vectors are elements of $\displaystyle F$) we're left with a vector space, call it $\displaystyle \mathcal{F}$. It's fairly easy to see though that $\displaystyle \{1_F\}$ is linearly independent and $\displaystyle \text{span}\{1_F\}=\mathcal{F}$ and thus $\displaystyle \dim_F \mathcal{F}=1$. Note though that with this in mind we can see that any linear functional $\displaystyle \varphi:V\to F$ is, in fact, also a linear transformation $\displaystyle \varphi:V\to\mathcal{F}$. But, the Rank-Nullity theorem then states that $\displaystyle \dim_F V=\dim_F\ker\varphi+\dim_F\varphi\left(V\right)$. Note though that by the surjectivity of $\displaystyle \varphi$ we have that $\displaystyle \varphi\left(V\right)=\mathcal{F}$ and so $\displaystyle \dim_F\varphi\left(V\right)=1$ and thus plugging this into our nice little formula gives $\displaystyle \dim_F \ker\varphi=\dim_F V-1$.
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$\displaystyle \varphi:V\to W:w+\alpha x_0\mapsto \alpha$?[/QUOTE] what is $\displaystyle \alpha x_0\$ and why it go to $\displaystyle \alpha\$? What does that mean???
THX
We know that if $\displaystyle W$ is a $\displaystyle n-1$ dimensional subspace that we can fix a complement $\displaystyle U$ of it so that $\displaystyle V=W\oplus U$. But, since $\displaystyle n=\dim_F W+\dim_F U$ we know that $\displaystyle \dim_F U=1$ and thus $\displaystyle U=\text{span}\{x_0\}$ for some $\displaystyle x_0\in V$. But, since $\displaystyle V=W\oplus U$ we know that every vector $\displaystyle v\in V$ can be represented uniquely as $\displaystyle w+\alpha x_0$ for some $\displaystyle w\in W$ and some $\displaystyle \alpha\in F$. So consider the function $\displaystyle \varphi$ that takes $\displaystyle w+\alpha x_0$ and maps it to $\displaystyle \alpha$. Show that $\displaystyle \varphi\in\text{Hom}\left(V,F\right)$ (if you aren't familiar with this notation, it just means $\displaystyle V^*$) and $\displaystyle \ker\varphi=W$.
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