Proof that two linear forms kernels are equal

[Portuga]

TL;DR Summary

Let $F$ and $G$ be non-zero linear forms in vector space $V$, linearly dependent. Prove that $\ker\left(F\right)=\ker\left(G\right)$ and that its dimension is $n-1$ if $\dim V=n$.

Attempt of a solution.
By the Rank–nullity theorem,
$$
\dim V=\dim Im_{F}+\dim\ker\left(F\right)
\Rightarrow n=1+\dim\ker\left(F\right)
\Rightarrow \dim\ker\left(F\right)=n-1.
$$
Similarly, it follows that $$\dim\ker\left(G\right)=n-1.$$

This first part, for obvious reasons, is very clear.
The next one offered me a terrible challenge.
I saw a solution, but it didn't convince me for sure.

Here it is.

As $F$ and $G$ are linearly dependent,
$$
\alpha F\left(u\right)+\beta G\left(u\right)=0
\Rightarrow G\left(u\right)=\frac{-\alpha}{\beta}F\left(u\right),
$$
since in this expression, $\alpha$ and $\beta$ have to be different from 0, because the two linear forms are non-zero. It follows that every vector in the image of $F$ is also a vector in the image of $G$, which implies that $Im_{F}=Im_{G}$. As for any subspace $V$, $V=\ker\left(F\right) \oplus Im_{F}$, where $Im_{F}=Im_{G}$, then
$$
\ker\left(F\right)\oplus Im_{F}=\ker\left(G\right)\oplus Im_{G}
\Rightarrow \ker\left(F\right)\oplus Im_{F}=\ker\left(G\right)\oplus Im_{F}
\Rightarrow \ker\left(F\right)=\ker\left(G\right),
$$
as there is no intersection between the two sets.
I would like to know if this solution is correct and if is the best one, because it seems not elegant for me.

 

[fresh_42]

You should type formulas as explained here:
https://www.physicsforums.com/help/latexhelp/

Your proof looks correct, from what I could read in this mess. The second part is definitely too long.
Since $F$ and $G$ are linear dependent, and non-zero, we immediately have $F=\alpha \cdot G$ for some scalar $\alpha \neq 0.$

Now it is easy to see that $\ker F \subseteq \ker G \subseteq \ker F$ and that the images are equal, too: $F(u)=\alpha \cdot G(u)=G(\alpha \cdot u).$

 

[Portuga]

Thank you very much! Sorry for the mess in latex.

 

[wrobel]

It will also be a good exercise for OP. Let $f_1,\ldots, f_n,f:\mathbb{R}^m\to\mathbb{R}$ be linear forms. Show that
$$\bigcap_{i=1}^n\ker f_i\subset \ker f\Leftrightarrow f=\lambda_1 f_1+\ldots+\lambda_n f_n$$

 

[Hall]

It will also be a good exercise for OP. Let $f_1,\ldots, f_n,f:\mathbb{R}^m\to\mathbb{R}$ be linear forms. Show that
$$\bigcap_{i=1}^n\ker f_i\subset \ker f\Leftrightarrow f=\lambda_1 f_1+\ldots+\lambda_n f_n$$

Here is my attempt:

WLOG let us assume that $f_i$ s are independent.

$dim ~R^{m*}=m$

If $m\leq n$ then it becomes quite easier. Therefore, let's assume $m \gt n$ and $n+r=m$. We can extend $f_i s$ to a basis of the dual space, let the basis be $\{f_1, f_2, \cdots , f_n , f_{n+1}, \cdots , f_{n+r} \}$.

We're given that
$$
A = \bigcap_{i=1}^n ker (f_i) \subset ker (f)$$

As $f$ belongs to the dual space, we have
$$
f(v) = \lambda_1 f_1(v) +\cdots + \lambda_n f_n(v) + \lambda_{n+1} f_{n+1} (v) \cdots +\lambda_{n+r} f_{n+r} (v)$$

For any vector x belonging to A, we have
$$0= \lambda_{n+1} f_{n+1} (x) \cdots +\lambda_{n+r} f_{n+r} (x)$$

That's a contradiction, because that's the case for all the values of $x\in \bigcap_{i=1}^n ker (f_i)$.

Hence, $f$ must be a Linear combination of $f_1, f_2, \cdots , f_n$ only.

The converse is very easy to prove:

If $f = \lambda_1 f_1 +\cdots + \lambda_n f_n$
Then, for all $x \in A=\bigcap_{i=1}^n ker (f_i)$ we would have
$$
f(x)=0$$
Hence, $A \subset ker(f)$.