I Proof that two linear forms kernels are equal

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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.
 
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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).##
 
Thank you very much! Sorry for the mess in latex.
 
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
$$\bigcap_{i=1}^n\ker f_i\subset \ker f\Leftrightarrow f=\lambda_1 f_1+\ldots+\lambda_n f_n$$
 
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wrobel said:
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)##.
 
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