# Proof that two linear forms kernels are equal

• I
• Portuga
In summary, we have shown that ##\bigcap_{i=1}^n\ker f_i\subset \ker f \Leftrightarrow f=\lambda_1 f_1+\ldots+\lambda_n f_n##.
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.

Last edited:
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.

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
$$\bigcap_{i=1}^n\ker f_i\subset \ker f\Leftrightarrow f=\lambda_1 f_1+\ldots+\lambda_n f_n$$

Last edited by a moderator:
Hall and Portuga
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)##.

## What is the definition of a linear form?

A linear form is a mathematical function that takes in one or more variables and outputs a value that is a linear combination of those variables. It can be represented as a sum of constants multiplied by the variables, with no higher powers or products of variables.

## What does it mean for two linear forms to have equal kernels?

Two linear forms have equal kernels if they have the same set of solutions or roots. In other words, they have the same inputs that result in an output of 0.

## How can you prove that two linear forms have equal kernels?

To prove that two linear forms have equal kernels, you can use the property that if two linear forms are equal, their coefficients must be equal. This means that you can set the two linear forms equal to each other and solve for the variables. If the resulting equations have the same solutions, then the two linear forms have equal kernels.

## What are some common applications of proving equal kernels of linear forms?

Proving equal kernels of linear forms is important in many areas of mathematics and science, including linear algebra, differential equations, and optimization problems. It is also used in various engineering fields, such as signal processing and control systems.

## Are there any limitations or exceptions to proving equal kernels of linear forms?

There are some cases where two linear forms may have equal kernels, but are not actually equal. This can happen if one of the linear forms is a scalar multiple of the other, or if one of the linear forms is a linear combination of the other. In these cases, the two linear forms may have the same set of solutions, but are not truly equal.

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