How Does the Non-Equality of Kernels Imply Their Sum Equals the Vector Space?

  • Context: Graduate 
  • Thread starter Thread starter Portuga
  • Start date Start date
  • Tags Tags
    Kernel
Click For Summary
SUMMARY

The discussion centers on the implications of the non-equality of kernels in linear transformations, specifically regarding the sum of kernels equating to the vector space. The kernel and image theorem establishes that for linear transformations F and G, the dimensions satisfy the equation ##\dim V=n=\dim\ker(F)+\dim \text{im}(F)=\dim\ker(G)+\dim \text{im}(G)##. Given that both transformations have images of dimension 1, it follows that the dimensions of their kernels are ##\dim\ker(F)=\dim\ker(G)=n-1##. The conclusion drawn is that the non-equality of the kernels implies that their sum equals the entire vector space, leading to the dimension of the intersection being ##\dim(\ker(F) \cap \ker(G))=n-2##.

PREREQUISITES
  • Understanding of linear transformations and their properties
  • Familiarity with the kernel and image theorem in linear algebra
  • Knowledge of vector space dimensions and their implications
  • Basic grasp of intersection and sum of subspaces
NEXT STEPS
  • Study the kernel and image theorem in detail
  • Explore the properties of vector space dimensions
  • Learn about subspace intersections and their dimensions
  • Investigate examples of linear transformations with non-equal kernels
USEFUL FOR

Students of linear algebra, mathematicians, and educators seeking to deepen their understanding of kernel properties in linear transformations.

Portuga
Messages
56
Reaction score
6
TL;DR
Let ##F## and ##G## be two non-zero linear functionals over a vector space ##V## of dimension ##n##. Assuming ##ker (F ) \neq \ker (G)##, determine the dimensions of the following subspaces: ##\ker (F )##, ##\ker (G)##, ##\ker (F ) + \ker (G)##, and ##\ker (F ) \cap \ker (G)##.
This is actually a solved exercise from a Brazilian book on Linear Algebra. The author presented the following solution:

The kernel and image theorem tells us that dimension ##\dim V=n=\dim\ker\left(F\right)+\dim \text{im}\left(F\right)=\dim\ker\left(G\right)+\dim\text{im}\left(G\right)##. As ##\text{im}\left(F\right)\subset R##, ##\dim\mathbb{R}=1## and ##F\neq0##, then ##\dim\text{im}\left(F\right)=1##. Similarly ##\dim\text{im}\left(G\right)=1##. Therefore, ##\dim\ker\left(F\right)=\dim\ker\left(G\right)=n-1##. On the other hand, the dimension theorem of the sum assures us that

$$
\dim\left(\ker\left(F\right)+\ker\left(G\right)\right)+\dim\left(\ker\left(F\right)\cap\ker\left(G\right)\right)=\dim\ker\left(F\right)+\dim\ker\left(G\right)=2n-2.
$$

In general, ##\ker\left(G\right)\subset\ker\left(F\right)+\ker\left(G\right)## and due to the hypothesis ##\ker\left(F\right)\neq\ker\left(G\right)##, we will have ##\ker\left(F\right)\begin{array}{c}

\subset \\ \neq \end{array}\ker\left(F\right)+\ker\left(G\right)##; then necessarily ##\ker\left(F\right)+\ker\left(G\right)=V##. So ##\dim\left(\ker\left(F\right)+\ker\left(G\right)\right)=n## and hence

$$
\dim\left(\ker\left(F\right)\cap\ker\left(G\right)\right)=\left(2n-2\right)=n-2.
$$

I am ok with almost everything he presented, but couldn't understand why

the hypothesis ##\ker\left(F\right)\neq\ker\left(G\right)## implies that ##\ker\left(F\right)+\ker\left(G\right)=V.##

Any ideas?

Thanks in advance.
 
Last edited:
Physics news on Phys.org
Do you understand why ##ker(F)\subsetneq ker(F)+ker(G)##?

Since the left hand side has dimension n-1, the right hand side must have dimension n or higher.
 
Office_Shredder said:
Do you understand why ##ker(F)\subsetneq ker(F)+ker(G)##?

Since the left hand side has dimension n-1, the right hand side must have dimension n or higher.
Oh my god! It was so simple!
Thank you very much!
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 24 ·
Replies
24
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
4
Views
2K
Replies
36
Views
7K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K