Understanding Linear Algebra Basics: Finding a Basis for a Subspace in F^3

Click For Summary

Homework Help Overview

The discussion revolves around understanding the concept of a basis in the context of linear algebra, specifically within the vector space F^3. The original poster expresses difficulty in grasping the idea of a basis and presents a specific problem involving finding a basis for a subspace defined by a linear equation.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the representation of elements in the subspace W and discuss the implications of the linear equation defining W. There is an attempt to express the basis in terms of specific vectors and to clarify the conditions under which a set of vectors can serve as a basis.

Discussion Status

Participants are actively engaging with the problem, with some providing insights into the nature of the basis and the geometric interpretation of the subspace. Questions about the uniqueness of the basis and the conditions for a plane to pass through the origin are also raised, indicating a productive exploration of the topic.

Contextual Notes

There are discussions about the definitions and properties of vector spaces, particularly concerning the conditions that must be met for a set of vectors to form a basis. Additionally, there is a mention of the general equation of a plane and its relationship to the origin, which is relevant to the understanding of subspaces in F^3.

rad0786
Messages
187
Reaction score
0
Hello... I am doing this problem with basis. Infact, I am having a lot of problems understanding basis, i did every question in the textbook and I still get seem to understand the idea of it.

So i was hoping somebody can help me with the whole idea about it.

Say, for example, how would i go abouts a question like this:

--Let F be a field and let V = F^3. Let
W = {(a1 a2 a3) E F^3 / 2a1 - a2 - a3 = 0 }

Find a basis for W --

If a question like that came on a test, id fail it - sad to say. It would also be good if somebody knows a good website or has sameple tests that covers this mataril so that i may get used to it.

Thanks
 
Physics news on Phys.org
so any element in W can be represented like so:

w = (a1, a2, a3), where a1, a2, and a3 are arbitrary.

but W has the additional restriction that a1 = 1/2 (a2 + a3).

so

w= ( 1/2 (a2+a3), a2, a3).

w = a2 ( 1/2, 1, 0) + a3 (1/2, 0, 1). (it's easy to see that this is the same as above.)

so

w = span{(1/2, 1, 0), (1/2, 0, 1)}.


and that set {(1/2, 1, 0), (1/2, 0, 1)} is our basis.


ah, i miss these problems!
 
Keep in mind that the above is not the only basis.

Also, notice that the given vector space is nothing but a (generalization of a) plane through the origin in [itex]\mathbb{R}^3[/itex]. Any pair of vectors in the plane will serve as a basis.
 
Hey.. i was just wondering about Brad Barkers post above...

he said that a1 = 1/2 (a2 + a3).

well.. shouldn't it be a1 = 1/2 (a2 - a3) ?

Does it make a difference?
 
No, you said "2a1 - a2 -a3 = 0"

That gives 2a1 = a2 + a3
 
oh that was my silly mistake.. but either way.. i still learned something :)
 
Hey... what about the subspace...

U = (a+b+c=0/ a b c is in the Real Numbers)

How would you show the span of that.

Also, the comment Gokul43201 made, about the basis being the plane through the origin, how did he know that? I mean its a plane because of the two vectors, but how did he know that its through the origin?
 
rad0786 said:
Also, the comment Gokul43201 made, about the basis being the plane through the origin, how did he know that? I mean its a plane because of the two vectors, but how did he know that its through the origin?

The general equation of a plane is [itex]Ax+By+Cz = D[/itex].
[itex]D = 0 \Longleftrightarrow[/itex] the plane goes through the origin.
 
"and that set {(1/2, 1, 0), (1/2, 0, 1)} is our basis." Can that be a Basis for F^3? what is F (Feild)? isn't that the same as R, like R^3
 
  • #10
iNCREDiBLE said:
The general equation of a plane is [itex]Ax+By+Cz = D[/itex].
[itex]D = 0 \Longleftrightarrow[/itex] the plane goes through the origin.
Also, for a plane to constitute a vector space with the usual vector addition, it must pass through the origin, since this point is the additive identity element.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
15
Views
3K
  • · Replies 15 ·
Replies
15
Views
3K
Replies
4
Views
2K
Replies
18
Views
2K
Replies
34
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
8
Views
2K