How can I find the intersection of two vector spaces with given basis vectors?

  • Thread starter Thread starter transgalactic
  • Start date Start date
  • Tags Tags
    parts
Click For Summary

Homework Help Overview

The discussion revolves around finding the intersection of two vector spaces defined by their basis vectors. The original poster expresses difficulty in solving the last part of a problem related to this topic and seeks feedback on their previous work.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the implications of the dimensions of the vector spaces involved and the nature of their intersection. There are attempts to clarify the conditions under which the intersection can be trivial or non-trivial. Questions arise regarding specific calculations and the correctness of the original poster's reasoning.

Discussion Status

The conversation is ongoing, with participants providing insights into the nature of the intersection and questioning the original poster's calculations. Some guidance has been offered regarding the interpretation of the basis vectors and the conditions for the intersection.

Contextual Notes

There is mention of specific matrices and row reduction processes, but details are not fully articulated, leading to some confusion about the original poster's approach. The discussion also highlights the lack of definitive answers, as participants explore various interpretations of the problem.

transgalactic
Messages
1,386
Reaction score
0
i have solved a full question
(except the last part which i tried to come up with some formula but that didnt work out)

i showed every step of the way

unfortunetly i don't have the answers to this question
so can you tell me if i solved wrong some subquestion
and how to solve the 5th
it a really challenge for me

1st page :

http://img260.imageshack.us/my.php?image=img8259zs8.jpg

2nd page:

http://img155.imageshack.us/my.php?image=img8261ge2.jpg

3rd page:

http://img401.imageshack.us/my.php?image=img8262iu8.jpg
 
Physics news on Phys.org
transgalactic said:
i have solved a full question
(except the last part which i tried to come up with some formula but that didnt work out)

i showed every step of the way

unfortunetly i don't have the answers to this question
so can you tell me if i solved wrong some subquestion
and how to solve the 5th
it a really challenge for me

1st page :

http://img260.imageshack.us/my.php?image=img8259zs8.jpg
This looks good.

Since V has dimension 1, and the intersection with U is a subspace of V, the intersection must be either the trivial subspace (consisting only of the 0 vector) or V itself.

In problem 4, You have an orthogonal basis which is what they ask for. If they had asked for an orthonormal basis, you would need to divide each vector by its length.

For problem 5, you are told that, for linear transformation, A, A2+ A= -I and are asked to find A-1 (so you may assume that A has an inverse).
You have factored A2+ A= A(A+ I)= -I. I recommend you multiply both sides of that equation by A-1 and then solve for A-1.

An equivalent way to do the same problem is to write the original equation as I= -A2- A and multiply both sides of that equation by A-1.
 
so regarding the second page

you say that if the the solution is one vector
it always must be (0,0,0)

so my answer is wrong

where is my mistake?
 
I'm not sure where you made a mistake because it's not clear to me what you did. You have several matrices you have row reduced but you don't say where you got those matrices or what you are trying to do with them.

You have, I think correctly, calculated that V has <4, -5, 2> as basis and you are told that U is spanned by <1, 1, 1> and <9, 0 7>. That means that any vector in their intersection, since it is in both U and V can be written both ways. If <4, -5, 2>a is a vector in the intersection, then it can also be written in the form <1, 1, 1>b+ <9, 0, 7>c. In other words, there must exist numbers, a, b, c, such that <4a, -5a, 2a>= <b+ 9c, b, b+ 7c>. We have the three equations 4a= b+ 9c, -5a= b, and 2a= b+ 7c. From the second equation, b= -5a. Putting that into the first equation, 4a= -5a+ 9c so 9a=9c and c= a. Putting that into the third equation, 2a= -5a+ 7a is true! Any vector of the form a<4, -5, 2> is a member of the intersection so <4, -5, 2> is a basis vector. (Of course, what we have shown is that U intersect V is just V itself.)
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
2
Views
2K
Replies
4
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
7
Views
7K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
12
Views
3K
Replies
11
Views
4K
Replies
4
Views
2K