Subspace & Basis: Proving A is a Subspace of R^3

  • Thread starter Thread starter tracedinair
  • Start date Start date
  • Tags Tags
    Basis Subspaces
Click For Summary

Homework Help Overview

The problem involves demonstrating that a set of vectors orthogonal to a given vector u = [4, 3, 1] forms a subspace of R^3 and finding a basis for that subspace.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the properties of the zero vector and the closure under addition and scalar multiplication for the set A. There is uncertainty about the notation used, particularly regarding the multiplication of vectors by the set A. Questions arise about how to determine if an arbitrary vector is orthogonal to u.

Discussion Status

Some participants have provided initial reasoning regarding the subspace properties, while others express confusion about the notation and the process of finding a basis. There is an ongoing exploration of the concept of orthogonality and hints about using the dot product.

Contextual Notes

Participants note a lack of clarity in the problem statement and the definitions being used, particularly regarding the set A and its properties. There is also mention of difficulties in finding a basis without further guidance.

tracedinair
Messages
47
Reaction score
0

Homework Statement



Let u be a vector where u = [4 3 1]. Let A be the set of all vectors orthogonal to u. Show that A is subspace of R^3. Then find the basis for A.

Homework Equations





The Attempt at a Solution



For showing that A is a subspace...

Zero vector is in A because A(0) = 0

For any u & v, u+v is in A because Au=0, Av=0, and A(u+v) = Au+Av = 0

And for any scalar c, A(cu) = c(Au) = c(o) = 0

As for the basis, I really have no idea where to even start with that.

Thanks for any help.
 
Physics news on Phys.org
tracedinair said:

Homework Statement



Let u be a vector where u = [4 3 1]. Let A be the set of all vectors orthogonal to u. Show that A is subspace of R^3. Then find the basis for A.

Homework Equations





The Attempt at a Solution



For showing that A is a subspace...

Zero vector is in A because A(0) = 0

For any u & v, u+v is in A because Au=0, Av=0, and A(u+v) = Au+Av = 0

And for any scalar c, A(cu) = c(Au) = c(o) = 0

As for the basis, I really have no idea where to even start with that.

Thanks for any help.

You should do the first part of this problem; namely, finding the set of vectors that are orthogonal to u = (4, 3, 1). How can you tell that an arbitrary vector (x, y, z) is orthogonal to a given vector?
Zero vector is in A because A(0) = 0
For any u & v, u+v is in A because Au=0, Av=0, and A(u+v) = Au+Av = 0

And for any scalar c, A(cu) = c(Au) = c(o) = 0
None of this makes any sense. A is a set, not a matrix, so it doesn't make any sense to multiply a vector by A.

As for finding a basis for A, if you do the first part you will be on your way toward a basis.
 
I'm not entirely sure how to show an arbitrary vector is orthogonal to a given vector. I've looked through my text for help, but it's not really helping.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
8
Views
2K
  • · Replies 5 ·
Replies
5
Views
7K
  • · Replies 8 ·
Replies
8
Views
4K
  • · Replies 18 ·
Replies
18
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K