Basis of a subspace and dimension question

  • Thread starter clurt
  • Start date
  • #1
clurt
28
0
How do i go about this?

Find a basis for the subspace W of R^5 given by...

W = {x E R^5 : x . a = x . b = x . c = 0}, where a = (1, 0, 2, -1, -1), b = (2, 1, 1, 1, 0) and c = (4, 3, -1, 5, 2).

Determine the dimension of W. (as usual, "x . a" denotes the dot (inner) product of the vecotrs x and a).

Please show working, your help is much appreciated.
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
43,021
971
Let x= (p, q, r, s, t). Then a.x = (1)p+ (0)q+ (2)r+ (-1)s+ (-1)t= p+ 2r- s- t= 0, b.x= (2)p+ (1)q+ (1)r+ (1)s+ (0)t= 2p+ q+ r+ s= 0, and c.x = (4)p+ (3)q+ (-1)r+ (5)s+ (2)t= 4p+ 3q- r+ 5s+ 2t= 0.

That gives you three equations in 5 "unknown" values:
p+2r- s- t= 0
2p+ q+ r+ s= 0
4p+ 3q- r+ 5s+ 2t= 0

You can solve for three of the values in terms of the other two. (Unless the equations are "dependent" in which case you can solve for two in terms of the other three or one in terms of the other four.) Write the vector (p, q, r, s, t), replacing the ones you have solved for in terms of the other two.

For example, suppose you had found that p= s- t, q= 2s+ t, and r= 3s- 2t. (Those are NOT the correct solutions. I did NOT solve the equations- that is for you to do.) Then we would have (p, q, r, s, t)= (s- t, 2s+t, 3s- 2t, s, t)= (s, 2s, 3s, s, 0)+ (-t, t, -2, 0, t)= s(1, 2, 3, 1, 0)+ t(-1, 1, 2, 0, 1) so you should be able to see that the subspace is two-dimensional and a basis for the subspace is {(1, 2, 3, 1, 0), (-1, 1, 2, 0, 1)}.
 
Last edited by a moderator:

Suggested for: Basis of a subspace and dimension question

  • Last Post
Replies
1
Views
554
Replies
0
Views
261
  • Last Post
Replies
22
Views
475
Replies
16
Views
405
  • Last Post
Replies
2
Views
753
Replies
5
Views
289
Replies
2
Views
398
Replies
2
Views
382
  • Last Post
2
Replies
42
Views
2K
Top