# Basis of a subspace and dimension question

1. Sep 6, 2012

### clurt

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.

2. Sep 6, 2012

### HallsofIvy

Staff Emeritus
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: Sep 6, 2012