# Am i correct?

• Apr 5th 2008, 10:04 PM
deragon999
Am i correct?
I have to find a basis for the vector space spanned by these 3 column vectors:
[1;-1;2] [2;-3;2] and [1;-3;2]

I have written the vectors as rows of an array and solved as far as possible:
pivot rows are in bold and -1--> means add to -1*pivot row

1 -1 2
|
2 -3 2| -2 -->
1 -3 2| -1 -->
-1 -2|
-2 -6| -2-->
0 0 -2

does this mean that one possible basis is: {[1;-2;2],[0;-1;-2],[0;0;-2]}?

Also does this mean that the dimension for the subspace is 3?
• Apr 5th 2008, 10:18 PM
mr fantastic
Quote:

Originally Posted by deragon999
I have to find a basis for the vector space spanned by these 3 column vectors:
[1;-1;2] [2;-3;2] and [1;-3;2]

I have written the vectors as rows of an array and solved as far as possible:
pivot rows are in bold and -1--> means add to -1*pivot row

1 -1 2
|
2 -3 2| -2 -->
1 -3 2| -1 -->
-1 -2|
-2 -6| -2-->
0 0 -2

does this mean that one possible basis is: {[1;-2;2],[0;-1;-2],[0;0;-2]}?

Also does this mean that the dimension for the subspace is 3?

It's enough to prove that the three vectors are independent. Then the three vectors themselves form a basis for the vector space.

Dimension of vector space = size of basis therefore dimension of the vector space is equal to 3.
• Apr 5th 2008, 10:23 PM
deragon999
Thanks
Ok i fully understand that..but just because i spent the time on it..is my basis also true..or does that way only work if the vectors are Dependant?