Can anyone verify that I am doing this correctly?
I have to find a basis for
So, does this mean that and can be any numbers?
I began by setting them both equal to one, giving me (0, 1, 0, 0 ,0) and (0, 0, 0, 0, 1)
Then, by letting , this gave me . Then choosing or , I obtain or , respectively.
So that gave me the vectors (1, 0, 0, 1, 0) and (1, 0, 1, 0, 0).
Repeating this process for a different gave me
(0, 0, 1, 1, 0).
That's a total of 5 vectors, which is how many the basis should have, correct? Can anyone verify this solution? Is this a correct method to use? I just feel weird "choosing" the a's to be 1 or 0 and stuff.
No. A basis for all of must have 5 vectors but this is a subspace, not all of . Typically each equation restricting values reduces the dimension by 1. Since this space is restricted by one equation, we would expect the dimension to be 4 and expect a basis to contain four vectors.
Yes, and have no restrictions so (0, 1, 0, 0, 0) and (0, 0, 0, 0, 1) are basis vectors for it. NonCommAlg solved the single equation for . You could, of course, solve for either or as functions of the other two.
For example, solving for , we can see that if and , then so (1, 0, 1, 0, 0) is another basis vector. If and , then so (0, 0, 1, 1, 0) is a fourth basis vector. Those four vectors, {(0, 1, 0, 0, 0), (0, 0, 0, 0, 1), (1, 0, 1, 0, 0), (0, 0, 1, 1, 0)} form a basis for this subspace slightly different from the one NonCommAlgebra gave.