Showing linear independence, correct logic?

1. Feb 17, 2009

1. The problem statement, all variables and given/known data

Let u and v be two nonzero vectors in R^2. If there is no c E R such that u = cv, show that {u, Bv} is a basis of R^2 and that R^2 is a direct sum of the subspaces generated by U = <u> and V = <v> respectively.

2. Relevant equations

Clearly, u and v are linearly independent. So...

3. The attempt at a solution

I want to show r1u + r2Bv = 0 implies r1 = r2 = 0. If B=0 (since v cannot equal 0), then the second term drops out and r1u =0. Because u is non-zero, r1=0. Back to the original statement, if r1=0, then the first term drops out and r2Bv = 0. Since v is non-zero, and assume B is non zero as well, r2 = 0. But what if B IS zero? I'm getting a little confused by this logic...

2. Feb 17, 2009

Dick

If B=0 then {u,Bv}={u,0} is NOT a basis. So, yes, you can assume B is not zero. So assume r1*u+r2*B*v=0. If r1=0 then r2*B*v=0. What can you conclude about r2? If r1 is NOT equal to zero then u=-r2*B*v/r1. What's wrong with that?

3. Feb 17, 2009

Then u is no longer non-zero! It makes sense, now. Thanks very much!

4. Feb 17, 2009

Dick

No, you aren't quite getting it. The problem with writing u=-r2*B*v/r1 is that you were given that there is no number c such that u=c*v.

5. Feb 17, 2009