Can you generate with 2 vectors of ?
If you can, when and why? Then generalize to your spaces.
Hello,
I'm trying to figure out if (m-1)vectors of R^m can generate R^(m-1)?
What I've tried to do so far is this:
Let's take u_1,u_2,u_3 element of R^4 (with m=4)
By definition (I think)
u = sum(lambda_i*u_i,k=1,k=n)
I suppose u = (x,y,z) element of R^3 (m-1), u_1=(1,0,0,0), u_2=(0,1,0,0), u_3=(0,0,1,0)
A=[u_1 u_2 u_3]=
1 0 0
0 1 0
0 0 1
0 0 0
We know A*lamda_matrix=u
[ 1 0 0 ][lambda_1]=[lambda_1]=?? [x]
[ 0 1 0 ][lambda_2] [lambda_2] [y]
[ 0 0 1 ][lambda_3] [lambda_3] [z]
[ 0 0 0 ]...............=.. [0]
(4x3 * 3x1 = 4x1)
(I used ..... so the [0] wouldn't go in the multiplication, forum escapes whitespace)
So can I say that (lambda_1,lambda_2,lambda_3,0)=(x,y,z)? I would doubt so...
If (m-1) vectors from R^m can generate R^(m-1) I would like a proof and if not a counter-example.
Is it the right way to go with this or should I use something else?
Thanks (:
No. Because vectors in has components. While is a set of ordered coordinates that have components, so the two sets can never be equal. However, it is possible to generate a subspace with vectors so that this subspace is isomorphism to . Just let where is in the -th location and . Then would generate a subspace isomorphic to . But your problem is asking if they can be equal. In that case the answer is no.