# [SOLVED] Can (m-1)vectors of R^m generate R^(m-1)?

• Apr 8th 2009, 11:38 PM
Lok
[SOLVED] Can (m-1)vectors of R^m generate R^(m-1)?
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 (:
• Apr 9th 2009, 04:48 AM
Ruun
Can you generate $\displaystyle \mathbb{R}^2$ with 2 vectors of $\displaystyle \mathbb{R}^3$?

If you can, when and why? Then generalize to your spaces.
• Apr 9th 2009, 10:18 AM
Lok
Well I have no clue :(

If (lambda_1,lambda_2,lambda_3,0) can be equal to (x,y,z), then yes. But I would doubt so... I think you can't get R^(m-1) with vectors from R^m since there is one extra coordinate.

Sound right?
• Apr 9th 2009, 11:46 AM
ThePerfectHacker
Quote:

Originally Posted by Lok
Hello,

I'm trying to figure out if (m-1)vectors of R^m can generate R^(m-1)?

No. Because vectors in $\displaystyle \mathbb{R}^m$ has $\displaystyle m$ components. While $\displaystyle \mathbb{R}^{m-1}$ is a set of ordered coordinates that have $\displaystyle m-1$ components, so the two sets can never be equal. However, it is possible to generate a subspace with $\displaystyle m-1$ vectors so that this subspace is isomorphism to $\displaystyle \mathbb{R}^{m-1}$. Just let $\displaystyle v_j = (0,...,1,...,0,0)$ where $\displaystyle 1$ is in the $\displaystyle j$-th location and $\displaystyle 1\leq j\leq m-1$. Then $\displaystyle \text{spam}\{v_1,...,v_{m-1}\}$ would generate a subspace isomorphic to $\displaystyle \mathbb{R}^{m-1}$. But your problem is asking if they can be equal. In that case the answer is no.
• Apr 9th 2009, 01:12 PM
Lok
Thanks a lot