# Basis of these vectors

• May 23rd 2012, 10:02 AM
zzizi
Basis of these vectors
Im trying to figure out the following: can someone tell me the process and I will try and work it on my own. I am just not sure how to work it out.

Thanks

Let S be the subspace of R^4 with the vectors as basis:

(1,1,0,-1) (1,3,0,1) (-1,0,1,8)

a) Show that the (6,9,-9,-9)^T is in S
• May 23rd 2012, 02:41 PM
Deveno
Re: Basis of these vectors
solve the equation a(1,1,0,-1) + b(1,3,0,1) + c(-1,0,1,8) = (6,9,-9,-9) for a,b, and c.

this leads to: (a+b-c,a+3b,c,-a+b+8c) = (6,9,-9,-9) or the 4 linear equations:

a+b-c = 6
a+3b = 9
c = -9
-a+b+8c = -9

or, if you prefer, the matrix equation:

$\begin{bmatrix}1&1&-1\\1&3&0\\0&0&1\\-1&1&8 \end{bmatrix} \begin{bmatrix}a\\b\\c \end{bmatrix} = \begin{bmatrix}6\\9\\-9\\-9 \end{bmatrix}$

i can tell you right now, however, that a better question to ask yourself is NOT:

"how do i prove (6,9,-9,-9) is in S?"

but rather:

"how do i tell IF (6,9,-9,-9) is in S?"

S is SUBSPACE of R4, of dimension 3 < 4, so there are going to be LOTS of vectors that AREN'T in S.