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:

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

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 R^{4}, of dimension 3 < 4, so there are going to be LOTS of vectors that AREN'T in S.