1. ## subsets

Consider the vectors:

[-1,1,3]

[2,-1,0]

[1,1,5]

[-5,1,1]

Find a subset of the vectors that forms a basis for the space spanned by the vectors.

How do i start?

i have tried to find vectors linearly independant from one other but failed

Consider the vectors:

[-1,1,3]

[2,-1,0]

[1,1,5]

[-5,1,1]

Find a subset of the vectors that forms a basis for the space spanned by the vectors.
Method: make the vectors into the rows of a matrix.
$\begin{bmatrix}-1&1&3\\ 2&-1&0\\ 1&1&5\\ -5&1&1\end{bmatrix}$
Carry out a Gaussian row-reduction on the matrix. Ignore any rows that end up as consisting of all zeros. The remaining (nonzero) rows tell you which of the original vectors form a basis for the space that they span. Namely, you take the vectors that were originally in those nonzero rows.

3. Originally Posted by Opalg
Method: make the vectors into the rows of a matrix.
$\begin{bmatrix}-1&1&3\\ 2&-1&0\\ 1&1&5\\ -5&1&1\end{bmatrix}$
Carry out a Gaussian row-reduction on the matrix. Ignore any rows that end up as consisting of all zeros. The remaining (nonzero) rows tell you which of the original vectors form a basis for the space that they span. Namely, you take the vectors that were originally in those nonzero rows.
Ive done this but there seems to be no nonzero rows, have i done it right, if so what does this show?

I have got

[INDENT] $\begin{bmatrix}-1&1&5\\ 0&1&6\\ 0&0&-4\\ 0&0&10\end{bmatrix}$

Ive done this but there seems to be no nonzero rows, have i done it right, if so what does this show?

I have got
$\begin{bmatrix}-1&1&5\\ 0&1&6\\ 0&0&-4\\ 0&0&10\end{bmatrix}$
Multiply row 3 by –1/4. Then subtract 10 times the new row 3 from row 4, to get $\begin{bmatrix}-1&1&5\\ 0&1&6\\ 0&0&1\\ 0&0&0\end{bmatrix}$. The bottom row now has all zeros. So you can ditch the vector [–5,1,1] that was originally in row 4, and the remaining three original vectors will form a basis.

5. Originally Posted by Opalg
Multiply row 3 by –1/4. Then subtract 10 times the new row 3 from row 4, to get $\begin{bmatrix}-1&1&5\\ 0&1&6\\ 0&0&1\\ 0&0&0\end{bmatrix}$. The bottom row now has all zeros. So you can ditch the vector [–5,1,1] that was originally in row 4, and the remaining three original vectors will form a basis.
aaahhh its that easy, thanks