Question

find the basis of the subspace of R^3 that's spanned by the vectors. v1=(1,0,0) v2=(1,1,0) v3=(2,1,0) v4=(0,-1,0)

Answer 1

now a basis is defined as a spanning set using a minimum of vectors, right?l

Answer 2

right

Answer 3

can you row reduce these? or would that take to long?

Answer 4

into row echelon form?

Answer 5

1 1 2 0
0 1 1-1
0 0 0 0

yeah

Answer 6

I'm confused on that..am i suppose to make the supper triangle all 0's or the lower triangle?

Answer 7

upper*

Answer 8

1 0 0
1 1 0
2 1 0
0 -1 0


1 0 0
0 1 0
1 1 0
0 -1 0


1 0 0
0 1 0
0 1 0
0 -1 0

1 0 0
0 1 0
0 -1 0


1 0 0
0 1 0
0 0 0

100 and 010 if i did the process roght

Answer 9

i dont think i did it right ...

Answer 10

isn't it 4 columns?

Answer 11

wasnt sure, its been awhile, was about to try it

1 1 2 0
0 1 1-1
0 0 0 0

Answer 12

1 0 1 1
0 1 1-1
0 0 0 0

i think this means v1 and v2 we can always check and see

Answer 13

so check for determinant not 0?

Answer 14

let A be the vector span v1 v2 v3 v4

we dont have a square matrix

Answer 15

see if for some vector a,b that

av1 + bv2 = v3 and v4

Answer 16

v1=(1,0,0) v2=(1,1,0) v3=(2,1,0) v4=(0,-1,0)

1a +1b = 2
0a +1b = 1, b=1, so a=1 works
0a +0b = 0


1a +1b = 0
0a +1b = -1, b=-1, so a=1 works
0a +0b = 0

Answer 17

so v1 and v2 form a basis for the span since we can find some vector a,b that creates a linear combination of all the vectors present

Answer 18

lol, i just remembered, tht the long version and the echelon version tell us the same thing

Answer 19

1 1 2 0
0 1 1-1
0 0 0 0

to

1 0 1 1
0 1 1-1
0 0 0 0
^ ^ 1,-1,0
^
^
1,1,0

so the first 2 vectors reduce to a square matrix, and the other 2 reduce to the linear solutions results

Answer 20

ok

Answer 21

i have another question. can i just type in here?

Answer 22

sure