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Mathematics 17 Online
OpenStudy (anonymous):

Does S={(1,-1)(2,1) span R^2?

OpenStudy (phi):

use elimination to see if there are 2 pivots

OpenStudy (anonymous):

elimination? how?

OpenStudy (anonymous):

What do u mean?

OpenStudy (anonymous):

I'm curious, is this linear algebra or multi at your school?

OpenStudy (anonymous):

my textbook never explained it to us using that method

OpenStudy (anonymous):

linear algebra

OpenStudy (anonymous):

neways i wld say that it does span R^2

OpenStudy (phi):

In that case you must show that all linear combinations of (1,-1)(2,1) correspond to all linear combinations reached by (1,0), (0,1) or a matrix made of these vectors has a non-zero determinant

OpenStudy (phi):

|dw:1328905293111:dw| or show they are not dependent (fall on the same line)

OpenStudy (anonymous):

ok got it so i was correct then?

OpenStudy (phi):

Definitely spans R^2, by the picture, you see you can reach any point by moving some distance (plus or minus) along one vector, and then some distance along the second vector.

OpenStudy (anonymous):

ok i think i get it

OpenStudy (anonymous):

so i guess S={(0,2)(1,4) also spans R^2

OpenStudy (turingtest):

yep

OpenStudy (anonymous):

ok so basically if the points are like a multiple of another then it wld form a line so it wldnt span R^2

OpenStudy (turingtest):

if they can be reduced to the identity matrix/ if they are linearly independent/ if they form a matrix with a non-zero determinant they span R2 all those above things imply each other

OpenStudy (turingtest):

(1,0) and (0,1) span R2, right?

OpenStudy (anonymous):

kk i got it thanks.

OpenStudy (turingtest):

|dw:1329769904058:dw|if you think of them in a linear combination you get the above matrix, which you can see has a non-zero determinant

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