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Mathematics 10 Online
OpenStudy (anonymous):

Show that U={u1, u2, u3}, where u1=(1,1,0), and u2=(0,1,0), u3=(-1,0,1) is a basis for R3.

OpenStudy (turingtest):

all you need to do is show that this is row equivalent to the identity matrix

OpenStudy (turingtest):

i.e. show that these three vectors are linearly independent

OpenStudy (turingtest):

You could also set up a 3x3 matrix and take the determinant. If it is non-zero then the set is linearly independent, and 3 linearly independent vectors form a basis for R3.

OpenStudy (anonymous):

if i wanted to put it in an augmented matrix would i put in the zero vector as well and then reduce, or just reduce the matrix with the 3 vectors.

OpenStudy (turingtest):

just reduce the matrix with the 3 vectors

OpenStudy (anonymous):

and if there are 3 pivots then it is independent right

OpenStudy (turingtest):

all those theorems about uniqueness and linear independence are related

OpenStudy (turingtest):

yes, or if the determinant is zero, or like ten other ways to check hold on let me find the list of all the things that are equivalent...

OpenStudy (anonymous):

sweet, thanks... that would be awesome

OpenStudy (turingtest):

Theorem 8 on here shows how all these theorems are connected http://tutorial.math.lamar.edu/Classes/LinAlg/FundamentalSubspaces.aspx proving any one of these proves all the others

OpenStudy (anonymous):

okay, that you. i appreciate your help again.

OpenStudy (turingtest):

I suggest you go with (c) (getting the 3 pivots as you call it) or (g) showing the determinant is non-zero welcome!

OpenStudy (turingtest):

anytime

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