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.
all you need to do is show that this is row equivalent to the identity matrix
i.e. show that these three vectors are linearly independent
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.
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.
just reduce the matrix with the 3 vectors
and if there are 3 pivots then it is independent right
all those theorems about uniqueness and linear independence are related
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...
sweet, thanks... that would be awesome
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
okay, that you. i appreciate your help again.
I suggest you go with (c) (getting the 3 pivots as you call it) or (g) showing the determinant is non-zero welcome!
anytime
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