Determining if the following sets span R^3 and creating a basis, wee matrices

mr_coffee
Messages
1,613
Reaction score
1
Hello everyone, The problem says:Determine which, if, any, of the following sets span R^3?
b. {[1 1 -1], [-1 1 1]} note: I'm just transposing them, they should be verical.
c. {[1 1 0], [0 1 1], [1 0 0], [1 0 1]}
Those are the 2 sets, he showed us how to do them but i got lost on his steps:
he writes:
Intially b does not span R^3, which makes sense, because there are only 2 vector sets in b. He then goes on and writes
[1 1 -1] [-1 1 1] * [x y z];
and says use the dot product and name the first column which is, [1 1 -1] x1, then the 2nd column [-1 1 1], x2 and use the dot product.
he comes out with:
x+y-z = 0;
-x + y + z = 0;
he then writes:
y = a;
2y = 0;
y = 0;
x = z;

he totally lost me and now your probably lost as well. Anywho from that he got:
[1 0 1] which says will make b span R^3 and you would get:
{[1 1 -1], [-1 1 1], [1 0 1]}
Any ideas on how he got that last vector? Also i know if the dot porduct of 2 vectors is 0, then its linear indepdant.

He then says, create 3 bases for R^3 using the above sets. he showed us this:
[1 1 0] [0 1 1] [1 0 0] [1 0 1]
[1 0 1] = -[1 1 0] + [ 0 1 1] + 2[1 0 0]
so he said u don't need
[1 0 1], but he kind of did this by looking at it, is there a systematic approach to solving it? Thanks.
 
Last edited:
Physics news on Phys.org
{[1 1 -1], [-1 1 1]} is not a basis of R^3 as it has only two vectors. You can check they are linearly independant, so to get a basis for R^3 you can add any vector not in span{[1 1 -1], [-1 1 1]} as this will be an independant set (why?) of 3 vectors in R^3 and hence is a basis.

To find a vector not in span{[1 1 -1], [-1 1 1]}, he's finding a vector that is orthogonal to every vector in span{[1 1 -1], [-1 1 1]}.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top