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Determining if the following sets span R^3 and creating a basis, wee! matrices |
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| Dec11-05, 07:56 PM | #1 |
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Determining if the following sets span R^3 and creating a basis, wee! matrices
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. |
| Dec11-05, 08:03 PM | #2 |
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Recognitions:
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{[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]}. |
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