Determining Basis for a Vector SpaceDate: 11/15/96 at 22:00:45 From: Mike Webb Subject: Linear algebra Which of the following sets of vectors form a basis for |R^3? (a) (3,1,2) (b) (3,1,2) (c) (1,2,3) (d) (2,1,3) (1,1,2) (1,1,2) (0,2,7) (1,1,2) (7,1,2) (0,0,9) (2,2,1) (1,4,2) I get stuck dealing with bases in |R^3 because there are three components in each vector. My teacher tries to explain it to me but will not do the work so I have no clue what is going on. Date: 11/18/96 at 12:04:59 From: Doctor Brian Subject: Re: Linear algebra I guess the first thing to remember is the definition of a basis. A basis is a linearly independent set of vectors that "span" a space. In other words, for a set of vectors to form a basis, you have to be able to express any ordered triple (x,y,z) in R^3 as a linear combination of those vectors, and you can't be able to express a vector in the set as a combination of other vectors in the set. For (a) you can't express every possible ordered triple as a combination of those two vectors. A counterexample is (0,0,1). Imagine that you could write (0,0,1) as a combination: a(3,1,2) + b(1,1,2) = (0,0,1). This leads to the following three equations: 3a + b = 0, a + b = 0, and 2a + 2b = 1. That's an inconsistent system (compare the second and third equations). The two vectors don't span all of R^3, so they don't make a basis. In fact, two vectors alone can't ever span R^3, since you always end up with three equations in two variables. For (b) there are three vectors, so there's hope that it might be a basis since taking a linear combination will give three equations in three variables. However, the set is not linearly independent. We can solve the system: a(3,1,2) + b(1,1,2) = (7,1,2) to give a=3, b=-2. That means that (7,1,2) doesn't generate any triples that we couldn't already get by using combinations of (3,1,2) and (1,1,2). Since these three vectors are not linearly independent, they aren't a basis. For (c) we have three vectors. It's impossible to write (1,2,3) as a linear combination of (0,2,7) and (0,0,9)...any combo would have 0 in the first coordinate position. Also, there's no scalar that you can multiply by (0,2,7) to get (0,0,9) and vice versa, so those two vectors are independent. So we have three linearly independent vectors. Can we get any possible (x,y,z)? We'd have a(1,2,3) + b(0,2,7) + c(0,0,9). That's three equations in three variables. A solution does exist, and it's easy to find, by solving first for c and substituting back for b and then a. So our set does span R^3, which means it forms a basis. For (d) it might be intuitive that four vectors will be too many to form a basis. The easiest way to show that these vectors do not form a basis is to show that the set is not linearly independent...Try solving: a(2,1,3) + b(1,1,2) + c(2,2,1) = (1,4,2) It solves to a=-3,b=5,c=1. One of the vectors is a linear combination of the others, so this is not a basis. Hope this was of some use! -Doctor Brian, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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