Associated Topics || Dr. Math Home || Search Dr. Math

### Determining Basis for a Vector Space

```
Date: 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/
```
Associated Topics:
College Linear Algebra

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search