12
12.1
Basis and dimension of subspaces
The concept of basis
Example: Consider the set
    


 1
2
0






,
.
,
S=
 2
−1 
1
Then span(S) = R2 . (Exercise). In fact, any two of the elements of S span R2 . (Exercise).
So we can throw out any one of them, for example, the second one, obtaining the set
  

 1

2
 .
Sb =   , 
 2
−1 
And this smaller set Sb also spans R2 . (There are two other possibilities for subsets of S that
also span R2 .) But we can’t discard an element of Sb and still span R2 with the remaining
one vector.
b leaving us with the set
(Why not? Suppose we discard the second vector of S,
 
 1 
S̃ =   .
 2 
Now span(S̃) consists of all scalar multiples of this single vector (a line through 0). But
anything not on this line, for instance the vector
 
1
v= 
0
is not in the span. So S̃ does not span R2 .)
What’s going on here is simple: in the first instance, the three vectors in S are linearly
dependent, and any one of them can be expressed as a linear combination of the remaining
b we have a linearly independent set, and
two. Once we’ve discarded one of these to obtain S,
if we throw away one of these, the span changes.1
1
We use the word “span” in two ways: if V = spanS, then we say that S spans the subspace V .
1
This gives us a way, starting with a more general set S, to discard “redundant” vectors one
by one until we’re left with a set of linearly independent vectors which still spans the original
set: If S = {e1 , . . . , em } spans the subspace V but is linearly dependent, we can express one
of the elements in S as a linear combination of the others. By relabeling if necessary, we
suppose that em can be written as a linear combination of the others. Then
span(S) = span(e1 , . . . , em−1 ). Why?
If the remaining m − 1 vectors are still linearly dependent, we can repeat the process, writing
one of them as a linear combination of the remaining m − 2, relabeling, and then
span(S) = span(e1 , . . . , em−2 ).
We can continue this until we arrive finally at a “minimal” spanning set, say {e1 , . . . , ek }
which is linearly independent. Such a set will be called a basis for V :
Definition: The set B = {e1 , . . . , ek } is a basis for the subspace V if
• span(B) = V .
• The set B is linearly independent.
Remark: In definitions like that given above, we really should put ”iff” (if and only if) instead
of just ”if”, and that’s the way you should read it. More precisely, if B is a basis, then B
spans V and is linearly independent. Conversely, if B spans V and is linearly independent,
then B is a basis.
Examples:
• In R3 , the set
   


1
0

   
   
B =  0 ,  1 ,

   


 0
0
2


0 
 

 
 0  = {e1 , e2 , e3 }
 


1 

is a basis..
Why? (a) Any vector


a
 
 
v= b 
 
c
in R3 can be written as v = ae1 + be2 + ce3 , so B spans R3 . And (b): if c1 e1 + c2 e2 +
c3 e3 = 0, then

c1


0


  

  
 c2  =  0  ,

  
c3
0
which means that c1 = c2 = c3 = 0, so the set is linearly independent.
Definition: The set {e1 , e2 , e3 } is called the standard basis for R3 .
• The set

   


1
3
−1 






S=
,
,
 −2
1
1 
is linearly dependent. Any two elements of S are linearly dependent and form a basis
for R2 . Verify this!
Exercises:
1. Any 4 vectors in R3 are linearly dependent and therefore do not form a basis. You
should be able to supply the argument, which amounts to showing that a certain
homogeneous system of equations has a nontrivial solution.
2. No 2 vectors can span R3 . Why not?
3. The vector 0 is never part of a basis.
4. If a set B is a basis for R3 , then it contains exactly 3 elements. This has mostly been
done in the first two parts, but put it all together.
5. (**) Prove that any basis for Rn has precisely n elements.
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Example: Find a basis for the null space of the matrix

1 0 0 3
2


A =  0 1 0 1 −1

0 0 1 2
3



.

Solution: Since A is already in Gauss-Jordan form, we can just write down the general
solution to the homogeneous equation. These vectors are precisely the elements of the null
space of A. We have, setting x4 = s, and x5 = t,
x1 = −3s − 2t
−s + t
x2 =
x3 = −2s − 3t ,
x4 =
s
x5 =
t
so the general solution to Ax = 0 is given by Null(A) = {sv1 + tv2 s, t ∈ R}, where




−2
−3








 1 
 −1 








v1 =  −2  , and v2 =  −3  .








 0 
 1 




1
0
It is obvious
2
by inspection of the last two entries in each that the set B = {v1 , v2 } is
linearly independent. Furthermore, by construction, the set B spans the null space. So B is
a basis.
12.2
Dimension
As we’ve seen above, any basis for Rn has precisely n elements. Although we’re not going to
prove it here, the same property holds for any subspace of Rn : the number of elements
in any basis for the subspace is the same. Given this, we make the following
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When we say it’s “obvious” or that something is “clear”, we mean that it can easily be proven; if you
can do the proof in your head, fine. Otherwise, write it out.
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Definition: Let V 6= {0} be a subspace of Rn for some n. The dimension of V , written
dim(V ), is the number of elements in any basis of V .
Examples:
• dim(Rn ) = n. Why?
• For the matrix A above, the dimension of the null space of A is 2.
• The subspace V = {0} is a bit peculiar: it doesn’t have a basis according to our
definition, since any subset of V is linearly independent. We extend the definition of
dimension to this case by defining dim(V ) = 0.
Exercises:
1. Show that the dimension of the null space of any matrix R in reduced echelon form is
equal to the number of free variables in the echelon form.
2. Show that the dimension of the set
{(x, y, z) such that 2x − 3y + z = 0}
is two.
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