Download The Dimension of a Vector Space

These notes closely follow the presentation of the material given in David C. Lay’s
textbook Linear Algebra and its Applications (3rd edition). These notes are intended
primarily for in-class presentation and should not be regarded as a substitute for
thoroughly reading the textbook itself and working through the exercises therein.
The Dimension of a Vector Space
Recall that a basis for a vector space, V, is a set of vectors in V that is linearly
independent and spans V.
Theorem If a vector space, V, has a basis that contains exactly n vectors, then every
basis for V contains exactly n vectors.
Proof
Suppose that V is a vector space and that B v 1 , v 2 ,  , v n  is a basis for
V (containing exactly n vectors). We will first prove that any set of vectors in V
that contains more than n vectors cannot be a basis for V.
Let C w 1 , w 2 ,  , w n ,  , w m  be a set of vectors in V that contains more
than n vectors. (We are assuming that C contains exactly m vectors where m  n.).
Since B is a basis for V, every vector in C can be written (in a unique way) as a
linear combination of the vectors in B. That is,
w 1  a 11 v 1  a 12 v 2    a 1n v n
w 2  a 21 v 1  a 22 v 2    a 2n v n

w n  a n1 v 1  a n2 v 2    a nn v n

w m  a m1 v 1  a m2 v 2    a mn v n .
Now consider the equation
c 1w1  c 2w2    c nwn    c mwm  0V.
This equation can be written as
c 1 a 11 v 1  a 12 v 2    a 1n v n 
 c 2 a 21 v 1  a 22 v 2    a 2n v n 

 c n a n1 v 1  a n2 v 2    a nn v n 

 c m a m1 v 1  a m2 v 2    a mn v n 
 0V
or as
1
c 1 a 11  c 2 a 21    c n a n1    c m a m1 v 1
 c 1 a 12  c 2 a 22    c n a n2    c m a m2 v 2

 c 1 a 1n  c 2 a 2n    c n a nn    c m a mn v n
 0V.
Since the set B is linearly independent, all of the coefficients (weights) in the
above equation must be zero. Thus,
a 11 c 1  a 21 c 2    a n1 c n    a m1 c m  0
a 12 c 1  a 22 c 2    a n2 c n    a m2 c m  0

a 1n c 1  a 2n c 2    a nn c n    a mn c m  0.
The above system is a homogeneous linear system with more unknowns (m) than
equations (n). Such a system must have non–trivial solutions. Thus, there exist
scalars c 1 , c 2 ,  c n ,  , c m , not all zero, that satisfy the above system. These same
scalars also satisfy the vector equation
c 1w1  c 2w2    c nwn    c mwm  0V,
which shows that the set C is linearly dependent. Thus, C cannot be a basis for V.
We have proved that if V has a basis containing exactly n vectors, then no
basis for V can contain more than n vectors – but, if we think about it, we have
actually also proved that no basis for V can contain fewer than n vectors. The
reason is as follows: If we have a basis for V that contains exactly m vectors where
m  n, then the argument given above shows that no basis for V can contain more
than m vectors. In particular, no basis for V can contain n vectors, but this
contradicts our assumption that B is a basis for V.
Definition The dimension of a vector space is defined to be the number of vectors in any
basis of V. If each basis of V contains only a finite number, n, of vectors, then we
say that V is an n–dimensional vector space. If V does not have a basis consisting
of a finite number of vectors, then we say that V is an infinite–dimensional vector
space. The trivial vector space, V  0, is said to have dimension zero (even
though it actually has no basis).
Example For each positive integer n, the vector space  n has dimension n.
Example For each pair of positive integers m and n, the vector space M mn , consisting of
all m  n matrices, has dimension m  n.
Example For each positive integer n, the vector space P n , which consists of all
polynomial functions p :    that have degree less than or equal to n, has
dimension n  1.
2
Example The vector space P, which consists of all polynomial functions p :   , is
infinite–dimensional.
Example The vector space C 0 , which consists of all continuous functions f :   , is
infinite dimensional.
Example If V is any non–trivial vector space and S  v 1 ,v 2 ,,v n  is any linearly
independent set of vectors in V, then the vector space SpanS (which is a subspace
of V) has dimension n.
Example If A is an m  n matrix, then nulA, the nullspace of A, has dimension equal to
the number of non–pivot columns of A. Also, colA, the column space of A has
dimension equal to the number of pivot columns of A. This tells us that
dimnulA  dimcolA  n.
As a specific example of this, we can immediately observe for the matrix
1
A
2 3 4
4 3 0 6
0
,
1 0 3
that
dimnulA  dimcolA  4.
To verify that this is correct, find bases for nulA and colA.
3