These notes closely follow the presentation of the material given in David C. Laystextbook Linear Algebra and its Applications (3rd edition). These notes are intendedprimarily for in-class presentation and should not be regarded as a substitute forthoroughly reading the textbook itself and working through the exercises therein.

The Dimension of a Vector SpaceRecall 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 everybasis for V contains exactly n vectors.

Proof Suppose that V is a vector space and that B v1,v2, ,vn is a basis forV (containing exactly n vectors). We will first prove that any set of vectors in Vthat contains more than n vectors cannot be a basis for V.

Let C w1,w2, ,wn, ,wm be a set of vectors in V that contains morethan 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 alinear combination of the vectors in B. That is,

w1 a11v1 a12v2 a1nvnw2 a21v1 a22v2 a2nvn

wn an1v1 an2v2 annvn

wm am1v1 am2v2 amnvn.

Now consider the equationc1w1 c2w2 cnwn cmwm 0V.

This equation can be written asc1a11v1 a12v2 a1nvn c2a21v1 a22v2 a2nvn cnan1v1 an2v2 annvn cmam1v1 am2v2 amnvn

0Vor as

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c1a11 c2a21 cnan1 cmam1v1 c1a12 c2a22 cnan2 cmam2v2 c1a1n c2a2n cnann cmamnvn

0V.Since the set B is linearly independent, all of the coefficients (weights) in theabove equation must be zero. Thus,

a11c1 a21c2 an1cn am1cm 0a12c1 a22c2 an2cn am2cm 0

a1nc1 a2nc2 anncn amncm 0.

The above system is a homogeneous linear system with more unknowns (m) thanequations (n). Such a system must have nontrivial solutions. Thus, there existscalars c1,c2,cn, ,cm, not all zero, that satisfy the above system. These samescalars also satisfy the vector equation

c1w1 c2w2 cnwn cmwm 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 nobasis for V can contain more than n vectors but, if we think about it, we haveactually also proved that no basis for V can contain fewer than n vectors. Thereason is as follows: If we have a basis for V that contains exactly m vectors wherem n, then the argument given above shows that no basis for V can contain morethan m vectors. In particular, no basis for V can contain n vectors, but thiscontradicts 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 anybasis of V. If each basis of V contains only a finite number, n, of vectors, then wesay that V is an ndimensional vector space. If V does not have a basis consistingof a finite number of vectors, then we say that V is an infinitedimensional vectorspace. The trivial vector space, V 0, is said to have dimension zero (eventhough 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 Mmn, consisting of

all m n matrices, has dimension m n.Example For each positive integer n, the vector space Pn, which consists of all

polynomial functions p : that have degree less than or equal to n, hasdimension n 1.

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Example The vector space P, which consists of all polynomial functions p : , isinfinitedimensional.

Example The vector space C0, which consists of all continuous functions f : , isinfinite dimensional.

Example If V is any nontrivial vector space and S v1,v2,,vn is any linearlyindependent set of vectors in V, then the vector space SpanS (which is a subspaceof V) has dimension n.

Example If A is an m n matrix, then nulA, the nullspace of A, has dimension equal tothe number of nonpivot columns of A. Also, colA, the column space of A hasdimension equal to the number of pivot columns of A. This tells us that

dimnulA dimcolA n.As a specific example of this, we can immediately observe for the matrix

A 1 2 3 44 3 0 60 1 0 3

,

thatdimnulA dimcolA 4.

To verify that this is correct, find bases for nulA and colA.

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