FFinite dimensional vector space and basess

Finite Dimensional Vector Spaces and Bases

If a vector space V is spanned by a finite number of vectors, we say that it is finite dimensional. Most of the vector spaces we treat in this course are finite dimensional.

Examples: For any positive integer n,

Rn is a finite dimensional vector space. Indeed, the set of vectors

{E1 = (1,0,0,,0),

E2 = (0,1,0,,0),L

En = (0,0,0,,1)}

is a spanning set for the vector space, since every vector X = (

x1, x2,, xn )

Rn is expressible as a linear combination of these vectors:

X =

x1E1 + x2E2 +L + xn En

For any positive integer k, Pk(R) is finite dimensional. Every polynomial

p(x ) Pk(R) is clearly a linear combination of the k + 1 polynomials in the set

{1, x, x2,, xk } .

6. Finite Dimesnional Vector Spaces and Bases 1

For any finite set S, the vector space Fun(S,R) is finite dimensional since the set of characteristic functions

{s |s S} (one characteristic function for each element of S) is a finite set with the property that any function f in Fun(S,R) is a linear combination of characteristic functions:

f (t) = f (s) s (t)sS

A direct corollary to the theorem we last proved is the important

Theorem If V is a finite dimensional vector space, then there is a finite set B of vectors in V that(1) spans V (that is, V = L(B) ), and (2) is linearly independent. //

Any set of linearly independent vectors in a vector space which span the space is called a basis for V.

Thus, the examples above all describe bases for their respective vector spaces. Note that B is not uniquely determined; there are in general many different bases for the same vector space.


However, while there may be more than one basis for a given vector space, there is one invariant that stays the same from basis to basis in the same vector space: its size. To prove this, we first work through a long but technically useful result.

Proposition Let the n vectors X1, X2,, Xn from the vector space V determine the subspace L(X1, X2,, Xn). If S is any set of linearly independent vectors in U = L(X1, X2,, Xn), then S can contain no more than n vectors.

Proof Suppose S =

{S1,S2 ,,Sk } ; we need to show that k n. So consider this set of vectors in U:

T1 = {Sk , X1 , X 2,, Xn } .

Since

Sk U = L(X1, X2,, Xn), we can express

Sk as a linear combination of the Xs:

(*)

Sk = a1X1 + a2X 2 +L + an Xn .

At least one of the as must be nonzero, for otherwise,

Sk = 0, which contradicts the assumption that S is a linearly independent set. By renumbering the Xs if necessary, we may


assume that

an 0. So we can solve for Xn in (*). This implies that any vector in U = L(X1, X2,, Xn) is also a vector in L(

Sk, X1, X2,, Xn1); since the opposite is true, too, then U = L(

Sk, X1, X2,, Xn1).

Next, consider the set

T2 = {Sk1,Sk , X1, X 2,, Xn1 } .

Arguing as above, we know that

Sk1 L(

Sk, X1, X2,, Xn1), so we can find a linear relation of the form

(**)

Sk1 = rkSk + a1X1 + a2X 2 +L + an1Xn1

for appropriately chosen scalars. Again, at least one of the as must be nonzero, otherwise there is a linear relation amongst vectors of S in violation of the assumption that S is a linearly independent set. So by renumbering the Xs, we may assume now that

an1 0. We can then solve for Xn1 in (**). This implies that any vector in U = L(

Sk, X1, X2,, Xn1) is also a vector in L(

Sk1,Sk , X1 , X 2,, Xn 2); since the converse is also true, then U = L(

Sk1,Sk , X1 , X 2,, Xn 2).


Each time we repeat this argument, we express U in terms of a new set of n vectors, replacing one of the Xs with one of the Ss. If k > n, then at the nth stage of this argument, we would have replaced all of the Xs with Ss, showing that

U = L(

Skn +1,Skn +2,K,Sk ).

But then

S1 U, but

S1 is not among the vectors

Skn +1,Skn +2,K,Sk , implying that there is a linear dependence amongst the vectors in S, again a violation of the assumption that S is a linearly independent set. This impossibility means that k cannot be larger than n, so k n, completing the proof. //

Theorem Every finite dimensional vector space V has a basis, and any two bases for V have the same number of vectors.

Proof The first theorem above shows why V has a basis, so we need only prove the second statement here. Suppose then that

S = {S1,S2 ,,Sm } and

T = {T1, T2,, Tn } are both bases for V. Since T is a basis, V is spanned by the vectors in T. But then S is a linearly independent set of vectors in


V = L(T1, T2,, Tn), so by the previous proposition, we conclude that m n. Then, by reversing the roles of S and T in this argument, we conclude that n m. So m = n, and we are done. //

An important consequence of this theorem is that while a finite dimensional vector space can have many bases, they must all have the same size. We can then define the dimension of the vector space V to be the size of any basis for it. It is denoted dimV.

Examples:

dim Rn = n since

{E1 = (1,0,0,,0),

E2 = (0,1,0,,0),L

En = (0,0,0,,1)}

is a basis, as we saw earlier; it is called the canonical or standard basis for

Rn .

dimPk(R)

= k + 1 since

{1, x, x2,, xk } is a basis, the canonical basis for Pk(R).


If S is a finite set with |S| elements, then dim Fun(S,R) = |S|, for we saw above that

{s |s S} is a basis for this vector space.

Now not all vector spaces are finite dimensional.

Proposition P(R) is infinite dimensional.

Proof The infinite set

{1, x, x2, x3 ,} of powers of the variable x is a linearly independent set in P(R), for there can be no nontrivial linear combination of any finitely many of these powers of x that is identically equal to the zero polynomial. Indeed, any finite subset of this infinite set is linearly dependent as well. So if P(R) were finite dimensional, it would have a basis B of finite size, say n. Simply by taking m > n vectors from

{1, x, x2, x3 ,} , we would then have more than n linearly dependent vectors in the vector space P(R) = L(B). This contradicts the conclusion of the proposition we proved earlier, so this situation is not possible. Thus, P(R) must be infinite dimensional. //


Proposition If S is an infinite set, then Fun(S,R) is infinite dimensional.

Proof

{s |s S} is a linearly independent set in Fun(S,R) having infinitely many elements. //

Henceforth, we will make a standing assumption (unless otherwise explicitly mentioned) that all vector spaces we study are finite dimensional.


The importance of finding a basis s described in the following theorem.

Theorem Let B = {

A1, A 2,, An } be a basis for the n-dimensional vector space V. Then every

X V is uniquely expressible as a linear combination of the vectors in B, that is, there exists a unique collection of scalars

x1, x2,, xn depending only on X, called the coordinates of X relative to the basis B, so that

X = x1A1 + x2A 2 +L + xn An .

Proof Since B spans V, it is always possible to express X as a linear combination of the vectors in B; our task then is to show that this can be done in only one way. But if we could express X in more than one way as a linear combination of the vectors in B, then we could write

x1A1 + x2A 2 +L + xn An = x 1A1 + x 2A 2 +L + x n Anwhere

x1, x2,, xn and

x 1, x 2,, x n are two sets of coordinates for X relative to the basis B. But then

(x1 x 1 )A1 + (x2 x 2 )A 2 +L + (xn x n )An = 0

would be a nontrivial linear combination of the vectors in B equal to the zero vector, in violation of the fact that B is linearly independent. //


Examples:

B = {(1,1,0), (0,1,1), (1,0,1)} is a basis for

R3 different from the canonical basis

{E1,E2,E3 }. Every vector in

R3 is expressible uniquely in coordinates relative to B, as follows: to write

X = (x1, x2, x3) = a1(1,1,0)+ a2(0,1,1) +a3(1,0,1)

we expand to the system of equations

x1 = a1 + a3

x2 = a1 + a2

x3 = a2 + a3

and find the unique solution for the as:

a1 = 12 (x1 + x2 x3),

a2 = 12 (x1 + x2 + x3), and

a3 = 12 (x1 x2 + x3 ).

B = {1, 1+ x, 1+ x + x2 } is a basis for P2(R) different from the canonical basis

{1, x,x2 }. Every vector in P2(R) is expressible uniquely in coordinates relative to B, as follows: to write

p(x) = p0 + p1x + p2x2 = a1 1+ a2(1+ x) +a3(1+ x + x

2)


we expand and collect terms to obtain the system of equations

p0 = a1 + a2 + a3

p1 = a2 +a3

p2 = a3

and find the unique solution for the as:

a1 = p0 p1,

a2 = p1 p2, and

a3 = p2.

At the end of the previous section, we proved a theorem that showed that if S is a set of vectors that spans a finite dimensional vector space V, then there is a subset B of S that forms a basis for V, that is, for which V = L(B). In other words, every spanning set of vectors in a vector space can be trimmed to form a basis. The following theorem is a companion result. It says that any linearly independent set of vectors can be extended to form a basis.


Theorem If S is a set of linearly independent vectors in a finite dimensional vector space V, then S is a subset of a basis B for V; that is, V = L(B).

Proof Let

S = {S1,S2 ,,Sm } . If V = L(S), we can take B = S and there is nothing to prove. Otherwise, S does not span V, so there is some vector X1

V which is not in L(S). Then

T1 = {S1,S2 ,,Sm , X1 } is a linearly independent set, for if there are scalars for which

r1S1 + r2S2 +L + rmSm + a1X1 = 0 ,

then we must have

a1 = 0 (else we can solve for X1 and show that X1

L(S), a contradiction), which then gives a linear relation amongst the linearly independent vectors in S, another contradiction. If V = L(

T1), then B =

T1 is a basis for V. If not, there is some vector X2

V which is not in L(

T1). So

T2 = {S1,S2 ,,Sm , X1, X 2 } is a linearly independent set (why?). If V = L(

T2), then B =

T2 is a basis for V. Otherwise, we can continue expanding our set. Eventually, this process must end, for when the size of the set reaches dimV, it must span V. //


Proposition Suppose V is a vector space of dimension n. Then(1) any set of n linearly independent vectors in V spans the space, so is a basis for V, and(2) any set of n vectors in V that span the space is linearly independent, so is a basis for V. //

Proof (1) Since dimV = n, no set S of linearly independent vectors can have size greater than n. Thus, expanding S by including any other vector X

V must produce a linearly dependent set. That is, where

S = {S1,S2 ,,Sn } , there are scalars that satisfy

r1S1 + r2S2 +L + rnSn + aX = 0 . But a 0 (else S is not linearly independent), so we can solve the equation for X and show that X

L(S). This argument shows that V = L(S). So S spans the space and is therefore a basis for V.

(2) If S is a spanning set for V , then by the previous theorem, it can be expanded to form a basis; but it already has size n, and no basis for V can have more than n vectors. Thus S must already be a basis and it is linearly independent. //


These ideas can now be used to describe the relationship between a vector space and its subspaces with respect to dimension.

Theorem If V is a finite dimensional vector space and U is a subspace, then U is finite dimensional and dimU dimV .

Proof If U were not finite dimensional, then it would contain a set of m linearly independent vectors for aribitrarily large values of m, in particular, for m > dimV . But since U is a subspace of V, this would give a set of more than dimV linearly independent vectors in V, which is not possible. Thus, not only is U finite dimensional, but any set of linearly independent vectors in U has size dimV . Thus any basis for U has size dimU dimV . //

Corollary If V is a (finite dimensional) vector space and U is a subspace of the same dimension, then U = V . //


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FFinite dimensional vector space and basess