Finite Dimensional Vector Spaces and Bases If a vector space V is ...

6. Finite Dimesnional Vector Spaces and Bases

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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, R n 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 , x 2 ,…, x n ) ∈ R n is expressible as a linear combination of these vectors: €

X = x1€ E1 + x 2 E2 + L + x n 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, x 2 ,…, x k } . € €

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• For any finite set S, the vector space Fun(S,R) is finite dimensional since the set of characteristic functions { χ 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) ⋅ χ s (t) s∈S

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.

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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 , S 2 ,…, Sk } ; we need to show that k ≤ n. So consider this set of vectors in U: € T1 = {Sk , X 1 , X 2 ,…, X n } . Since Sk ∈ U = L(X1, X2,…, Xn), we can express Sk as a linear combination of the X’s: € €

(*)

Sk = a1 X 1 + a 2 X 2 + L + a n X n . €

At least one of the a’s must be nonzero, for otherwise, Sk = 0, which contradicts the € assumption that S is a linearly independent set. By renumbering the X’s if necessary, we may €

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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,…, Xn–1); since the opposite € is true, too, then U = L( Sk , X1, X2,…, Xn–1). Next, consider € the set € T2 = {Sk − 1 , Sk , X 1 , X 2 ,…, X n − 1 } . Arguing as above, we know that Sk − 1 ∈ L(€Sk , X1, X2,…, Xn–1), so we can find a linear relation of the form €

€ (**) Sk − 1 = rk Sk + a1 X 1 + a 2 X 2 + L + a n − 1 X n − 1 for appropriately chosen scalars. Again, at least one of the a’s 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 X’s, we may assume now that an − 1 ≠ 0 . We can then solve for Xn–1 in (**). This implies that any vector in U = L( Sk , X1, X2,…, Xn–1) is also a vector in € L( Sk − 1 , Sk , X 1 , X 2 ,…, X n − 2 ); since the converse is also true, then U = L( Sk − 1 , Sk , X 1 , X 2 ,…, X n − 2 ).

€ € €

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Each time we repeat this argument, we express U in terms of a new set of n vectors, replacing one of the X’s with one of the S’s. If k > n, then at the nth stage of this argument, we would have replaced all of the X’s with S’s, showing that U = L( Sk − n +1 , Sk − n + 2 ,K, Sk ).

€

But then S1 ∈ U, but S1 is not among the vectors Sk − n +1 , Sk −€ n + 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 , S 2 ,…, 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 €

€

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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 R n = 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 n canonical or standard basis for R . € dimPk(R) = k + 1 since {1, x, x 2 ,…, x k } is a basis, the canonical basis for Pk(R). € €

€

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• If S is a finite set with |S| elements, then dim Fun(S,R) = |S|, for we saw above that { χ 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, x 2 , x 3 ,…} 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, x 2 , x 3 ,…} , 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. //

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

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The importance of finding a basis s described in the following theorem. Theorem Let B = { A 1 , A 2 ,…, A n } 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 , x 2 ,…, x n depending only on € X, called the coordinates of X relative to the basis B, so that X = x1 A 1 + x 2 A 2 + L + x n A n . € 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 x1 A 1 + x 2 A 2 + L + x n A n = x1′ A 1 + x ′2 A 2 + L + xn′ A n where x1 , x 2 ,…, x n and x1′ , x ′2 ,…, x ′n are two sets of coordinates for X relative to the basis B. But then € €

€

(x1 − x ′1 )A 1 + (x 2 − x ′2 )A 2 + L + (x n − xn′ )A n = 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. //

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Examples: • B = {(1,1,0), (0,1,1), (1, 0,1)} is a basis for R 3 different from the canonical basis {E1, E2 ,E3 }. Every vector in R 3 is expressible uniquely in coordinates relative to B, as follows: to write €

€

€ X = (x€ 1, 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 a’s: a1 = 12 (x1 + x 2 − x3 ), a2 = 12 (−x1 + x2 + x 3 ), and € 1 a3 = 2 (x1 − x 2 + x3 ).

€ € €

• B = {1, 1 + x, 1 + x + x 2 } is a basis for P2(R) different € from the canonical basis {1, x, x 2 }. Every vector in P2(R) is expressible uniquely in coordinates relative to B, as follows: to write €2 p(x) = p 0 + p1x + p2 x = a1 ⋅1 + a2 (1 + x) + a3 (1 + x + x 2 )

€

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we expand and collect terms to obtain the system of equations p0 = a1 + a2 + a3 p1 =

a2 + a 3

p2 =

a3

and find the unique solution for the a’s: a1 = p0 − p1, a2 = p1 − p 2 , and a3 = p2 . €

€

At the end of the previous section, we proved a € that showed that € if S is a set of vectors that theorem 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.

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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 , S 2 ,…, 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 , S 2 ,…, Sm , X 1 } is a linearly independent set, for if there are scalars for which € r1S1 + r2S 2 + L + rm Sm + a1 X 1 = 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 , S 2 ,…, Sm , X 1 , 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. //

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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 , S 2 ,…, Sn } , there are scalars that satisfy r1S1 + r2S 2 + L + rn Sn + 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. //

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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 . //