Vector Spaces and Dimension. Subspaces of. R n. addition and scalar mutiplication. That is, if u, v in V and alpha in R then ( u + v) Exercise: x

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1 Vector Spaces and Dimension Subspaces of Definition: A non-empty subset V is a subspace of if V is closed under addition and scalar mutiplication. That is, if u, v in V and alpha in R then ( u + v) V and α v V Exercise: x a. Let V = { y z in R 3 such that x 2 + y 2 z 2 = }. Prove that V is not a subspace of b. Let S= V, V 2, V 3, V 4 be any subset of, prove that the linear span of S is a subspace of c. Prove that and O are subspaces of d. Prove that O is contained in every subspace of. e. Prove that the intersection of two subspaces of is a subspace of Definition. If V is a subspace of and M is a subset of V then M is a minimal spanning set of V if (a) M is a spanning set of V (b) No proper subset of M is a spanning set for V. Example: { e, e 2, e 3 } is a minimal spanning set for R 3 Example: { { e, e 2, e 3, e + e 2 } is a spanning set for R 3 which is not a minimal spanning set R 3 for. Theorem: If V is a subspace of and M = { V V m } is a finite spanning set for V then M contains a minimal spanning set for V. proof: If M is not a minimal spanning set then there is a proper subset M M such that is also spanning set for V. M

2 M has n < m elements and is a spanning set. If it is not minimal then there is a spanning set M 2 contained in M with n 2 elements and n 2 < n. This process must end with a minimal spanning set in at most m steps. Definition. If V is a subspace of and M is a subset of V then M is a maximal independent subset of V if (a) M is a linearly independent subset of V (b) Any subset of V which properly contains M is dependent. Another way to say that M is a maximal independent subset of V is to say that M is independent and if v is any element of V which is not in M then {M, v} is dependent. Example: B = { e, e 2, e 3 } is a maximal independent subset of R 3. a proof: If v = b is any member of R 3 which is not in B then { e, e 2, e 3, v} is c dependent since v = a e + b e 2 + c e 3 Observation: No independent subset of can have more than n elememts. proof: Let A A k be independent elements in. Let A be the matrix [ A A k ] Then A has rank at most n since A has n rows. If the colums of A are to be independent then their number must be at most the rank. That is k n Theorem: If M is any linearly independent subset of the vector subspace V can be extended to a maximal independent subset of V. That is if M = { v v k } independent subset of V then there are v k + v s in V such that then M is an v v k, v k + v s is a maximal linearly independent subset of V. Proof: If M is not a maximal independent subset of V then there is w in V such that M 2 = {M, w} is independent. If M 2 is not maximal then repeat the process. In at most {n-k} steps one must have produced a maximal independent set. Theorem: If V is a subspace of then any maximal independent set in V is a minimal spanning set for V and every minimal spanning set for V is a maxumal independent subset of V. Proof: Suppose M = { v v k } is a minimal spanning set of V. Then M is linearly independent. Otherwise some element of M, say is a linear combination of the other v k

3 elements of M. But then, by substituting for v k, evey linear combinaion of { v v k } is a linear combination of { v v k } which would say that M is not a minimal spanning set. To see that M must be a maximal independent set, choose any v in V which is not in M. Since M is a spanning set, v is a linear combination of the elements of M which means that {M,v} is linearly dependent. Conversely, suppose M = { v v k } is a maximal independent set. Then if v is any element of V the set {M, v} is dependent. From this it follows that v is in the linear span of M so M is a spaning set. Since M is independent, no member of M is in the linear span of the other elements. This means that if any element of M is omitted then it is no longer a spanning set. Hence it is a minmal spanning set. Definition: If V is a subspace of then B subset V is a basis for V if it is a minimal spanning sets of V. Equivalently, by the above, B is a basis for V if it is a maximal independent subset of V. Corollary: If V is a subspace and B is a basis for B then every element of V has a unique representation as a linear combination of the elements of B. > <a,b,c>; > > a b c More on Bases and Dimension If V is a vector space then for V. B V is a basis for V if it is a linearly independent spanning set Theorem : Every vector space has a basis. 2Definition: A vector space with a finite basis is said to be a finite dimensional vector space. All other vector spaces are said to be infinte dimensonal.

4 Theorem 3: Suppose B and C are finite bases for the same vector space V. Then the number of elements of B is equal to the number of elements of C. proof: Suppose B has m elements, C n elements and m<n. b c Then there is an m by n matrix A such that A... =... b m c n Since A as more columns than rows an REF, R for A will have a zero last row. There is an invertible matrix H such that HA = R. This means that if H n is the last row of H then H n A = [ ] but H n = [ h h n ] cannot be all zeroes since it is a row of an invertible matrix. This implies that = H n A b c... = H n... b m c n = h c h n c n which implies that C is dependent. 4Definition: If V has a finite basis B then the number of elements in B is called the dimension of V Corollary: If V is an n-dimensional vector space then any linearly independent set in V has at most n elements. Any linearly independent set in V which has n elements is a basis. proof: The previous argument shows that n+ linear combinations of any n vectors (independent or not) has to be linearly dependent. Thus if we take any n+ elements in an n- dimensionl vector space since they are all linear combinations of any basis (which must have n elements), they have to be linearly dependent. If B = { b b n } is an independent set in an n-dimensional vector space V then B must be a basis. If B is not a spanning set then there is b n + in V which is not in the linear span of B. But this means that b b n, b n + is independent since if a b + a 2 b a n b n + a n + b n + = O then all a i =. For if a i + we can solve for b n + as a linear combination of { b b n } so a i + = but that leaves a linear combinaion of { b b n } = O so the other must be since B is independent. 6 Corollary: The possible dimensions of subspaces of are {,,2,... n} Moreover each is possible since if we take a basis B = { b b n } them the linear span of { b b k } will be spanned by an independent set with k elements and therefore have dimension = k. ). Recall by convention the zero subspace has dimension = and the empty set is a basis for it. a i

5 7 Remark: There are infinite dimensional vector spaces.. For instance P, the set of all polynomials is a vector space. It has a basis B = {, x, x 2 x n,...}. However it could not possibly have a finite spanning set S = { f, f 2 f k } because there would have to be an integer w such that w is greater than the degree of any member of S. That would say S P [ w ] which is closed under linear combinations. That would say that the linear span of S = P is contained in P [. But x w P w ] and x w is not in P w We will see below any subspace of a finite dimensional vector space is finite dimensional. Thus if V is a vector space that contains an infinite dimension subspace then it is infinite dimensional too. For example, the set of all power series ( convergent or not), R[[x]] = { a i x i R } is easily seen to be a vector space. Unlike P, there is no simple basis to be written down. However we obviously have P R [[x]] so P is a subspace of R[[x]]. Similarly, if C is all the continuous functions then P subset C so C is infinite dimensional. Theorem 8: Suppose V has a finite spanning set S. Then there is a subset of S which is a basis for V. proof: If V = {} then by convention the empty set is a basis for V. So we can assume V { } Suppose S = { s, s 2 s k } is a spanning set for V. If S is linearly independent then S is a basis for B. If not then S is linearly dependent so some element of S is a linear combination of the others. We can assume s k is a linear combination of S k = { s, s 2 s k } in which is a spanning set for V. So we relace S by and do the same thing. If is S k S k S k S k 2 dependent we have a spanning set with k-2 elements. We have to reach an independent spanning set in at most k steps lest we end up with the empty set as a spanning set for V which can't happen since V { } i = a i 9 Calculation: Theorem 2 says that any spanning set for a vector space V contains a basis for V. Since any set of vectors is a spanning set for its linear span this can also be said as: Any set contains a basis for its linear span. If V = R m then any (finite) set of vectors in R m can be be the columns of a matrix and their linear span is then the column space of the matrix. Thus if V = R m then Theorem 2 says: If A is an m by n matrix then a subset of the columns of A will be a basis for the columm space of A. Finding a spanning set for the column space of a matrix A Theorem 2 : The pivot columns of A are a basis for the column space of A.

6 P3 roblem: Find a basis for the linear span of A = An REF for A is R = The pivot columns of R are numbers,2, 4 so columns,2,4 of A are a basis for the column space of A. Note the RREF is particularly useful is we want to write any column of A as a linear combination of the pivot columns. The RREF of A is 7 Suppose we want to write column of A as a linear combination of the pivot columns of A. Looking at the RREF we see that column is -7*(column ) +*(column 2) + (-2)*(column 4) That is the pivot row entries of any column are of the RREF are the coordinates of that column relative to the basis of pivot columns. Observation: Since the pivot columns are a basis fo the column space we have 4 T heorem If A is a matrix then the dimension of the column space of A is the number of pivots in any REF of A i,e, The dimension of the column space of A is the rank of A Corollary: If A is a matrix with n columns then since n = # pivot variables + number of free variables n = (dimension of the column space of A) + dimension of the null space of A) 6 Note: The fact that the number of pivots is the dimension of the column space for any REF is the proof that the rank is well-defined. The column space is a vector space and its dimension is a fixed integer that depends only on the linear span of the columns and not on any REF. Theorem 7. If V is a vector space of dimension n then any linearly independent set in V can be extended to a basis for V. proof: Let S = { v v t } be an independent set in V. If S is a spanning set then S is a basis and we are done. If S is not a spanning set then there is v t + V which is not in the linear span of S. We claim that = { v v t, v t + } is linearly independent. If not then

7 a v + a 2 v a t v t + a t + v t + = O with not all a i =. This is impossible since if a t + = then that means { v v t } is dependent but if a[t+] <> then we can write solve for v[t+] in the form of a linear combination of { v v t }. However v t + is not a linear combination of { v v t }. Thus { v v t, v t + } is independent. If it is not a basis then we can find a. This must terminate since a vector space of dimension n v t + 8. Extending a linearly independent set to a basis. If B = { v[], `...`, v[s] } is a linearly independent set in then a matrix A with these as columns will have rank s (since its columns are independent). Any basis of has n elements so the problem of "filling B out to a basis" is the same as the follwing: 9. Theorem: If A is any matrix with m rows and s columns and A has rank s then we can add m-s columns to A to produce a matrix of rank m. This is equivalent to saying that A can be completed to a square matrix with non-zero determinant. Such a matrix will be square and will therefore be invertible since its RREF will be the n by n identity matrix. I For any particular matrix there may well be an ad hoc way to add vectors to produce an invertible matrix, or one of non-zero determinant, or one with the right rank (these all say the same thing). However here is a calculation that will always work: a. Calculate G, the consistency matrix of A. b. Form the matrix A* = A G t. That is, augment A but G t If A is m by t and has rank t then an REF of A will have m-t zero rows. This means that the consistency matrix G will be an m-t by m matrix. Thus its transpose, G t will be m by m-t. That is G t has exactly the number of columns that we need to append to A. The fact that the columns of G are rows of an invertible matrix says that they are linearly independent. We will see later that the fact that GA = O implies that the columns of G t and those of A are linearly independent. 2 Example: Complete the matrix to an 3 by 3 matrix of rank 3 2 Calculate an REF of A augmented by the identity 2 2 -> 2 3 G = [ 3 ]

8 G t = 3 2 Let A* = A G t = 3 then the columns of A* are a basis for 2 R 3 ( REF(A*) = Example 2: Fill out to a matrix whose columns are a basis for R > 4 2 G =, A* = REF(A*) = 4 3 Review of the Consistency Matrix:

9 If A is a matrix and <A I> is A augmented by the identity and M = H<A I> = <HA H> where H is an invertible matrix, the product of the elmentary matrices corresponding to the sequence of elementary row operations that convert A to an REF of A. If AX=B is not consistent for all B then A has rank less than the number of its rows and the last m-r rows of HA are all zero where m = row num (A) and r = rank A. Since r = rank A none of the other rows of HA are all. If G is the sub matrix of H whose rows are the last m-r rows of H then G is called the consistency matrix of A because GB = O if and only if the matrix equation AX = B is consistent which is true if and only if the B is in the column space of A. That is the following are equivalent: i. GB = O ii. AX = B is consistent iii. B is in the column space of A 2 Example: Let A = Calculate the consistency matrix of A <A I> -> M = <HA H> > G = is the consistency matrix of A. 3 2 Question: Is AX = solvable? ( equivalently, is in the column space of A?) Answer: G. = is not the zero vector so the answer is no. 2

10 Question: is AX = solvable? ( equivalently, is in the column space of A?) Answer: G = so the answer is yes. Solve AX = <A > = From the above H = so H<A > = = Taking variables x,y,z,w we have x - y +2z-w =, 4y - 4z +8w = -2 with pivot variables y,z. Solving for the pivot variables in terms of the free we have = + z + w = x y z w + z w 2 z 2 w 3 z w 2