Solution: (a) S 1 = span. (b) S 2 = R n, x 1. x 1 + x 2 + x 3 + x 4 = 0. x 4 Solution: S 5 = x 2. x 3. (b) The standard basis vectors

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1 .. Dimension In this section, we introduce the notion of dimension for a subspace. For a finite set, we can measure its size by counting its elements. We are interested in a measure of size on subspaces as well. Counting the number of elements will not suffice here since all nonzero subspaces have infinitely many elements. However, bases of subspaces are finite sets, of which we can measure the size. Definition... Suppose S R n is a subspace. The dimension of S, denoted dim(s), is the size of any basis of S. Example..2. What are the dimensions of the subspaces in Example.2.2? (a) S = span (b) S 2 = R n, (c) S = { }, ([ 2]), (d) S 4 = span( u,..., u k ), where ( u,..., u k ) is linearly independent, x (e) S 5 = x 2 x x + x 2 + x + x 4 =. x 4 ([ (a) is a basis for S 2]), so dim(s ) =. (b) The standard basis vectors.,.,...,. in Rn form a basis of R n, so dim(r n ) = n. You likely could have guessed this, given that R n is called n-dimensional Euclidean space. (c) ( ) is a basis for { }, so dim({ }) = the size of ( ) =. (d) ( u,..., u k ) is a basis for S 4, so dim(s 4 ) = k. (e) We can describe the solution to x + 2x 2 + x + 4x 4 = as x = 2x 2 x 4x 4 with x 2, x, x 4 free, so 2x 2 x 4x S 5 = x 2 x = x 2 + x + x 4 x = span,,. 2 4 Thus, dim(s 5 ) = because,, is linearly independent by Proposition.2.8.

2 2 Do you see any potential issues with Definition..? What if a subspace S had a basis of size 2 and a basis of size? Should dim(s) be 2 or? Could we compromise and say dim(s) = 2.5? What if S had a bases of size 2,, and 7? The point is that if a subspace S had two bases of different sizes, Definition.. would not make sense. Fortunately, this is never an issue. Theorem... Suppose S R n be a subspace. Any two bases of S have the same size. Proof. See Section.2, Problem 9b or Problem 5. Algorithms.2.4 and.2. give us a method for building up a basis for subspace S starting from the empty list and for removing vectors from a spanning list for S to get a basis for S. We now formalize the implications of these algorithms in Proposition..4 and Corollaries..6 and..5. In addition, we can say how many vectors we need to add or remove because we know each basis for S has size dim(s). Proposition..4. Suppose S R n is a subspace and u,..., u k S. () If ( u,..., u k ) is linearly independent, then dim(s) k vectors can be added to ( u,..., u k ) to make a basis for S. (2) If ( u,..., u k ) spans S, then k dim(s) vectors can be removed from ( u,..., u k ) to make a basis for S. Proof. These facts follows immediately from Proposition.2.6, Algorithm.2., and the fact that any basis of S must have size dim(s). Corollary..5. Any two of the following are sufficient for B S to be a basis for S: (i) span(b) = S. (ii) B is linearly independent. (iii) B = dim(s). Proof. Given (i) and (ii), B is a basis for S by definition. Given (i) and (iii), by Proposition..4, we can remove vectors from B to get a basis for S, so B is a basis for S. Given (ii) and (iii), by Proposition..4, we can add vectors to B to get a basis for S, so B is a basis for S. Corollary..6. Suppose S, S 2 are subspaces of R n with S S 2, then dim(s ) dim(s 2 ), with equality if and only if S = S 2. In particular, each subspace of R n except R n has dimension less than n. Proof. Let B be a basis for S, so dim(s ) = B. Since B is a linearly independent list of vectors in S S 2, we can add dim(s 2 ) B vectors to B to get a basis for S 2. By definition of dimension, dim(s ) dim(s 2 ). If dim(s ) = dim(s 2 ), then B is linearly independent and B = dim(s 2 ), so B is also a basis for S 2 by Corollary..5, forcing S 2 = span(b) = S. On the other hand, if S = S 2 then clearly, dim(s ) = dim(s 2 ). The final statement follows from the first with S 2 = R n. Now that we have the necessary definitions, we can state and prove one of the most important Theorems in linear algebra, the Rank-Nullity Theorem. For any linear map T : R n R m, intuitively, the more vectors that T sends to, the less vectors T hits. In other words, the larger ker(t ), the smaller range(t ) is. The Rank-Nullity Theorem makes this more precise using our measurement of size for a subspace, dimension. We begin by defining rank and nullity for maps and matrices and going through an example in order to motivate the result and clarify the proof.

3 Definition..7. For any linear map T : R n R m, define rank(t ) = dim(range(t )), nullity(t ) = dim(ker(t )), rank([t ]) = rank(t ), nullity([t ]) = nullity(t ). Example..8. Suppose T : R 5 R 4 is linear with [T ] = A = [ ] a... a 5 with 2 5 RREF(A) = 4 (a) Find a basis for range(a), and then find rank(t ), rank(a). (b) Find a basis for ker(a), and then find nullity(t ), nullity(a). (c) What is rank(t ) + nullity(t )? (a) The first two columns of RREF(A) are the pivot columns, so by Algorithm.2., ( a, a 2 ) is a basis for range(a). Hence, rank(t ) = rank(a) = 2. (b) Columns, 4, 5 are free, which means a solution description to A x = has x, x 4, x 5 free, and then x = 2x x 4 5x 5, x 2 = x x 4 5x 5. Thus, 2x x 4 5x x x 4 5x 5 ker(t ) = x = x x = span,, 5. x Since,, 5 is linearly independent - in fact, this list must be linearly independent by Proposition.2.8, it forms a basis for ker(t ). Hence, nullity(t ) = nullity(a) =. (c) Notice rank(t )+nullity(t ) = 2+ = 5, which is the dimension of the domain. The Rank-Nullity Theorem says that this is no coincidence. Theorem..9 (Rank-Nullity). Suppose T : R n R m is linear. Then, dim(range(t )) + dim(ker(t )) = n. Proof. Let [T ] = [ a... a n ] and R = RREF(A), which are both m n matrices. We have range(t ) = span( a,..., a n ) and ker(t ) = ker([t ]) = ker(r). Then, by Algorithm.2., range(t ) has a basis whose size is the number of pivot columns of R, so dim(range(t )) = # of pivot columns of R In addition, by Proposition.2.8, ker(t ) = ker(r) has a basis whose size is the number of free columns of R, so dim(ker(t )) = # of free columns of R

4 4 Since each column of R is either pivot or free, dim(range(t )) + dim(ker(t )) = # of columns of R = n. Example... Find linear maps T : R 4 R with each of the possible ranks and nullities. rank(t ) nullity(t ) Example [T ] range(t ) ker(t ) 4 { } R 4 span span,, 2 2 span, span, R span 4 Does not exist There does not exist a linear map T : R 4 R with rank(t ) = 4 because there is no 4-dimensional subspace of R. The Rank-Nullity Theorem limits the possibilities for range(t ) and ker(t ) for linear maps T : R n R m. In fact, it is the only limit on what range(t ) and ker(t ) can be, provided that they are subspaces, Proposition... Proposition... Given any subspaces S R n, S 2 R m satisfying dim(s ) + dim(s 2 ) = n, there exists a linear map T : R n R m so that Proof. Problem 5. ker(t ) = S, range(t ) = S 2. Exercises:. Determine the ranks and nullities of the following matrices, given in RREF. 2 (a) (b) 2 2. What are the possible ranks and nullities for linear maps T : R R 4? Give an example for each possibility.

5 . Given an positive integers m, n, what are minimum and maximum possible values of rank(a) and nullity(a) for all m n matrices A? Describe examples to show these minimum and maximum values are possible. 4. Determine if a linear map T : R n R m with ker(t ) = S and range(t ) = S 2, with appropriate m, n exists. (a) S = span ([ 2, S 2 = span. 2]) (b) S = span 2, S 2 = span,. (c) S = span,, (d) S = span, 2, S 2 = span 2. S 2 = span,. (e) S = span,, S 2 = span,. 5 (f) S = span,, S 2 = span 2, (g) S = span 2, S 2 = span,, For each of the parts in Exercise (4) for which such a map exists, find T : R n R m with ker(t ) = S and range(t ) = S 2. I recommend finding a general construction first, Problem 5. Problems:. (2) Suppose S R n is a subspace. Show that the minimum size of a list of vectors which spans S is dim(s). This requires showing size dim(s) is possible, but any smaller size is impossible. 2. (2) Suppose S R n is a subspace. Show that the maximum size of a linearly independent list of vectors in S is dim(s). This requires showing size dim(s) is possible, but any larger size is impossible.. () Suppose A is an m n matrix. Show that every solution description to A x = has the same number of free variables. 4. () Suppose A is an m n matrix, and b R m. Show that if A x = b has a solution, then every solution description to A x = b has the same number of free variables. 5

6 6 5. () Suppose T : R n R m is linear map and S is a subspace of R n. Recall that T (S) = {T ( x) x S}. Show that dim(t (S)) dim(s). 6. () Suppose T : R n R m is an injective linear map and S is a subspace of R n. Recall that T (S) = {T ( x) x S}. Show that dim(t (S)) = dim(s). 7. () Suppose T : R n R m is a linear map satisfying dim(t (S)) = dim(s) for all subspaces S of R n. Show that T is injective. 8. Suppose S R m is a subspace. Show that there exists a linear map T : R n R m with range(t ) = S if and only if n dim(s). 9. () Suppose S R m is a subspace with dim(s) = n. Show that there exists an injective linear map T : R n R m with range(t ) = S.. () If S R n and S 2 R m are subspaces with dim(s ) = dim(s 2 ), show that there exists a linear map T : R n R m so that T (S ) = S 2.. The purpose of this problem is to give another proof of the Rank-Nullity Theorem. Suppose T : R n R m is linear. Let ( u,..., u k ) be a basis for ker(t ), and extend it to a basis ( u,..., u k, v,..., v r ) of R n. (a.) (2) Show that r + k = n. (b.) (2) Show that range(t ) = span(t ( v ),..., T ( v r )). (c.) () Show that (T ( v ),..., T ( v r )) is linearly independent. (d.) (2) Conclude that dim(ker(t )) + dim(range(t )) = n. 2. (2+) Suppose T : R n R m is linear. Let ( u,..., u k ) be a basis for ker(t ), and extend it to a basis ( u,..., u k, v,..., v r ) of R n. Use the Rank-Nullity Theorem to show that (T ( v ),..., T ( v r )) is linearly independent.. (2) Suppose A, C are m n matrices with A C. Show that rank(a) = rank(c). 4. (2+) For what values of n does there exists a map T : R n R n with ker(t ) = range(t )? Justify your answer. 5. Prove Proposition (2) Prove or give a counterexample: For any m n matrix A, rank(a) is the number of nonzero rows in RREF(A). 7. (2) Prove or give a counterexample: For any m n matrix A, nullity(a) is the number of rows of all zeros in RREF(A). 8. (+) Suppose S R n is a subspace. Show that the minimum number of relations for a relation form of S is n dim(s). This requires showing that n dim(s) relations is possible, but any less is impossible.

7 9. (4) If T, T 2 : R n R m are linear maps with range(t ) range(t 2 ), show that there exists a linear map U : R n R n so that T = T 2 U. 2. () If T, T 2 : R n R m are linear maps with range(t ) = range(t 2 ), show that there exists an invertible linear map U : R n R n so that T 2 = T U. 2. (4) If T, T 2 : R n R m are linear maps with ker(t ) ker(t 2 ), show that there exists a linear map U : R m R m so that T 2 = U T. 22. () If T, T 2 : R n R m are linear maps with ker(t ) = ker(t 2 ), show that there exists an invertible linear map U : R m R m so that T 2 = U T. 2. () If T, T 2 : R n R m are linear maps with range(t ) = range(t 2 ) and ker(t ) = ker(t 2 ), show that there exists an invertible linear map U : range(t ) range(t ) so that T 2 = U T. 24. () If T, T 2 : R n R m are linear maps with rank(t ) = rank(t 2 ), show that there exists an invertible linear map U : R m R m so that range(t 2 ) = range(u T ). 25. () If T, T 2 : R n R m are linear maps with rank(t ) = rank(t 2 ), show that there exists an invertible linear map W : R m R m so that ker(t 2 ) = ker(t W ). 26. () If T, T 2 : R n R m are linear maps with rank(t ) = rank(t 2 ), show that there exists invertible linear maps U : R n R n and W : R m R m so that T 2 = W T U. 27. (5) Suppose T : R n R n is a nilpotent linear map. Show that T n = O. 7