Sec 4.1 Vector Spaces and Subspaces

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1 Sec 4. Vector Spaces and Subspaces Motivation Let S be the set of all solutions to the differential equation y + y =. Let T be the set of all 2 3 matrices with real entries. These two sets share many common properties: S = the set of all solutions to y + y = T = the set of all 2 3 matrices The sum of two solutions y (x) = sin x and y 2 (x) = cos x to the differential equation, say y 3 (x) = sin x + cos x, is also a solution to the equation. [ [ and [ so is their sum. 6 2 are in T and The zero function is a solution to the equation. y (x) = sin x is a solution to the equation and so is any constant multiple y c (x) = c sin x. In particular y (x) = sin x is also a solution. The zero matrix [ is in T. [ 2 3 is in T and so is [ [ 2 3 c 2c 3c c = for every c 3c 4c constant [ c. In particular [ = is in T It is be- Even though the sets S and T are totally different objects, they resemble each other. cause they are both vector spaces. We give the definition below. Abstract Definition of the Vector Space A vector space (V, +, ) is a nonempty set V of elements, called vectors, together with two operations + and, called addition and scalar multiplication, subject to the following conditions. For all u, v, w V and α, β R,. The sum u + v is in V. 2. u + v = v + u. 3. u + (v + w) = (u + v) + w. 4. There is a special element in V such that u + = + u = u for all u in V. 5. For each u in V, there is a vector u in V such that u + ( u) = ( u) + u =. 6. The scalar multiple of u by α, denoted by α u or simply αu, is again in V. 7. α(u + v) = αu + αv. 8. (α + β)u = αu + βu. 9. α(βu) = (αβ)u.. u = u.

2 2 Remark One can show that the special element in 4 above is actually unique. From now on that element will be called the zero vector. Similarly, for each u in V, u mentioned in 5 above is uniquely determined and is called the additive inverse of u. Remark 2 Instead of writing (V, +, ) all the time, many people simply call V a vector space when the operations + and are obvious from the context. Ex. Consider (R n, +, ). This triple is a vector space. What is the meaning of + in this setting? Ex.2 Show that, for any c R, c =. Ex.3 Is (Z, +, ) a vector space, where + and are usual addition and multiplication of real numbers? Here Z denotes the set of integers. Ex.4 How about (I, +, ), where I is the set of all irrational numbers? Ex.5 Let P n denote the set of all polynomials in t of degree at most n (A typical element in P n looks like p(t) = a + a t + a 2 t a n t n, where a, a, a 2,, a n are in R). Let s define + on P n as follows: for two elements p(t) = a + a t + a 2 t a n t n and q(t) = b + b t + b 2 t b n t n in P n, we define p(t) + q(t) to be p(t) + q(t) = (a + b ) + (a + b )t + (a 2 + b 2 )t (a n + b n )t n. Let s consider scalar multiplication on P n. For α R and p(t) = a + a t + a 2 t a n t n in P n, we define α p(t) by α p(t) = αa + αa t + αa 2 t αa n t n.

3 3 Then it is easy to show that (P n, +, ) is a vector space. If p(t) P n is of the form p(t) a (i.e., this polynomial is constantly a as a function, without depending on t), then p(t) is said to be a constant polynomial. In particular, if z(t), then z(t) is called the zero polynomial. What is the role of the zero polynomial in P n? What is the additive inverse of p(t) = 3 + 2t 4t 2 in P 2? Other Examples of Vector Spaces. C[a, b, the set of all continuous functions on the closed interval [a, b. 2. Diff[a, b, the set of all differentiable functions on [a, b. 3. M m,n, the set of all m n matrices. M n is used to denote M n,n. 4. the set of all solutions to a homogeneous differential equation. Suppose we have a subset H of a vector space (V, +, ). H, combined with + and which were already defined on V, may be a vector space, satisfying all conditions described above. However, to show that H is a vector space itself, it suffices to show that H satisfies three key properties. Definition A subspace of a vector space V is a subset H of V that has three properties: a. The zero vector of V is in H. b. H is closed under vector addition. That is, for each u and v in H, the sum u + v is in H. c. H is closed under multiplication by scalars. That is, for each u in H and each scalar α, the vector αu is in H. Remark 3 Properties a,b, and c above guarantee that a subspace H of V is itself a vector space, under the vector space operations +, already defined in V. In short, a subspace of a vector space is a smaller vector space sitting inside the original one. Ex.6 For a given vector space (V, +, ), consider H = { }, the set consisting only of the zero vector. Then this satisfies the conditions a,b, and c above, so H is a (rather uninteresting) subspace of V, called the zero subspace or trivial subspace. {[ } x Ex.7 Consider (R 2, +, ). Let H = : x R. Clearly H is a subset of R 2. Let s check whether H satisfies the conditions to be a subspace of R 2. Note that R 2 can be identified with the plane. To what in the plane does H correspond?

4 4 Ex.8 Again consider (R 2, +, ). Let K = {[ x x 2 } : x R. Is K a subspace of R 2? Ex.9 Let L = {[ x y } : x and y. Is this a subspace of R 2? Recall that we defined the terminologies like linear dependence/independence, linear combination and span for column vectors. We can now extend these definitions for general vectors (i.e., elements) of a vector space. That is to say, when S = {u, u 2,, u p } is a subset of a vector space V, then a vector v of the form v = α u + α 2 u α p u p, where α, α 2,, α p are scalars, is called a linear combination of u, u 2,, u p with weights (or coefficients) α, α 2,, α p. the span of u, u 2,, u p, denoted by Span{u, u 2,, u p } or simply SpanS, is the set of all linear combinations of u, u 2,, u p. the set S is said to be linearly dependent if there are weights c, c 2,, c p, not all zero, such that c u +c 2 u 2 + +c p u p =. Otherwise, S is said to be linearly independent (i.e., u +u 2 + +u p is the only linear combination of u, u 2,, u p that becomes ). Digression To show that X Y (X is a subset of Y, the notation X Y is also used), we need to show that every element x of X belongs to Y. To show that two sets A = B (A and B coincide), we need to show that A B and B A. Problem Let A be the set of all odd integers. Let B be the set of all integers which can be written as the difference of the squares of two successive numbers. Show that A = B. A B B A

5 5 Ex. What is the span of T = 2 in R3? To justify this claim, we need two inclu- We claim that SpanT = sion relations: Note that SpanT in Ex. is a subspace R 3. This is true in more general setting. In fact, we have the following Theorem If u, u 2,, u p are vectors in a vector space V, then Span{u, u 2,, u p } is a subspace of V. Remark 4 This subspace is called the subspace (of V ) generated (or spanned) by u, u 2,, u p. Proof

6 6 Sec 4.2 Null Spaces and Column Spaces Definition The null space of an m n matrix A, written as NulA, is the set of all solutions to the homogeneous equation A x =. In set notations, [ Ex. Let A = NulA = { x R n : A x = }. 5. Is u = 3 in NulA? 2 Theorem The null space of an m n matrix is a subspace of R n. Proof Ex.2 Let A = Find NulA. Describe NulA as the span of a set of vectors. Definition For an m n matrix A, the column space of A, written as ColA, is Span{ v, v 2,, v n }, where v, v 2,, v n are columns of A. 5 4 Ex.3 Let A =. What is ColA? Is in ColA?

7 7 Theorem 2 The column space of an m n matrix is a subspace of R m. Proof It follows immediately from the theorem stated right before Remark 4 in Sec 4.. Ex.4 Suppose that A is an invertible n n matrix. What is NulA? What is ColA? Summary Let A be an m n matrix. Then u R n is in NulA if and only if A u =. u R m is in ColA if and only if the system A x = u is consistent. Note The table in p.232 of the textbook might help you understand the contrast between NulA and ColA. Ignore statement 8 in both columns of the table. Remark For a matrix A, the row space of A, denoted by Row A, is defined as Row A = Col(A T ). Ex.5 Find Row A, where A is the matrix in Ex.3 above.

8 8 Sec 4.3 Linearly Independent Sets: Bases Review Linear dependence/independence. Let V be a vector space and S = {v,, v p } be a subset of V. Let s consider a linear combination c v + + c p v p of v,, v p. We can ask ourselves the following question: for which values of c,, c p does the linear combination c v + + c p v p become, the zero vector in V? Of course, if we take c = c 2 = = c p =, then clearly c v + + c p v p =. So we have two possibilities. Case : Taking c = c 2 = = c p = is the ONLY way to make c v + + c p v p =. Case 2: There are other choices of c, c 2,, c p, NOT ALL ZERO, such that c v + + c p v p =. When Case happens, the set S = {v,, v p } is said to be linearly independent. When Case 2 happens, the set S = {v,, v p } is said to be linearly dependent. Remark If S = {v,, v p } is linearly dependent, then we can find a vector v i such that v i can be expressed as a linear combination of remaining p vectors. This is why we have the terminology linearly dependent set. If S = {v,, v p } is linearly independent, no vector in S can be written as a linear combination of remaining p vectors. Definition Let H be a subspace of a vector space V (H could be the whole V ). A set B = {u,, u p } in V is said to be a basis for H if (i) B is linearly independent, and (ii) SpanB = H. Ex. Let B = {[ [, Claim : the set B is linearly independent. }. We will show that this is a basis for the vector space R 2. Claim 2: SpanB = R 2. Ex.2 Let B = {[ [, }. Show that B is another basis for R 2.

9 9 Note Ex. and Ex.2 show that a vector space can have more than one basis. 2 Ex.3 Is the set S = a basis for R3? Ex.4 Is the set T = {[ [, [, } a basis for R 2? Ex.5 Find a basis for P 3. Do the same for P n. Ex.6 Is S always a basis for SpanS? Remark 2 The trivial subspace { } in V does NOT have a basis at all. The Spanning Set Theorem Let S = {v,, v p } be a set in a vector space V, and let H = Span{v,, v p }. If one of vectors in S, say v k, is a linear combination of the remaining vectors in S, then the set formed from S by removing v k still spans H. That is, H = SpanS = Span(S \ {v k }) = Span{v, v 2,, v k, v k+,, v p }. Remark 3 If v k in S is a linear combination of the remaining vectors in S, then S must be linearly dependent. So S cannot be a basis for H = SpanS. However, S \ {v k } still spans H. Keep removing vectors in S which are unnecessary to span H until no element can be written as a linear combination of others. This procedure produces a basis for H = SpanS. Ex.7 Let s consider a set S = 3 in R3. Find a basis for SpanS.

10 For an m n matrix A, we learned that ColA is a subspace of R m. How can we find a basis for ColA? Theorem The pivot columns of a matrix A form a basis for ColA. 4 2 Ex.8 Let A = then it row-reduces to SpanS = ColA is a basis for ColA Therefore S = In other words, S is linearly independent and Ex.9 How to find a basis for NulA? First of all, to find a basis for NulA, one need to find NulA first. For example, for the matrix in Ex.8 above, we see that the general solution is given by x = 4x 2 2x 4 x 2 is free x 3 = x 4. x 4 is free x 5 = That is, NulA = x 2 show that the set basis for NulA x : x 2, x 4 are free = Span 4 is linearly independent and hence 2. We can 4 2 is a Summary Let A be an m n matrix. To get a basis for ColA, find the rref of A and collect all pivot columns. To get a basis for NulA, find the general solution of the system A x = (To do this, you will probably also need to use the rref of A) and express the general solution as the span of as many vectors as free variables. Those vectors form a basis for NulA.

11 Sec 4.5 The Dimension of a Vector Space Let H be a subspace of a vector space V. In general, H has more than one basis. For example, look at Ex. and Ex.2 in Sec 4.3. However, we have the following Theorem If a vector space V has a basis of n vectors, then every basis of V must consist of exactly n vectors. Ex. We observed that R 2 has a basis consisting of 2 vectors. Thus every basis of R 2 must have exactly vectors. Definition If a vector space V is spanned by a finite set, then V is said to be finite-dimensional, and the dimension of V, written as dim V, is the number of vectors in a (so every) basis for V. Remark Some vector spaces are not finite-dimensional. In this case, the vector spaces are said to be infinite-dimensional. Examples are Diff[a, b, C[a, b, etc. Remark 2 The dimension of the trivial vector space (=zero vector space) { } is defined to be zero. Ex.2 dim R n = and dim P n =. Ex.3 Let H = Span. What is the dimension of H? Ex.4 Let K = Span 2. What is the dimension of K? In general, dim Span{v, v 2,, v p } is at most p. Moreover, dim Span{v, v 2,, v p } = p if and only if

12 2 Theorem 2 Let H be a subspace of a finite-dimensional vector space V. Then dim H dim V and the equality holds if and only if H = V. Ex.5 Suppose that V is an n-dimensional vector space and S is a subset of V containing m vectors with m > n. Then S must be linearly. Ex.6 (Theorem 2 in Sec.7 revisited) Let S = { v, v 2,, v p } R n with p > n, then S is linearly dependent. The Basis Theorem Let V be a p-dimensional vector space. (i) If S = {v, v 2,, v p } is a linearly independent subset of V, then SpanS = V, i.e., S is a basis for V. (ii) If T = {w, w 2,, w p } spans V, then T is linearly independent, i.e., T is a basis for V. Ex.7 Show that S = is a basis for R3. Ex.8 Show that T = is a basis for R 4.

13 3 Sec 4.6 Rank Definition Let A be an m n matrix. The rank of A, denoted by ranka, is the dimension of the column space of A. The nullity of A, denoted by nullitya, is the dimension of the null space of A Ex. Let A = then it row-reduces to. The null space of A is given by NulA = Span. Therefore ranka = dim ColA = and nullitya = dim NulA = Remark Recall that pivot columns of A form a basis for ColA, so the rank of A is the number of pivots of A. Since the dimension of NulA equals the number of free variables in the solutions to the homogeneous system A x =, we obtain the following The Rank Theorem Let A be an m n matrix. Then ranka + nullitya = n. The Invertible Matrix Theorem, continued Let A be an n n square matrix. Then the following statements are each equivalent to the statement that A is an invertible matrix. (k) det A. (l) The columns of A form a basis for R n. (m) ColA = R n. (n) dim ColA = n. (o) ranka = n. (p) NulA = { }. (q) nullitya =.

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