LINEAR ALGEBRA W W L CHEN
|
|
- Ralf Owens
- 5 years ago
- Views:
Transcription
1 LINEAR ALGEBRA W W L CHEN c W W L Chen, 994, 28. This chapter is available free to all individuals, on the understanding that it is not to be used for financial gain, and may be downloaded and/or photocopied, with or without permission from the author. However, this document may not be kept on any information storage and retrieval system without permission from the author, unless such system is not accessible to any individuals other than its owners. Chapter 5 INTRODUCTION TO VECTOR SPACES 5.. Real Vector Spaces Before we give any formal definition of a vector space, we shall consider a few concrete examples of such an abstract object. We first study two examples from the theory of vectors which we first discussed in Chapter 4. Example 5... Consider the set R 2 of all vectors of the form u = (u, u 2 ), where u, u 2 R. Consider vector addition and also multiplication of vectors by real numbers. It is easy to check that we have the following properties: (.) For every u, v R 2, we have u + v R 2. (.2) For every u, v, w R 2, we have u + (v + w) = (u + v) + w. (.3) For every u R 2, we have u + = + u = u. (.4) For every u R 2, we have u + ( u) =. (.5) For every u, v R 2, we have u + v = v + u. (2.) For every c R and u R 2, we have cu R 2. (2.2) For every c R and u, v R 2, we have c(u + v) = cu + cv. (2.3) For every a, b R and u R 2, we have (a + b)u = au + bu. (2.4) For every a, b R and u R 2, we have (ab)u = a(bu). (2.5) For every u R 2, we have u = u. Example Consider the set R 3 of all vectors of the form u = (u, u 2, u 3 ), where u, u 2, u 3 R. Consider vector addition and also multiplication of vectors by real numbers. It is easy to check that we have properties analogous to (.) (.5) and (2.) (2.5) in the previous example, with reference to R 2 being replaced by R 3. We next turn to an example from the theory of matrices which we first discussed in Chapter 2. Chapter 5 : Introduction to Vector Spaces page of 7
2 Example Consider the set M 2,2 (R) of all 2 2 matrices with entries in R. Consider matrix addition and also multiplication of matrices by real numbers. Denote by O the 2 2 null matrix. It is easy to check that we have the following properties: (.) For every P, Q M 2,2 (R), we have P + Q M 2,2 (R). (.2) For every P, Q, R M 2,2 (R), we have P + (Q + R) = (P + Q) + R. (.3) For every P M 2,2 (R), we have P + O = O + P = P. (.4) For every P M 2,2 (R), we have P + ( P ) = O. (.5) For every P, Q M 2,2 (R), we have P + Q = Q + P. (2.) For every c R and P M 2,2 (R), we have cp M 2,2 (R). (2.2) For every c R and P, Q M 2,2 (R), we have c(p + Q) = cp + cq. (2.3) For every a, b R and P M 2,2 (R), we have (a + b)p = ap + bp. (2.4) For every a, b R and P M 2,2 (R), we have (ab)p = a(bp ). (2.5) For every P M 2,2 (R), we have P = P. We also turn to an example from the theory of functions. Example Consider the set A of all functions of the form f : R R. For any two functions f, g A, define the function f + g : R R by writing (f + g)(x) = f(x) + g(x) for every x R. For every function f A and every number c R, define the function cf : R R by writing (cf)(x) = cf(x) for every x R. Denote by λ : R R the function where λ(x) = for every x R. Then it is easy to check that we have the following properties: (.) For every f, g A, we have f + g A. (.2) For every f, g, h A, we have f + (g + h) = (f + g) + h. (.3) For every f A, we have f + λ = λ + f = f. (.4) For every f A, we have f + ( f) = λ. (.5) For every f, g A, we have f + g = g + f. (2.) For every c R and f A, we have cf A. (2.2) For every c R and f, g A, we have c(f + g) = cf + cg. (2.3) For every a, b R and f A, we have (a + b)f = af + bf. (2.4) For every a, b R and f A, we have (ab)f = a(bf). (2.5) For every f A, we have f = f. There are many more examples of sets where properties analogous to (.) (.5) and (2.) (2.5) in the four examples above hold. This apparent similarity leads us to consider an abstract object which will incorporate all these individual cases as examples. We say that these examples are all vector spaces over R. Definition. A vector space V over R, or a real vector space V, is a set of objects, known as vectors, together with vector addition + and multiplication of vectors by element of R, and satisfying the following properties: (VA) For every u, v V, we have u + v V. (VA2) For every u, v, w V, we have u + (v + w) = (u + v) + w. (VA3) There exists an element V such that for every u V, we have u + = + u = u. (VA4) For every u V, there exists u V such that u + ( u) =. (VA5) For every u, v V, we have u + v = v + u. (SM) For every c R and u V, we have cu V. (SM2) For every c R and u, v V, we have c(u + v) = cu + cv. (SM3) For every a, b R and u V, we have (a + b)u = au + bu. (SM4) For every a, b R and u V, we have (ab)u = a(bu). (SM5) For every u V, we have u = u. Remark. The elements a, b, c R discussed in (SM) (SM5) are known as scalars. Multiplication of vectors by elements of R is sometimes known as scalar multiplication. Chapter 5 : Introduction to Vector Spaces page 2 of 7
3 Example Let n N. Consider the set R n of all vectors of the form u = (u,..., u n ), where u,..., u n R. For any two vectors u = (u,..., u n ) and v = (v,..., v n ) in R n and any number c R, write u + v = (u + v,..., u n + v n ) and cu = (cu,..., cu n ). To check (VA), simply note that u +v,..., u n +v n R. To check (VA2), note that if w = (w,..., w n ), then u + (v + w) = (u,..., u n ) + (v + w,..., v n + w n ) = (u + (v + w ),..., u n + (v n + w n )) = ((u + v ) + w,..., (u n + v n ) + w n ) = (u + v,..., u n + v n ) + (w,..., w n ) = (u + v) + w. If we take to be the zero vector (,..., ), then u + = + u = u, giving (VA3). Next, writing u = ( u,..., u n ), we have u + ( u) =, giving (VA4). To check (VA5), note that u + v = (u + v,..., u n + v n ) = (v + u,..., v n + u n ) = v + u. To check (SM), simply note that cu,..., cu n R. To check (SM2), note that c(u + v) = c(u + v,..., u n + v n ) = (c(u + v ),..., c(u n + v n )) To check (SM3), note that To check (SM4), note that = (cu + cv,..., cu n + cv n ) = (cu,..., cu n ) + (cv,..., cv n ) = cu + cv. (a + b)u = ((a + b)u,..., (a + b)u n ) = (au + bu,..., au n + bu n ) = (au,..., au n ) + (bu,..., bu n ) = au + bu. (ab)u = ((ab)u,..., (ab)u n ) = (a(bu ),..., a(bu n )) = a(bu,..., bu n ) = a(bu). Finally, to check (SM5), note that u = (u,..., u n ) = (u,..., u n ) = u. It follows that R n is a vector space over R. This is known as the n-dimensional euclidean space. Example Let k N. Consider the set P k of all polynomials of the form p(x) = p + p x p k x k, where p, p,..., p k R. In other words, P k is the set of all polynomials of degree at most k and with coefficients in R. For any two polynomials p(x) = p + p x p k x k and q(x) = q + q x q k x k in P k and for any number c R, write p(x) + q(x) = (p + q ) + (p + q )x (p k + q k )x k and cp(x) = cp + cp x cp k x k. To check (VA), simply note that p + q,..., p k + q k R. To check (VA2), note that if we write r(x) = r + r x r k x k, then we have p(x) + (q(x) + r(x)) = (p + p x p k x k ) + ((q + r ) + (q + r )x (q k + r k )x k ) = (p + (q + r )) + (p + (q + r ))x (p k + (q k + r k ))x k = ((p + q ) + r ) + ((p + q ) + r )x ((p k + q k ) + r k )x k = ((p + q ) + (p + q )x (p k + q k )x k ) + (r + r x r k x k ) = (p(x) + q(x)) + r(x). Chapter 5 : Introduction to Vector Spaces page 3 of 7
4 If we take to be the zero polynomial + x x k, then p(x) + = + p(x) = p(x), giving (VA3). Next, writing p(x) = p p x... p k x k, we have p(x) + ( p(x)) =, giving (VA4). To check (VA5), note that p(x) + q(x) = (p + q ) + (p + q )x (p k + q k )x k = (q + p ) + (q + p )x (q k + p k )x k = q(x) + p(x). To check (SM), simply note that cp,..., cp k R. To check (SM2), note that To check (SM3), note that To check (SM4), note that c(p(x) + q(x)) = c((p + q ) + (p + q )x (p k + q k )x k ) = c(p + q ) + c(p + q )x c(p k + q k )x k = (cp + cq ) + (cp + cq )x (cp k + cq k )x k = (cp + cp x cp k x k ) + (cq + cq x cq k x k ) = cp(x) + cq(x). (a + b)p(x) = (a + b)p + (a + b)p x (a + b)p k x k = (ap + bp ) + (ap + bp )x (ap k + bp k )x k = (ap + ap x ap k x k ) + (bp + bp x bp k x k ) = ap(x) + bp(x). (ab)p(x) = (ab)p + (ab)p x (ab)p k x k = a(bp ) + a(bp )x a(bp k )x k Finally, to check (SM5), note that = a(bp + bp x bp k x k ) = a(bp(x)). p(x) = p + p x p k x k = p + p x p k x k = p(x). It follows that P k is a vector space over R. Note also that the vectors are the polynomials. There are a few simple properties of vector spaces that we can deduce easily from the definition. PROPOSITION 5A. Suppose that V is a vector space over R, and that u V and c R. (a) We have u =. (b) We have c =. (c) We have ( )u = u. (d) If cu =, then c = or u =. Proof. (a) By (SM), we have u V. Hence It follows that u + u = ( + )u (by (SM3)), = u (since R). u = u + (by (VA3)), = u + (u + ( (u))) (by (VA4)), = (u + u) + ( (u)) (by (VA2)), = u + ( (u)) (from above), = (by (VA4)). Chapter 5 : Introduction to Vector Spaces page 4 of 7
5 (b) By (SM), we have c V. Hence c + c = c( + ) (by (SM2)), = c (by (VA3)). It follows that c = c + (by (VA3)), = c + (c + ( (c))) (by (VA4)), = (c + c) + ( (c)) (by (VA2)), = c + ( (c)) (from above), = (by (VA4)). (c) We have ( )u = ( )u + (by (VA3)), = ( )u + (u + ( u)) (by (VA4)), = (( )u + u) + ( u) (by (VA2)), = (( )u + u) + ( u) (by (SM5)), = (( ) + )u + ( u) (by (SM3)), = u + ( u) (since R), = + ( u) (from (a)), = u (by (VA3)). (d) Suppose that cu = and c. Then c R and u = u (by (SM5)), = (c c)u (since c R \ {}), = c (cu) (by (SM4)), = c (assumption), = (from (b)), as required Subspaces Example Consider the vector space R 2 of all points (x, y), where x, y R. Let L be a line through the origin = (, ). Suppose that L is represented by the equation αx + βy = ; in other words, L = {(x, y) R 2 : αx + βy = }. Note first of all that = (, ) L, so that (VA3) and (VA4) clearly hold in L. Also (VA2) and (VA5) clearly hold in L. To check (VA), note that if (x, y), (u, v) L, then αx + βy = and αu + βv =, so that α(x + u) + β(y + v) =, whence (x, y) + (u, v) = (x + u, y + v) L. Next, note that (SM2) (SM5) clearly hold in L. To check (SM), note that if (x, y) L, then αx + βy =, so that α(cx) + β(cy) =, whence c(x, y) = (cx, cy) L. It follows that L forms a vector space over R. In fact, we have shown that every line in R 2 through the origin is a vector space over R. Chapter 5 : Introduction to Vector Spaces page 5 of 7
6 Definition. Suppose that V is a vector space over R, and that W is a subset of V. Then we say that W is a subspace of V if W forms a vector space over R under the vector addition and scalar multiplication defined in V. Example We have just shown in Example 5.2. that every line in R 2 through the origin is a subspace of R 2. On the other hand, if we work through the example again, then it is clear that we have really only checked conditions (VA) and (SM) for L, and that = (, ) L. PROPOSITION 5B. Suppose that V is a vector space over R, and that W is a non-empty subset of V. Then W is a subspace of V if the following conditions are satisfied: (SP) For every u, v W, we have u + v W. (SP2) For every c R and u W, we have cu W. Proof. To show that W is a vector space over R, it remains to check that W satisfies (VA2) (VA5) and (SM2) (SM5). To check (VA3) and (VA4) for W, it clearly suffices to check that W. Since W is non-empty, there exists u W. Then it follows from (SP2) and Proposition 5A(a) that = u W. The remaining conditions (VA2), (VA5) and (SM2) (SM5) hold for all vectors in V, and hence also for all vectors in W. Example Consider the vector space R 3 of all points (x, y, z), where x, y, z R. Let P be a plane through the origin = (,, ). Suppose that P is represented by the equation αx + βy + γz = ; in other words, P = {(x, y, z) R 2 : αx + βy + γz = }. To check (SP), note that if (x, y, z), (u, v, w) P, then αx + βy + γz = and αu + βv + γw =, so that α(x + u) + β(y + v) + γ(z + w) =, whence (x, y, z) + (u, v, w) = (x + u, y + v, z + w) P. To check (SP2), note that if (x, y, z) P, then αx + βy + γz =, so that α(cx) + β(cy) + γ(cz) =, whence c(x, y, z) = (cx, cy, cz) P. It follows that P is a subspace of R 3. Next, let L be a line through the origin = (,, ). Suppose that (α, β, γ) R 3 is a non-zero point on L. Then we can write L = {t(α, β, γ) : t R}. Suppose that u = t(α, β, γ) L and v = s(α, β, γ) L, and that c R. Then u + v = t(α, β, γ) + s(α, β, γ) = (t + s)(α, β, γ) L, giving (SP). Also, cu = c(t(α, β, γ)) = (ct)(α, β, γ) L, giving (SP2). It follows that L is a subspace of R 3. Finally, it is not difficult to see that both {} and R 3 are subspaces of R 3. Example Note that R 2 is not a subspace of R 3. First of all, R 2 is not a subset of R 3. Note also that vector addition and scalar multiplication are different in R 2 and R 3. Example Suppose that A is an m n matrix and is the m zero column matrix. Consider the system Ax = of m homogeneous linear equations in the n unknowns x,..., x n, where x x =. x n is interpreted as an element of the vector space R n, with usual vector addition and scalar multiplication. Let S denote the set of all solutions of the system. Suppose that x, y S and c R. Then A(x + y) = Ax + Ay = + =, giving (SP). Also, A(cx) = c(ax) = c =, giving (SP2). It follows that S is a subspace of R n. To summarize, the space of solutions of a system of m homogeneous linear equations in n unknowns is a subspace of R n. Chapter 5 : Introduction to Vector Spaces page 6 of 7
7 Example As a special case of Example 5.2.5, note that if we take two non-parallel planes in R 3 through the origin = (,, ), then the intersection of these two planes is clearly a line through the origin. However, each plane is a homogeneous equation in the three unknowns x, y, z R. It follows that the intersection of the two planes is the collection of all solutions (x, y, z) R 3 of the system formed by the two homogeneous equations in the three unknowns x, y, z representing these two planes. We have already shown in Example that the line representing all these solutions is a subspace of R 3. Example We showed in Example 5..3 that the set M 2,2 (R) of all 2 2 matrices with entries in R forms a vector space over R. Consider the subset of M 2,2 (R). Since W = { } a a 2 : a a 2, a 2, a 2 R a a 2 + a 2 b b 2 = b 2 ( a + b ) a 2 + b 2 a 2 + b 2 and c a a 2 = a 2 ca ca 2, ca 2 it follows that (SP) and (SP2) are satisfied. Hence W is a subspace of M 2,2 (R). Example We showed in Example 5..4 that the set A of all functions of the form f : R R forms a vector space over R. Let C denote the set of all functions of the form f : R R which are continuous at x = 2, and let C denote the set of all functions of the form f : R R which are differentiable at x = 2. Then it follows from the arithmetic of limits and the arithmetic of derivatives that C and C are both subspaces of A. Furthermore, C is a subspace of C (why?). On the other hand, let k N. Recall from Example 5..6 the vector space P k of all polynomials of the form p(x) = p + p x p k x k, where p, p,..., p k R. In other words, P k is the set of all polynomials of degree at most k and with coefficients in R. Clearly P k is a subspace of C Linear Combination In this section and the next two, we shall study ways of describing the vectors in a vector space V. Our ultimate goal is to be able to determine a subset B of vectors in V and describe every element of V in terms of elements of B in a unique way. The first step in this direction is summarized below. Definition. Suppose that v,..., v r are vectors in a vector space V over R. By a linear combination of the vectors v,..., v r, we mean an expression of the type where c,..., c r R. c v c r v r, Example In R 2, every vector (x, y) is a linear combination of the two vectors i = (, ) and j = (, ), for clearly (x, y) = xi + yj. Example In R 3, every vector (x, y, z) is a linear combination of the three vectors i = (,, ), j = (,, ) and k = (,, ), for clearly (x, y, z) = xi + yj + zk. Chapter 5 : Introduction to Vector Spaces page 7 of 7
8 Example In R 4, the vector (, 4, 2, 6) is a linear combination of the two vectors (, 2,, 4) and (,,, 3), for we have (, 4, 2, 6) = 3(, 2,, 4) 2(,,, 3). On the other hand, the vector (2, 6,, 9) is not a linear combination of the two vectors (, 2,, 4) and (,,, 3), for would lead to the system of four equations (2, 6,, 9) = c (, 2,, 4) + (,,, 3) c + = 2, 2c + = 6, =, 4c + 3 = 9. It is easily checked that this system has no solutions. Example In the vector space A of all functions of the form f : R R described in Example 5..4, the function cos 2x is a linear combination of the three functions cos 2 x, cosh 2 x and sinh 2 x. It is not too difficult to check that cos 2x = 2 cos 2 x + sinh 2 x cosh 2 x, noting that cos 2x = 2 cos 2 x and cosh 2 x sinh 2 x =. We observe that in Example 5.3., every vector in R 2 is a linear combination of the two vectors i and j. Similarly, in Example 5.3.2, every vector in R 3 is a linear combination of the three vectors i, j and k. On the other hand, we observe that in Example 5.3.3, not every vector in R 4 is a linear combination of the two vectors (, 2,, 4) and (,,, 3). Let us therefore investigate the collection of all vectors in a vector space that can be represented as linear combinations of a given set of vectors in V. Definition. Suppose that v,..., v r are vectors in a vector space V over R. The set span{v,..., v r } = {c v c r v r : c,..., c r R} is called the span of the vectors v,..., v r. We also say that the vectors v,..., v r span V if span{v,..., v r } = V ; in other words, if every vector in V can be expressed as a linear combination of the vectors v,..., v r. Example The two vectors i = (, ) and j = (, ) span R 2. Example The three vectors i = (,, ), j = (,, ) and k = (,, ) span R 3. Example The two vectors (, 2,, 4) and (,,, 3) do not span R 4. PROPOSITION 5C. Suppose that v,..., v r are vectors in a vector space V over R. (a) Then span{v,..., v r } is a subspace of V. (b) Suppose further that W is a subspace of V and v,..., v r W. Then span{v,..., v r } W. Proof. (a) Suppose that u, w span{v,..., v r } and c R. There exist a,..., a r, b,..., b r R such that u = a v a r v r and w = b v b r v r. Chapter 5 : Introduction to Vector Spaces page 8 of 7
9 Then and u + w = (a v a r v r ) + (b v b r v r ) = (a + b )v (a r + b r )v r span{v,..., v r } cu = c(a v a r v r ) = (ca )v (ca r )v r span{v,..., v r }. It follows from Proposition 5B that span{v,..., v r } is a subspace of V. (b) Suppose that c,..., c r R and u = c v c r v r span{v,..., v r }. If v,..., v r W, then it follows from (SM) for W that c v,..., c r v r W. It then follows from (VA) for W that u = c v c r v r W. Example In R 2, any non-zero vector v spans the subspace {cv : c R}. This is clearly a line through the origin. Also, try to draw a picture to convince yourself that any two non-zero vectors that are not on the same line span R 2. Example In R 3, try to draw pictures to convince yourself that any non-zero vector spans a subspace which is a line through the origin; any two non-zero vectors that are not on the same line span a subspace which is a plane through the origin; and any three non-zero vectors that do not lie on the same plane span R Linear Independence We first study two simple examples. Example Consider the three vectors v = (, 2, 3), v 2 = (3, 2, ) and v 3 = (3, 3, 3) in R 3. Then span{v, v 2, v 3 } = {c (, 2, 3) + (3, 2, ) + c 3 (3, 3, 3) : c,, c 3 R} = {(c c 3, 2c c 3, 3c + + 3c 3 ) : c,, c 3 R}. Write (x, y, z) = (c c 3, 2c c 3, 3c + + 3c 3 ). Then it is not difficult to see that x y = c, z 3 3 c 3 and so (do not worry if you cannot understand why we take this next step) ( 2 ) x y z = ( 2 ) c = ( ) c = ( ), 3 3 c 3 so that x 2y + z =. It follows that span{v, v 2, v 3 } is a plane through the origin and not R 3. Note, in fact, that 3v + 3v 2 4v 3 =. Note also that det =. 3 3 c 3 Chapter 5 : Introduction to Vector Spaces page 9 of 7
10 Example Consider the three vectors v = (,, ), v 2 = (5,, 3) and v 3 = (2, 7, 4) in R 3. Then span{v, v 2, v 3 } = {c (,, ) + (5,, 3) + c 3 (2, 7, 4) : c,, c 3 R} = {(c c 3, c + + 7c 3, 3 + 4c 3 ) : c,, c 3 R}. Write (x, y, z) = (c c 3, c + + 7c 3, 3 + 4c 3 ). Then it is not difficult to see that x y = c, z 3 4 c 3 so that x y z = c c 3 = c = c. c 3 c 3 It follows that for every (x, y, z) R 3, we can find c,, c 3 R such that (x, y, z) = c v + v 2 + c 3 v 3. Hence span{v, v 2, v 3 } = R 3. Note that and that the only solution for det 5 2 7, 3 4 is c = = c 3 =. (,, ) = c v + v 2 + c 3 v 3 Definition. Suppose that v,..., v r are vectors in a vector space V over R. (LD) We say that v,..., v r are linearly dependent if there exist c,..., c r R, not all zero, such that c v c r v r =. (LI) We say that v,..., v r are linearly independent if they are not linearly dependent; in other words, if the only solution of c v c r v r = in c,..., c r R is given by c =... = c r =. Example Let us return to Example 5.4. and consider again the three vectors v = (, 2, 3), v 2 = (3, 2, ) and v 3 = (3, 3, 3) in R 3. Consider the equation c v + v 2 + c 3 v 3 =. This can be rewritten in matrix form as c = 3 3 c 3. Since det =, 3 3 the system has non-trivial solutions; for example, (c,, c 3 ) = (3, 3, 4), so that 3v + 3v 2 4v 3 =. Hence v, v 2, v 3 are linearly dependent. Chapter 5 : Introduction to Vector Spaces page of 7
11 Example Let us return to Example and consider again the three vectors v = (,, ), v 2 = (5,, 3) and v 3 = (2, 7, 4) in R 3. Consider the equation c v + v 2 + c 3 v 3 =. This can be rewritten in matrix form as c = 3 4 c 3. Since det 5 2 7, 3 4 the only solution is c = = c 3 =. Hence v, v 2, v 3 are linearly independent. Example In the vector space A of all functions of the form f : R R described in Example 5..4, the functions x, x 2 and sin x are linearly independent. To see this, note that for every c,, c 3 R, the linear combination c x + x 2 + c 3 sin x is never identically zero unless c = = c 3 =. Example In R n, the vectors e,..., e n, where are linearly independent (why?). e j = (,...,,,,..., ) for every j =,..., n, }{{}}{{} j n j We observe in Examples that the determination of whether a collection of vectors in R 3 are linearly dependent is based on whether a system of homogeneous linear equations has non-trivial solutions. The same idea can be used to prove the following result concerning R n. PROPOSITION 5D. Suppose that v,..., v r are vectors in the vector space R n. If r > n, then v,..., v r are linearly dependent. Proof. For every j =,..., r, write v j = (a j,..., a nj ). Then the equation c v c r v r = can be rewritten in matrix form as a... a r c... =.. a n... a nr If r > n, then there are more variables than equations. It follows that there must be non-trivial solutions c,..., c r R. Hence v,..., v r are linearly dependent. Remarks. () Consider two vectors v = (a, a 2 ) and v 2 = (a 2, a 22 ) in R 2. To study linear independence, we consider the equation c v + v 2 =, which can be written in matrix form as a a 2 c =. a 2 a 22 The vectors v and v 2 are linearly independent precisely when a a det 2. a 2 a 22 c r Chapter 5 : Introduction to Vector Spaces page of 7
12 This can be interpreted geometrically in the following way: The area of the parallelogram formed by the two vectors v and v 2 is in fact equal to the absolute value of the determinant of the matrix formed with v and v 2 as the columns; in other words, a a det 2. a 2 a 22 It follows that the two vectors are linearly dependent precisely when the parallelogram has zero area; in other words, when the two vectors lie on the same line. On the other hand, if the parallelogram has positive area, then the two vectors are linearly independent. (2) Consider three vectors v = (a, a 2, a 3 ), v 2 = (a 2, a 22, a 32 ), and v 3 = (a 3, a 23, a 33 ) in R 3. To study linear independence, we consider the equation c v + v 2 + c 3 v 3 =, which can be written in matrix form as a a 2 a 3 a 2 a 22 a 23 c = a 3 a 32 a 33 c 3 The vectors v, v 2 and v 3 are linearly independent precisely when det a a 2 a 3 a 2 a 22 a 23. a 3 a 32 a 33 This can be interpreted geometrically in the following way: The volume of the parallelepiped formed by the three vectors v, v 2 and v 3 is in fact equal to the absolute value of the determinant of the matrix formed with v, v 2 and v 3 as the columns; in other words, det a a 2 a 3 a 2 a 22 a 23. a 3 a 32 a 33 It follows that the three vectors are linearly dependent precisely when the parallelepiped has zero volume; in other words, when the three vectors lie on the same plane. On the other hand, if the parallelepiped has positive volume, then the three vectors are linearly independent. (3) What is the geometric interpretation of two linearly independent vectors in R 3? Well, note that if v and v 2 are non-zero and linearly dependent, then there exist c, R, not both zero, such that c v + v 2 =. This forces the two vectors to be multiples of each other, so that they lie on the same line, whence the parallelogram they form has zero area. It follows that if two vectors in R 3 form a parallelogram with positive area, then they are linearly independent Basis and Dimension In this section, we complete the task of describing uniquely every element of a vector space V in terms of the elements of a suitable subset B. To motivate the ideas, we first consider an example. Example Let us consider the three vectors v = (,, ), v 2 = (5,, 3) and v 3 = (2, 7, 4) in R 3, as in Examples and We have already shown that span{v, v 2, v 3 } = R 3, and that the vectors v, v 2, v 3 are linearly independent. Furthermore, we have shown that for every u = (x, y, z) R 3, we can write u = c v + v 2 + c 3 v 3, where c,, c 3 R are determined uniquely by c = c x y z. Chapter 5 : Introduction to Vector Spaces page 2 of 7
13 Definition. Suppose that v,..., v r are vectors in a vector space V over R. We say that {v,..., v r } is a basis for V if the following two conditions are satisfied: (B) We have span{v,..., v r } = V. (B2) The vectors v,..., v r are linearly independent. Example Consider two vectors v = (a, a 2 ) and v 2 = (a 2, a 22 ) in R 2. Suppose that a a det 2 ; a 2 a 22 in other words, suppose that the parallelogram formed by the two vectors has non-zero area. Then it follows from Remark () in Section 5.4 that v and v 2 are linearly independent. Furthermore, for every u = (x, y) R 2, there exist c, R such that u = c v + v 2. Indeed, c and are determined as the unique solution of the system a a 2 c = a 2 a 22 x. y Hence span{v, v 2 } = R 2. It follows that {v, v 2 } is a basis for R 2. Example Consider three vectors of the type v = (a, a 2, a 3 ), v 2 = (a 2, a 22, a 32 ) and v 3 = (a 3, a 23, a 33 ) in R 3. Suppose that det a a 2 a 3 a 2 a 22 a 23 ; a 3 a 32 a 33 in other words, suppose that the parallelepiped formed by the three vectors has non-zero volume. Then it follows from Remark (2) in Section 5.4 that v, v 2 and v 3 are linearly independent. Furthermore, for every u = (x, y, z) R 3, there exist c,, c 3 R such that u = c v + v 2 + c 3 v 3. Indeed, c, and c 3 are determined as the unique solution of the system a a 2 a 3 a 2 a 22 a 23 c = x y. a 3 a 32 a 33 c 3 z Hence span{v, v 2, v 3 } = R 3. It follows that {v, v 2, v 3 } is a basis for R 3. Example In R n, the vectors e,..., e n, where e j = (,...,,,,..., ) for every j =,..., n, }{{}}{{} j n j are linearly independent and span R n. Hence {e,..., e n } is a basis for R n. This is known as the standard basis for R n. Example In the vector space M 2,2 (R) of all 2 2 matrices with entries in R as discussed in Example 5..3, the set is a basis. {,,, } Example In the vector space P k of polynomials of degree at most k and with coefficients in R as discussed in Example 5..6, the set {, x, x 2,..., x k } is a basis. Chapter 5 : Introduction to Vector Spaces page 3 of 7
14 PROPOSITION 5E. Suppose that {v,..., v r } is a basis for a vector space V over R. Then every element u V can be expressed uniquely in the form u = c v c r v r, where c,..., c r R. Proof. Since u V = span{v,..., v r }, there exist c,..., c r R such that u = c v c r v r. Suppose now that b,..., b r R such that Then c v c r v r = b v b r v r. (c b )v (c r b r )v r =. Since v,..., v r are linearly independent, it follows that c b =... = c r b r =. Hence c,..., c r are uniquely determined. We have shown earlier that a vector space can have many bases. For example, any collection of three vectors not on the same plane is a basis for R 3. In the following discussion, we attempt to find out some properties of bases. However, we shall restrict our discussion to the following simple case. Definition. A vector space V over R is said to be finite-dimensional if it has a basis containing only finitely many elements. Example The vector spaces R n, M 2,2 (R) and P k that we have discussed earlier are all finitedimensional. Recall that in R n, the standard basis has exactly n elements. On the other hand, it follows from Proposition 5D that any basis for R n cannot contain more than n elements. However, can a basis for R n contain fewer than n elements? We shall answer this question by showing that all bases for a given vector space have the same number of elements. As a first step, we establish the following generalization of Proposition 5D. PROPOSITION 5F. Suppose that {v,..., v n } is a basis for a vector space V over R. Suppose further that r > n, and that the vectors u,..., u r V. Then the vectors u,..., u r are linearly dependent. Proof. Since {v,..., v n } is a basis for the vector space V, we can write u = a v a n v n,. u r = a r v a nr v n, where a ij R for every i =,..., n and j =,..., r. Let c,..., c r R. Since v,..., v n are linearly independent, it follows that if c u c r u r = c (a v a n v n ) c r (a r v a nr v n ) = (a c a r c r )v (a n c a nr c r )v n =, then a c +...+a r c r =... = a n c +...+a nr c r = ; in other words, we have the homogeneous system a... a r c... =.. a n... a nr c r Chapter 5 : Introduction to Vector Spaces page 4 of 7
15 If r > n, then there are more variables than equations. It follows that there must be non-trivial solutions c,..., c r R. Hence u,..., u r are linearly dependent. PROPOSITION 5G. Suppose that V is a finite-dimensional vector space V over R. Then any two bases for V have the same number of elements. Proof. Note simply that by Proposition 5F, the vectors in the basis with more elements must be linearly dependent, and so cannot be a basis. We are now in a position to make the following definition. Definition. Suppose that V is a finite-dimensional vector space over R. Then we say that V is of dimension n if a basis for V contains exactly n elements. Example The vector space R n has dimension n. Example The vector space M 2,2 (R) of all 2 2 matrices with entries in R, as discussed in Example 5..3, has dimension 4. Example The vector space P k of all polynomials of degree at most k and with coefficients in R, as discussed in Example 5..6, has dimension (k + ). Example Recall Example 5.2.5, where we showed that the set of solutions to a system of m homogeneous linear equations in n unknowns is a subspace of R n. Consider now the homogeneous system The solutions can be described in the form x x x =. x x 5 2 x = c + where c, R (the reader must check this). It can be checked that (, 2,,, ) and (, 3,, 5, ) are linearly independent and so form a basis for the space of solutions of the system. It follows that the space of solutions of the system has dimension 2. Suppose that V is an n-dimensional vector space over R. Then any basis for V consists of exactly n linearly independent vectors in V. Suppose now that we have a set of n linearly independent vectors in V. Will this form a basis for V? We have already answered this question in the affirmative in the cases when the vector space is R 2 or R 3. To seek an answer to the general case, we first establish the following result. PROPOSITION 5H. Suppose that V is a finite-dimensional vector space over R. Then any finite set of linearly independent vectors in V can be expanded, if necessary, to a basis for V. Proof. Let S = {v,..., v k } be a finite set of linearly independent vectors in V. If S spans V, then the proof is complete. If S does not span V, then there exists v k+ V that is not a linear combination 3 5, Chapter 5 : Introduction to Vector Spaces page 5 of 7
16 of the elements of S. The set T = {v,..., v k, v k+ } is a finite set of linearly independent vectors in V ; for otherwise, there exist c,..., c k, c k+, not all zero, such that c v c k v k + c k+ v k+ =. If c k+ =, then c v c k v k =, contradicting the assumption that S is a finite set of linearly independent vectors in V. If c k+, then v k+ = c v... c k v k, c k+ c k+ contradicting the assumption that v k+ is not a linear combination of the elements of S. We now study the finite set T of linearly independent vectors in V. If T spans V, then the proof is complete. If T does not span V, then we repeat the argument. Note that the number of vectors in a linearly independent expansion of S cannot exceed the dimension of V, in view of Proposition 5F. So eventually some linearly independent expansion of S will span V. PROPOSITION 5J. Suppose that V is an n-dimensional vector space over R. Then any set of n linearly independent vectors in V is a basis for V. Proof. Let S be a set of n linearly independent vectors in V. By Proposition 5H, S can be expanded, if necessary, to a basis for V. By Proposition 5F, any expansion of S will result in a linearly dependent set of vectors in V. It follows that S is already a basis for V. Example Consider the three vectors v = (, 2, 3), v 2 = (3, 2, ) and v 3 = (3, 3, 3) in R 3, as in Examples 5.4. and We showed that these three vectors are linearly dependent, and span the plane x 2y + z =. Note that v 3 = 3 4 v v 2, and that v and v 2 are linearly independent. Consider now the vector v 4 = (,, ). Note that v 4 does not lie on the plane x 2y + z =, so that {v, v 2, v 4 } form a linearly independent set. It follows that {v, v 2, v 4 } is a basis for R 3. Chapter 5 : Introduction to Vector Spaces page 6 of 7
17 Problems for Chapter 5. Determine whether each of the following subsets of R 3 is a subspace of R 3 : a) {(x, y, z) R 3 : x = } b) {(x, y, z) R 3 : x + y = } c) {(x, y, z) R 3 : xz = } d) {(x, y, z) R 3 : y } e) {(x, y, z) R 3 : x = y = z} 2. For each of the following collections of vectors, determine whether the first vector is a linear combination of the remaining ones: a) (, 2, 3); (,, ), (2,, ) in R 3 b) x 3 + 2x 2 + 3x + ; x 3, x 2 + 3x, x 2 + in P 4 c) (, 3, 5, 7); (,,, ), (,,, ), (,,, ) in R 4 3. For each of the following collections of vectors, determine whether the vectors are linearly independent: a) (, 2, 3), (,, ), (2,, ) in R 3 b) (, 2), (3, 5), (, 3) in R 2 c) (2, 5, 3, 6), (,,, ), (4,, 9, 6) in R 4 d) x 2 +, x +, x 2 + x in P 3 4. Find the volume of the parallelepiped in R 3 formed by the vectors (, 2, 3), (,, ) and (3,, 2). 5. Let S be the set of all functions y that satisfy the differential equation 2 d2 y dx 2 3 dy dx + y =. Show that S is a subspace of the vector space A described in Example For each of the sets in Problem which is a subspace of R 3, find a basis for the subspace, and then extend it to a basis for R 3. Chapter 5 : Introduction to Vector Spaces page 7 of 7
LINEAR ALGEBRA W W L CHEN
LINEAR ALGEBRA W W L CHEN c W W L Chen, 1997, 2008. This chapter is available free to all individuals, on the understanding that it is not to be used for financial gain, and may be downloaded and/or photocopied,
More informationChapter 2: Linear Independence and Bases
MATH20300: Linear Algebra 2 (2016 Chapter 2: Linear Independence and Bases 1 Linear Combinations and Spans Example 11 Consider the vector v (1, 1 R 2 What is the smallest subspace of (the real vector space
More informationLINEAR ALGEBRA W W L CHEN
LINEAR ALGEBRA W W L CHEN c W W L Chen, 1982, 28. This chapter originates from material used by the author at Imperial College, University of London, between 1981 and 199. It is available free to all individuals,
More informationEXERCISE SET 5.1. = (kx + kx + k, ky + ky + k ) = (kx + kx + 1, ky + ky + 1) = ((k + )x + 1, (k + )y + 1)
EXERCISE SET 5. 6. The pair (, 2) is in the set but the pair ( )(, 2) = (, 2) is not because the first component is negative; hence Axiom 6 fails. Axiom 5 also fails. 8. Axioms, 2, 3, 6, 9, and are easily
More informationVector Spaces. (1) Every vector space V has a zero vector 0 V
Vector Spaces 1. Vector Spaces A (real) vector space V is a set which has two operations: 1. An association of x, y V to an element x+y V. This operation is called vector addition. 2. The association of
More informationMAT 242 CHAPTER 4: SUBSPACES OF R n
MAT 242 CHAPTER 4: SUBSPACES OF R n JOHN QUIGG 1. Subspaces Recall that R n is the set of n 1 matrices, also called vectors, and satisfies the following properties: x + y = y + x x + (y + z) = (x + y)
More informationChapter 3. More about Vector Spaces Linear Independence, Basis and Dimension. Contents. 1 Linear Combinations, Span
Chapter 3 More about Vector Spaces Linear Independence, Basis and Dimension Vincent Astier, School of Mathematical Sciences, University College Dublin 3. Contents Linear Combinations, Span Linear Independence,
More informationChapter 3. Vector spaces
Chapter 3. Vector spaces Lecture notes for MA1111 P. Karageorgis pete@maths.tcd.ie 1/22 Linear combinations Suppose that v 1,v 2,...,v n and v are vectors in R m. Definition 3.1 Linear combination We say
More informationSolutions for Math 225 Assignment #4 1
Solutions for Math 225 Assignment #4 () Let B {(3, 4), (4, 5)} and C {(, ), (0, )} be two ordered bases of R 2 (a) Find the change-of-basis matrices P C B and P B C (b) Find v] B if v] C ] (c) Find v]
More informationMTH 464: Computational Linear Algebra
MTH 464: Computational Linear Algebra Lecture Outlines Exam 2 Material Prof. M. Beauregard Department of Mathematics & Statistics Stephen F. Austin State University March 2, 2018 Linear Algebra (MTH 464)
More informationENGINEERING MATH 1 Fall 2009 VECTOR SPACES
ENGINEERING MATH 1 Fall 2009 VECTOR SPACES A vector space, more specifically, a real vector space (as opposed to a complex one or some even stranger ones) is any set that is closed under an operation of
More information2.3. VECTOR SPACES 25
2.3. VECTOR SPACES 25 2.3 Vector Spaces MATH 294 FALL 982 PRELIM # 3a 2.3. Let C[, ] denote the space of continuous functions defined on the interval [,] (i.e. f(x) is a member of C[, ] if f(x) is continuous
More informationMATH 115A: SAMPLE FINAL SOLUTIONS
MATH A: SAMPLE FINAL SOLUTIONS JOE HUGHES. Let V be the set of all functions f : R R such that f( x) = f(x) for all x R. Show that V is a vector space over R under the usual addition and scalar multiplication
More informationAbstract Vector Spaces and Concrete Examples
LECTURE 18 Abstract Vector Spaces and Concrete Examples Our discussion of linear algebra so far has been devoted to discussing the relations between systems of linear equations, matrices, and vectors.
More informationChapter 1 Vector Spaces
Chapter 1 Vector Spaces Per-Olof Persson persson@berkeley.edu Department of Mathematics University of California, Berkeley Math 110 Linear Algebra Vector Spaces Definition A vector space V over a field
More informationDefinition 1. A set V is a vector space over the scalar field F {R, C} iff. there are two operations defined on V, called vector addition
6 Vector Spaces with Inned Product Basis and Dimension Section Objective(s): Vector Spaces and Subspaces Linear (In)dependence Basis and Dimension Inner Product 6 Vector Spaces and Subspaces Definition
More informationFIRST YEAR CALCULUS W W L CHEN
FIRST YER CLCULUS W W L CHEN c W W L Chen, 994, 28. This chapter is available free to all individuals, on the understanding that it is not to be used for financial gain, and may be downloaded and/or photocopied,
More informationMath 4377/6308 Advanced Linear Algebra I Dr. Vaughn Climenhaga, PGH 651A HOMEWORK 3
Math 4377/6308 Advanced Linear Algebra I Dr. Vaughn Climenhaga, PGH 651A Fall 2013 HOMEWORK 3 Due 4pm Wednesday, September 11. You will be graded not only on the correctness of your answers but also on
More informationLinear Algebra, Summer 2011, pt. 2
Linear Algebra, Summer 2, pt. 2 June 8, 2 Contents Inverses. 2 Vector Spaces. 3 2. Examples of vector spaces..................... 3 2.2 The column space......................... 6 2.3 The null space...........................
More informationVector Spaces ปร ภ ม เวกเตอร
Vector Spaces ปร ภ ม เวกเตอร 5.1 Real Vector Spaces ปร ภ ม เวกเตอร ของจ านวนจร ง Vector Space Axioms (1/2) Let V be an arbitrary nonempty set of objects on which two operations are defined, addition and
More informationVector Spaces ปร ภ ม เวกเตอร
Vector Spaces ปร ภ ม เวกเตอร 1 5.1 Real Vector Spaces ปร ภ ม เวกเตอร ของจ านวนจร ง Vector Space Axioms (1/2) Let V be an arbitrary nonempty set of objects on which two operations are defined, addition
More information2 so Q[ 2] is closed under both additive and multiplicative inverses. a 2 2b 2 + b
. FINITE-DIMENSIONAL VECTOR SPACES.. Fields By now you ll have acquired a fair knowledge of matrices. These are a concrete embodiment of something rather more abstract. Sometimes it is easier to use matrices,
More informationChapter 2: Matrix Algebra
Chapter 2: Matrix Algebra (Last Updated: October 12, 2016) These notes are derived primarily from Linear Algebra and its applications by David Lay (4ed). Write A = 1. Matrix operations [a 1 a n. Then entry
More informationLinear Algebra 1 Exam 2 Solutions 7/14/3
Linear Algebra 1 Exam Solutions 7/14/3 Question 1 The line L has the symmetric equation: x 1 = y + 3 The line M has the parametric equation: = z 4. [x, y, z] = [ 4, 10, 5] + s[10, 7, ]. The line N is perpendicular
More informationLinear Algebra Lecture Notes-I
Linear Algebra Lecture Notes-I Vikas Bist Department of Mathematics Panjab University, Chandigarh-6004 email: bistvikas@gmail.com Last revised on February 9, 208 This text is based on the lectures delivered
More informationMath 24 Spring 2012 Questions (mostly) from the Textbook
Math 24 Spring 2012 Questions (mostly) from the Textbook 1. TRUE OR FALSE? (a) The zero vector space has no basis. (F) (b) Every vector space that is generated by a finite set has a basis. (c) Every vector
More informationAugust 23, 2017 Let us measure everything that is measurable, and make measurable everything that is not yet so. Galileo Galilei. 1.
August 23, 2017 Let us measure everything that is measurable, and make measurable everything that is not yet so. Galileo Galilei 1. Vector spaces 1.1. Notations. x S denotes the fact that the element x
More informationMath Linear Algebra II. 1. Inner Products and Norms
Math 342 - Linear Algebra II Notes 1. Inner Products and Norms One knows from a basic introduction to vectors in R n Math 254 at OSU) that the length of a vector x = x 1 x 2... x n ) T R n, denoted x,
More informationAbstract Vector Spaces
CHAPTER 1 Abstract Vector Spaces 1.1 Vector Spaces Let K be a field, i.e. a number system where you can add, subtract, multiply and divide. In this course we will take K to be R, C or Q. Definition 1.1.
More informationGENERAL VECTOR SPACES AND SUBSPACES [4.1]
GENERAL VECTOR SPACES AND SUBSPACES [4.1] General vector spaces So far we have seen special spaces of vectors of n dimensions denoted by R n. It is possible to define more general vector spaces A vector
More information1. What is the determinant of the following matrix? a 1 a 2 4a 3 2a 2 b 1 b 2 4b 3 2b c 1. = 4, then det
What is the determinant of the following matrix? 3 4 3 4 3 4 4 3 A 0 B 8 C 55 D 0 E 60 If det a a a 3 b b b 3 c c c 3 = 4, then det a a 4a 3 a b b 4b 3 b c c c 3 c = A 8 B 6 C 4 D E 3 Let A be an n n matrix
More information4.3 - Linear Combinations and Independence of Vectors
- Linear Combinations and Independence of Vectors De nitions, Theorems, and Examples De nition 1 A vector v in a vector space V is called a linear combination of the vectors u 1, u,,u k in V if v can be
More informationMath113: Linear Algebra. Beifang Chen
Math3: Linear Algebra Beifang Chen Spring 26 Contents Systems of Linear Equations 3 Systems of Linear Equations 3 Linear Systems 3 2 Geometric Interpretation 3 3 Matrices of Linear Systems 4 4 Elementary
More informationCambridge University Press The Mathematics of Signal Processing Steven B. Damelin and Willard Miller Excerpt More information
Introduction Consider a linear system y = Φx where Φ can be taken as an m n matrix acting on Euclidean space or more generally, a linear operator on a Hilbert space. We call the vector x a signal or input,
More informationFurther Mathematical Methods (Linear Algebra) 2002
Further Mathematical Methods (Linear Algebra) 22 Solutions For Problem Sheet 3 In this Problem Sheet, we looked at some problems on real inner product spaces. In particular, we saw that many different
More informationOHSX XM511 Linear Algebra: Multiple Choice Exercises for Chapter 2
OHSX XM5 Linear Algebra: Multiple Choice Exercises for Chapter. In the following, a set is given together with operations of addition and scalar multiplication. Which is not a vector space under the given
More informationLinear Algebra (part 1) : Vector Spaces (by Evan Dummit, 2017, v. 1.07) 1.1 The Formal Denition of a Vector Space
Linear Algebra (part 1) : Vector Spaces (by Evan Dummit, 2017, v. 1.07) Contents 1 Vector Spaces 1 1.1 The Formal Denition of a Vector Space.................................. 1 1.2 Subspaces...................................................
More informationMath 3108: Linear Algebra
Math 3108: Linear Algebra Instructor: Jason Murphy Department of Mathematics and Statistics Missouri University of Science and Technology 1 / 323 Contents. Chapter 1. Slides 3 70 Chapter 2. Slides 71 118
More information2. Every linear system with the same number of equations as unknowns has a unique solution.
1. For matrices A, B, C, A + B = A + C if and only if A = B. 2. Every linear system with the same number of equations as unknowns has a unique solution. 3. Every linear system with the same number of equations
More informationCSL361 Problem set 4: Basic linear algebra
CSL361 Problem set 4: Basic linear algebra February 21, 2017 [Note:] If the numerical matrix computations turn out to be tedious, you may use the function rref in Matlab. 1 Row-reduced echelon matrices
More information1. General Vector Spaces
1.1. Vector space axioms. 1. General Vector Spaces Definition 1.1. Let V be a nonempty set of objects on which the operations of addition and scalar multiplication are defined. By addition we mean a rule
More informationContents. 2.1 Vectors in R n. Linear Algebra (part 2) : Vector Spaces (by Evan Dummit, 2017, v. 2.50) 2 Vector Spaces
Linear Algebra (part 2) : Vector Spaces (by Evan Dummit, 2017, v 250) Contents 2 Vector Spaces 1 21 Vectors in R n 1 22 The Formal Denition of a Vector Space 4 23 Subspaces 6 24 Linear Combinations and
More informationProblem 1: (3 points) Recall that the dot product of two vectors in R 3 is
Linear Algebra, Spring 206 Homework 3 Name: Problem : (3 points) Recall that the dot product of two vectors in R 3 is a x b y = ax + by + cz, c z and this is essentially the same as the matrix multiplication
More informationMath 350 Fall 2011 Notes about inner product spaces. In this notes we state and prove some important properties of inner product spaces.
Math 350 Fall 2011 Notes about inner product spaces In this notes we state and prove some important properties of inner product spaces. First, recall the dot product on R n : if x, y R n, say x = (x 1,...,
More informationChapter 6 - Orthogonality
Chapter 6 - Orthogonality Maggie Myers Robert A. van de Geijn The University of Texas at Austin Orthogonality Fall 2009 http://z.cs.utexas.edu/wiki/pla.wiki/ 1 Orthogonal Vectors and Subspaces http://z.cs.utexas.edu/wiki/pla.wiki/
More informationLinear Algebra. Linear Algebra. Chih-Wei Yi. Dept. of Computer Science National Chiao Tung University. November 12, 2008
Linear Algebra Chih-Wei Yi Dept. of Computer Science National Chiao Tung University November, 008 Section De nition and Examples Section De nition and Examples Section De nition and Examples De nition
More informationPreliminary Linear Algebra 1. Copyright c 2012 Dan Nettleton (Iowa State University) Statistics / 100
Preliminary Linear Algebra 1 Copyright c 2012 Dan Nettleton (Iowa State University) Statistics 611 1 / 100 Notation for all there exists such that therefore because end of proof (QED) Copyright c 2012
More information1 Last time: inverses
MATH Linear algebra (Fall 8) Lecture 8 Last time: inverses The following all mean the same thing for a function f : X Y : f is invertible f is one-to-one and onto 3 For each b Y there is exactly one a
More informationMath 115 Midterm Solutions
Math 115 Midterm Solutions 1. (25 points) (a) (10 points) Let S = {0, 1, 2}. Find polynomials with real coefficients of degree 2 P 0, P 1, and P 2 such that P i (i) = 1 and P i (j) = 0 when i j, for i,
More informationVECTOR SPACES & SUBSPACES
VECTOR SPACES & SUBSPACES Definition: The real Vector Space " V " is the set of all entities called "vectors" with real entries that satisfy two closure properties and obey a set of eight rules. If "x"
More informationLinear Algebra. Preliminary Lecture Notes
Linear Algebra Preliminary Lecture Notes Adolfo J. Rumbos c Draft date May 9, 29 2 Contents 1 Motivation for the course 5 2 Euclidean n dimensional Space 7 2.1 Definition of n Dimensional Euclidean Space...........
More informationLinear Equation: a 1 x 1 + a 2 x a n x n = b. x 1, x 2,..., x n : variables or unknowns
Linear Equation: a x + a 2 x 2 +... + a n x n = b. x, x 2,..., x n : variables or unknowns a, a 2,..., a n : coefficients b: constant term Examples: x + 4 2 y + (2 5)z = is linear. x 2 + y + yz = 2 is
More informationAPPM 3310 Problem Set 4 Solutions
APPM 33 Problem Set 4 Solutions. Problem.. Note: Since these are nonstandard definitions of addition and scalar multiplication, be sure to show that they satisfy all of the vector space axioms. Solution:
More informationMATH 304 Linear Algebra Lecture 20: Review for Test 1.
MATH 304 Linear Algebra Lecture 20: Review for Test 1. Topics for Test 1 Part I: Elementary linear algebra (Leon 1.1 1.4, 2.1 2.2) Systems of linear equations: elementary operations, Gaussian elimination,
More informationFall 2016 MATH*1160 Final Exam
Fall 2016 MATH*1160 Final Exam Last name: (PRINT) First name: Student #: Instructor: M. R. Garvie Dec 16, 2016 INSTRUCTIONS: 1. The exam is 2 hours long. Do NOT start until instructed. You may use blank
More informationLinear Algebra (Math-324) Lecture Notes
Linear Algebra (Math-324) Lecture Notes Dr. Ali Koam and Dr. Azeem Haider September 24, 2017 c 2017,, Jazan All Rights Reserved 1 Contents 1 Real Vector Spaces 6 2 Subspaces 11 3 Linear Combination and
More informationStudy Guide for Linear Algebra Exam 2
Study Guide for Linear Algebra Exam 2 Term Vector Space Definition A Vector Space is a nonempty set V of objects, on which are defined two operations, called addition and multiplication by scalars (real
More informationFundamentals of Linear Algebra. Marcel B. Finan Arkansas Tech University c All Rights Reserved
Fundamentals of Linear Algebra Marcel B. Finan Arkansas Tech University c All Rights Reserved 2 PREFACE Linear algebra has evolved as a branch of mathematics with wide range of applications to the natural
More informationSolution to Homework 1
Solution to Homework Sec 2 (a) Yes It is condition (VS 3) (b) No If x, y are both zero vectors Then by condition (VS 3) x = x + y = y (c) No Let e be the zero vector We have e = 2e (d) No It will be false
More informationLinear Algebra. Preliminary Lecture Notes
Linear Algebra Preliminary Lecture Notes Adolfo J. Rumbos c Draft date April 29, 23 2 Contents Motivation for the course 5 2 Euclidean n dimensional Space 7 2. Definition of n Dimensional Euclidean Space...........
More informationConsequences of Orthogonality
Consequences of Orthogonality Philippe B. Laval KSU Today Philippe B. Laval (KSU) Consequences of Orthogonality Today 1 / 23 Introduction The three kind of examples we did above involved Dirichlet, Neumann
More informationLecture 16: 9.2 Geometry of Linear Operators
Lecture 16: 9.2 Geometry of Linear Operators Wei-Ta Chu 2008/11/19 Theorem 9.2.1 If T: R 2 R 2 is multiplication by an invertible matrix A, then the geometric effect of T is the same as an appropriate
More informationLECTURE 6: VECTOR SPACES II (CHAPTER 3 IN THE BOOK)
LECTURE 6: VECTOR SPACES II (CHAPTER 3 IN THE BOOK) In this lecture, F is a fixed field. One can assume F = R or C. 1. More about the spanning set 1.1. Let S = { v 1, v n } be n vectors in V, we have defined
More informationVector Space Basics. 1 Abstract Vector Spaces. 1. (commutativity of vector addition) u + v = v + u. 2. (associativity of vector addition)
Vector Space Basics (Remark: these notes are highly formal and may be a useful reference to some students however I am also posting Ray Heitmann's notes to Canvas for students interested in a direct computational
More informationMath 54 HW 4 solutions
Math 54 HW 4 solutions 2.2. Section 2.2 (a) False: Recall that performing a series of elementary row operations A is equivalent to multiplying A by a series of elementary matrices. Suppose that E,...,
More information1 Last time: multiplying vectors matrices
MATH Linear algebra (Fall 7) Lecture Last time: multiplying vectors matrices Given a matrix A = a a a n a a a n and a vector v = a m a m a mn Av = v a a + v a a v v + + Rn we define a n a n a m a m a mn
More informationLinear DifferentiaL Equation
Linear DifferentiaL Equation Massoud Malek The set F of all complex-valued functions is known to be a vector space of infinite dimension. Solutions to any linear differential equations, form a subspace
More informationSolutions to Selected Questions from Denis Sevee s Vector Geometry. (Updated )
Solutions to Selected Questions from Denis Sevee s Vector Geometry. (Updated 24--27) Denis Sevee s Vector Geometry notes appear as Chapter 5 in the current custom textbook used at John Abbott College for
More informationNONCOMMUTATIVE POLYNOMIAL EQUATIONS. Edward S. Letzter. Introduction
NONCOMMUTATIVE POLYNOMIAL EQUATIONS Edward S Letzter Introduction My aim in these notes is twofold: First, to briefly review some linear algebra Second, to provide you with some new tools and techniques
More informationMATH 315 Linear Algebra Homework #1 Assigned: August 20, 2018
Homework #1 Assigned: August 20, 2018 Review the following subjects involving systems of equations and matrices from Calculus II. Linear systems of equations Converting systems to matrix form Pivot entry
More information4.6 Bases and Dimension
46 Bases and Dimension 281 40 (a) Show that {1,x,x 2,x 3 } is linearly independent on every interval (b) If f k (x) = x k for k = 0, 1,,n, show that {f 0,f 1,,f n } is linearly independent on every interval
More informationLecture 17: Section 4.2
Lecture 17: Section 4.2 Shuanglin Shao November 4, 2013 Subspaces We will discuss subspaces of vector spaces. Subspaces Definition. A subset W is a vector space V is called a subspace of V if W is itself
More informationMATH 2331 Linear Algebra. Section 1.1 Systems of Linear Equations. Finding the solution to a set of two equations in two variables: Example 1: Solve:
MATH 2331 Linear Algebra Section 1.1 Systems of Linear Equations Finding the solution to a set of two equations in two variables: Example 1: Solve: x x = 3 1 2 2x + 4x = 12 1 2 Geometric meaning: Do these
More informationEigenvalues and Eigenvectors
LECTURE 3 Eigenvalues and Eigenvectors Definition 3.. Let A be an n n matrix. The eigenvalue-eigenvector problem for A is the problem of finding numbers λ and vectors v R 3 such that Av = λv. If λ, v are
More informationMATH 304 Linear Algebra Lecture 10: Linear independence. Wronskian.
MATH 304 Linear Algebra Lecture 10: Linear independence. Wronskian. Spanning set Let S be a subset of a vector space V. Definition. The span of the set S is the smallest subspace W V that contains S. If
More informationLINEAR ALGEBRA SUMMARY SHEET.
LINEAR ALGEBRA SUMMARY SHEET RADON ROSBOROUGH https://intuitiveexplanationscom/linear-algebra-summary-sheet/ This document is a concise collection of many of the important theorems of linear algebra, organized
More informationChapter 2. Vector Spaces
Chapter 2 Vector Spaces Vector spaces and their ancillary structures provide the common language of linear algebra, and, as such, are an essential prerequisite for understanding contemporary applied mathematics.
More informationDISCRETE MATHEMATICS W W L CHEN
DISCRETE MATHEMATICS W W L CHEN c W W L Chen, 1991, 2008. This chapter is available free to all individuals, on the understanding that it is not to be used for financial gain, and may be downloaded and/or
More informationOrthogonality. 6.1 Orthogonal Vectors and Subspaces. Chapter 6
Chapter 6 Orthogonality 6.1 Orthogonal Vectors and Subspaces Recall that if nonzero vectors x, y R n are linearly independent then the subspace of all vectors αx + βy, α, β R (the space spanned by x and
More informationSupplementary Material for MTH 299 Online Edition
Supplementary Material for MTH 299 Online Edition Abstract This document contains supplementary material, such as definitions, explanations, examples, etc., to complement that of the text, How to Think
More informationAlgebra II. Paulius Drungilas and Jonas Jankauskas
Algebra II Paulius Drungilas and Jonas Jankauskas Contents 1. Quadratic forms 3 What is quadratic form? 3 Change of variables. 3 Equivalence of quadratic forms. 4 Canonical form. 4 Normal form. 7 Positive
More informationMath 5520 Homework 2 Solutions
Math 552 Homework 2 Solutions March, 26. Consider the function fx) = 2x ) 3 if x, 3x ) 2 if < x 2. Determine for which k there holds f H k, 2). Find D α f for α k. Solution. We show that k = 2. The formulas
More informationMath Exam 2, October 14, 2008
Math 96 - Exam 2, October 4, 28 Name: Problem (5 points Find all solutions to the following system of linear equations, check your work: x + x 2 x 3 2x 2 2x 3 2 x x 2 + x 3 2 Solution Let s perform Gaussian
More informationChapter 1: Systems of Linear Equations
Chapter : Systems of Linear Equations February, 9 Systems of linear equations Linear systems Lecture A linear equation in variables x, x,, x n is an equation of the form a x + a x + + a n x n = b, where
More informationIRREDUCIBILITY OF ELLIPTIC CURVES AND INTERSECTION WITH LINES.
IRREDUCIBILITY OF ELLIPTIC CURVES AND INTERSECTION WITH LINES. IAN KIMING 1. Non-singular points and tangents. Suppose that k is a field and that F (x 1,..., x n ) is a homogeneous polynomial in n variables
More informationMath 322. Spring 2015 Review Problems for Midterm 2
Linear Algebra: Topic: Linear Independence of vectors. Question. Math 3. Spring Review Problems for Midterm Explain why if A is not square, then either the row vectors or the column vectors of A are linearly
More informationThe Theory of Second Order Linear Differential Equations 1 Michael C. Sullivan Math Department Southern Illinois University
The Theory of Second Order Linear Differential Equations 1 Michael C. Sullivan Math Department Southern Illinois University These notes are intended as a supplement to section 3.2 of the textbook Elementary
More informationMATH 223 FINAL EXAM APRIL, 2005
MATH 223 FINAL EXAM APRIL, 2005 Instructions: (a) There are 10 problems in this exam. Each problem is worth five points, divided equally among parts. (b) Full credit is given to complete work only. Simply
More informationMTH Linear Algebra. Study Guide. Dr. Tony Yee Department of Mathematics and Information Technology The Hong Kong Institute of Education
MTH 3 Linear Algebra Study Guide Dr. Tony Yee Department of Mathematics and Information Technology The Hong Kong Institute of Education June 3, ii Contents Table of Contents iii Matrix Algebra. Real Life
More informationa (b + c) = a b + a c
Chapter 1 Vector spaces In the Linear Algebra I module, we encountered two kinds of vector space, namely real and complex. The real numbers and the complex numbers are both examples of an algebraic structure
More informationReview of Some Concepts from Linear Algebra: Part 2
Review of Some Concepts from Linear Algebra: Part 2 Department of Mathematics Boise State University January 16, 2019 Math 566 Linear Algebra Review: Part 2 January 16, 2019 1 / 22 Vector spaces A set
More informationb 1 b 2.. b = b m A = [a 1,a 2,...,a n ] where a 1,j a 2,j a j = a m,j Let A R m n and x 1 x 2 x = x n
Lectures -2: Linear Algebra Background Almost all linear and nonlinear problems in scientific computation require the use of linear algebra These lectures review basic concepts in a way that has proven
More informationInner Product Spaces
Inner Product Spaces Introduction Recall in the lecture on vector spaces that geometric vectors (i.e. vectors in two and three-dimensional Cartesian space have the properties of addition, subtraction,
More informationELEMENTARY LINEAR ALGEBRA
ELEMENTARY LINEAR ALGEBRA K R MATTHEWS DEPARTMENT OF MATHEMATICS UNIVERSITY OF QUEENSLAND First Printing, 99 Chapter LINEAR EQUATIONS Introduction to linear equations A linear equation in n unknowns x,
More informationNOTES on LINEAR ALGEBRA 1
School of Economics, Management and Statistics University of Bologna Academic Year 207/8 NOTES on LINEAR ALGEBRA for the students of Stats and Maths This is a modified version of the notes by Prof Laura
More informationBASIC NOTIONS. x + y = 1 3, 3x 5y + z = A + 3B,C + 2D, DC are not defined. A + C =
CHAPTER I BASIC NOTIONS (a) 8666 and 8833 (b) a =6,a =4 will work in the first case, but there are no possible such weightings to produce the second case, since Student and Student 3 have to end up with
More informationDS-GA 1002 Lecture notes 0 Fall Linear Algebra. These notes provide a review of basic concepts in linear algebra.
DS-GA 1002 Lecture notes 0 Fall 2016 Linear Algebra These notes provide a review of basic concepts in linear algebra. 1 Vector spaces You are no doubt familiar with vectors in R 2 or R 3, i.e. [ ] 1.1
More informationMAT Linear Algebra Collection of sample exams
MAT 342 - Linear Algebra Collection of sample exams A-x. (0 pts Give the precise definition of the row echelon form. 2. ( 0 pts After performing row reductions on the augmented matrix for a certain system
More informationAlgebra Workshops 10 and 11
Algebra Workshops 1 and 11 Suggestion: For Workshop 1 please do questions 2,3 and 14. For the other questions, it s best to wait till the material is covered in lectures. Bilinear and Quadratic Forms on
More informationSolution to Set 7, Math 2568
Solution to Set 7, Math 568 S 5.: No. 18: Let Q be the set of all nonsingular matrices with the usual definition of addition and scalar multiplication. Show that Q is not a vector space. In particular,
More information