v = w if the same length and the same direction Given v, we have the negative v. We denote the length of v by v.

Transcription

1 Linear Algebra [1] 4.1 Vectors and Lines Definition scalar : magnitude vector : magnitude and direction Geometrically, a vector v can be represented by an arrow. We denote the length of v by v. zero vector 0 : 0 = 0 Given v, we have the negative v. v = w if the same length and the same direction

2 Linear Algebra [2] sum v + w v + w v w w v + w v

3 Linear Algebra [3] scalar multiplication av (a R) v 2v 1 2 v 2v subtraction u v = u + ( v)

4 Linear Algebra [4] Thm. u, v, w : vectors, k, p R 1. u + v = v + u, 2. u + (v + w) = (u + v) + w 3. 0 s.t. 0 + u = u for each u. 4. For each u, u s.t. u + ( u) = k(u + v) = ku + kv, (k + p)u = ku + pu 6. (kp)u = k(pu), 7. 1 u = u

5 Linear Algebra [5] Thm. A, B, C : matrices of the same size, k, p F 1. A + B = B + A, 2. A + (B + C) = (A + B) + C 3. O s.t. O + A = A for each A. 4. For each A, A s.t. A + ( A) = O. 5. k(a + B) = ka + kb, (k + p)a = ka + pa 6. (kp)a = k(pa), 7. 1 A = A

6 Linear Algebra [6] Thm. f, g, h : continuous functions on D, k, p F 1. f + g = g + f, 2. f + (g + h) = (f + g) + h 3. 0 s.t. 0 + f = f for each f. 4. For each f, f s.t. f + ( f) = k(f + g) = kf + kg, (k + p)f = kf + pf 6. (kp)f = k(pf), 7. 1 f = f

7 Linear Algebra [7] The notion of vector space! 1. The set of matrices of the same size 2. The set of vectors in R 3 3. The set of continuous functions on D and so on.

8 Linear Algebra [8] The theorem says that we can manipulate vectors as if they are variables w.r.t. addition and scalar multiplication. Eg. 5(u 2v) + 6(5u + 2v) = 5u 10v + 30u + 12v = 35u + 2v.

9 Linear Algebra [9] Coordinates Consider a point P = (x, y, z). Then we obtain a vector p = OP : the position vector. Conversely, a vector p determines a unique point P. Thus we identify each point with the corresponding position vector. P (x, y, z) O p

10 Linear Algebra [10] Given u = (x, y, z) and u1 = (x1, y1, z1), we have u + u1 = (x + x1, y + y1, z + z1), au = (ax, ay, az), u u1 = (x x1, y y1, z z1).

11 Linear Algebra [11] Lines d P0 P p0 p O Assume that p0 and d are given. Then p is the position vector of a point P on the line if and only if p = p0 + td (t R).

12 Linear Algebra [12] If p = (x, y, z), d = (a, b, c), p0 = (x0, y0, z0), then we have = z0 + tc, y = y0 + tb, x = x0 + ta, (t R). This is the equation of the line through p0 parallel to d. Planes Later... we need the notion of inner product and cross product of vectors.

13 Linear Algebra [13] 5.1 Subspaces and Dimension Subspaces of F n vector = point in R 3 (x, y, z) coordinates????? (a1, a2,, an) R n = {(a1, a2,, an) ai R} = a1 a2. an ai R

14 Linear Algebra [14] C n = {(a1, a2,, an) ai C} = a1 a2. an ai C F n = R n or C n The n-tuples in F n will be called vectors.

15 Linear Algebra [15] Subspaces A subset U of F n is called a subspace if it satisfies the following conditions. 1. If X, Y U, then X + Y U. 2. If X U, then rx U for r F. Eg. 1. F n 2. {0} : the zero subspace

16 Linear Algebra [16] 3. a line through the origin in R n : {td} If t1d and t2d on the line, then t1d + t2d = (t1 + t2)d and r(t1d) = (rt1)d. 4. Let A be an m n matrix. We define nulla = kera = {X F n AX = O} and ima = {Y F m Y = AX for some X F n }. If X1, X2 kera, then A(X1 +X2) = AX1 +AX2 = O and A(rX1) = r(ax1) = O. If Y1, Y2 ima, then X1, X2 s.t. AX1 = Y1 and AX2 = Y2. Now A(X1 + X2) = Y1 + Y2 and A(rX1) = ry1.

17 Linear Algebra [17] 5. U = {(x, y) R 2 x 2 + y 2 = 1}. We have (1, 0), (0, 1) U, but (1, 0) + (0, 1) = (1, 1) / U. Thus U is not a subspace of R 2.

18 Linear Algebra [18] Spanning sets Def. Assume that X1, X2,, Xk F n. An expression a1x1 + a2x2 + + akxk is called a linear combination of X1, X2,, Xk (ai F). The span of X1, X2,, Xk is the set of all linear combinations of X1, X2,, Xk. span{x1, X2,, Xk} = {a1x1+a2x2+ +akxk ai F}

19 Linear Algebra [19] Thm. Assume that X1, X2,, Xk F n. 1. The span{x1, X2,, Xk} is a subspace of F n. 2. If W is a subspace containing X1, X2,, Xk, then span{x1, X2,, Xk} W. Proof. 1. Let U = span{x1, X2,, Xk}. If Y = s1x1 + + skxk, Z = t1x1 + + tkxk U, then Y + Z = (s1 + t1)x1 + + (sk + tk)xk U and ry = rs1x1 + + rskxk U. 2. Clear!

20 Linear Algebra [20] The span{x1, X2,, Xk} is the smallest subspace containing X1,, Xk. If U = span{x1, X2,, Xk}, then {X1, X2,, Xk} is a spanning set of U, and U is spanned by the Xi s. Eg. Recall Thm. Given AX = O, every solution is a linear combination of the basic solutions. Equivalently, the kera is the span of the basic solutions.

21 Linear Algebra [21] Assume A = [ C1 C2 Cn] : m n matrix. Then ima = span{c1, C2,, Cn}. Proof. For X F n, x1 AX = [ ] x2 C1 C2 Cn = x1c1+x2c2+ +xncn. xn ima = {AX X F n } = {x1c1 + x2c2 + + xncn} = span{c1, C2,, Cn}

22 Linear Algebra [22] Independence Def. {X1, X2,, Xk} : linearly independent if t1x1 + t2x2 + + tkxk = 0 implies t1 = t2 = = tk = 0. Thm. If {X1, X2,, Xk} is linearly independent, X span{x1, X2,, Xk} has a unique representation as a linear combination of the Xi s. Proof. r1x1 + + rkxk = s1x1 + + skxk (r1 s1)x1 + + (rk sk)xk = 0 Thus we have ri = si for all i.

23 Linear Algebra [23] Eg. X1, X2, X1 + X2 2X1 + 2X2 = 2(X1 + X2) Eg ,, r s t = r s =, r = s = t = t

24 Linear Algebra [24] Eg , 2, 2, r r r r4 1 r4 2 0 = 0 r r = 0 r

25 Linear Algebra [25] , 2, Eg. {X, Y } : indep. {2X + 3Y, X 5Y }: indep. r(2x + 3Y ) + s(x 5Y ) = O (2r + s)x + (3r 5s)Y = O 2r + s = 0, 3r 5s = 0 r = s = 0

26 Linear Algebra [26] Eg. lin. dep. lin. indep. lin. dep. lin. indep.

27 Linear Algebra [27] Thm. TFAE 1. A is invertible. 2. The columns of A are linearly independent. 3. The columns of A span F n. 4. The rows of A are linearly independent. 5. The rows of A span F n. 6. ima = F n. 7. kera = O.

28 Linear Algebra [28] Proof. x1 AX = [ ] x2 C1 C2 Cn = x1c1+x2c2+ +xncn. xn AX = O x1c1 + x2c2 + + xncn = O 2 AX = O has only the trivial solution. kera = O AX = B x1c1 + x2c2 + + xncn = B 3 AX = B has a solution for every B F n ima = F n A is invertible A T is invertible

29 Linear Algebra [29] Eg. 1 0, ,, A = , det A =

30 Linear Algebra [30] Dimension Def. U F n : a subspace A set {X1, X2,, Xk} is a basis of U, if 1. {X1, X2,, Xk} is linearly independent, 2. U = span{x1, X2,, Xk}. Thm. If {X1, X2,, Xk} and {Y1, Y2,, Ym} are two bases of U, then k = m. Def. the number of vectors in a basis of U = the dimension of U = dim U

31 Linear Algebra [31] Eg. For F n, E1 = 1, E2 = ,, En = ; the standard basis of F n Eg. If {X1, X2,, Xn} is a basis of F n and A is invertible, then {AX1, AX2,, AXn} is also a basis of F n.

32 Linear Algebra [32] Eg. Consider AX = O. Recall that kera=the span of the basic solutions. In fact, the basic solutions are linearly independent. The basic solutions form a basis for kera. Eg. Subspaces of R If dim U = 3, then U = R If dim U = 2, then U is a plane through O. 3. If dim U = 1, then U is a line through O. 4. If dim U = 0, then U = {O}.

33 Linear Algebra [33] Thm. Assume that dim U = m = B. Then B is linearly independent B spans U; in either case, B is a basis of U.