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4.1Vectorsandlines • Definition – Scalar :Magnitude Vector :Magnitudeanddirection Geometrically,Avector V Canberepresentedby

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4.1Vectorsandlines • Definition – Scalar :Magnitude Vector :Magnitudeanddirection Geometrically,Avector V Canberepresentedby 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 kvk. { zero vector 0 : k0k = 0 { Given v, we have the negative −v. { v = w if the same length and the same direction Kyu-Hwan Lee Linear Algebra [2] { sum v + w v + w v w w v + w v Kyu-Hwan Lee Linear Algebra [3] { scalar multiplication av (a 2 R) v 1 2v 2v −2v { subtraction u − v = u + (−v) Kyu-Hwan Lee Linear Algebra [4] Thm. u, v, w : vectors, k; p 2 R 1. u + v = v + u, 2. u + (v + w) = (u + v) + w 3. 9 0 s.t. 0 + u = u for each u. 4. For each u, 9 −u s.t. u + (−u) = 0. 5. k(u + v) = ku + kv , (k + p)u = ku + pu 6. (kp)u = k(pu), 7. 1 · u = u Kyu-Hwan Lee Linear Algebra [5] Thm. A, B, C : matrices of the same size, k; p 2 F 1. A + B = B + A, 2. A + (B + C) = (A + B) + C 3. 9 O s.t. O + A = A for each A. 4. For each A, 9 −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 Kyu-Hwan Lee Linear Algebra [6] Thm. f, g, h : continuous functions on D, k; p 2 F 1. f + g = g + f, 2. f + (g + h) = (f + g) + h 3. 9 0 s.t. 0 + f = f for each f. 4. For each f, 9 −f s.t. f + (−f) = 0. 5. k(f + g) = kf + kg , (k + p)f = kf + pf 6. (kp)f = k(pf), 7. 1 · f = f Kyu-Hwan Lee Linear Algebra [7] The notion of vector space! 1. The set of matrices of the same size 2. The set of vectors in R3 3. The set of continuous functions on D 4. ... 5. ... and so on. Kyu-Hwan Lee 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: Kyu-Hwan Lee 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) p O Kyu-Hwan Lee 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): Kyu-Hwan Lee Linear Algebra [11] • Lines P0 d 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 2 R): Kyu-Hwan Lee Linear Algebra [12] If p = (x; y; z), d = (a; b; c), p0 = (x0; y0; z0), then we have 8x = x0 + ta; > <>y = y0 + tb; (t 2 R): z = z0 + tc; > :> 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. Kyu-Hwan Lee Linear Algebra [13] 5.1 Subspaces and Dimension • Subspaces of Fn vector = point in R3 $ (x; y; z) coordinates ????? $ (a1; a2; · · · ; an) n R = f(a1; a2; · · · ; an)jai 2 Rg a1 82 3 9 > a2 > ∼ > i R> = <>6 . 7 a 2 => 6 . 7 6 7 >4an5 > > > : ; Kyu-Hwan Lee Linear Algebra [14] n C = f(a1; a2; · · · ; an)jai 2 Cg a1 82 3 9 > a2 > ∼ > i C> = <>6 . 7 a 2 => 6 . 7 6 7 >4an5 > > > : ; Fn = Rn or Cn The n-tuples in Fn will be called vectors. Kyu-Hwan Lee Linear Algebra [15] • Subspaces A subset U of Fn is called a subspace if it satisfies the following conditions. 1. If X; Y 2 U, then X + Y 2 U. 2. If X 2 U, then rX 2 U for r 2 F. Eg. 1. Fn 2. f0g : the zero subspace Kyu-Hwan Lee Linear Algebra [16] 3. a line through the origin in Rn : ftdg 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 n nullA = kerA = fX 2 F jAX = Og and m n imA = fY 2 F jY = AX for some X 2 F g: If X1; X2 2 kerA, then A(X1 +X2) = AX1 +AX2 = O and A(rX1) = r(AX1) = O. If Y1; Y2 2 imA, then 9 X1; X2 s.t. AX1 = Y1 and AX2 = Y2. Now A(X1 + X2) = Y1 + Y2 and A(rX1) = rY1. Kyu-Hwan Lee Linear Algebra [17] 5. U = f(x; y) 2 R2jx2 + y2 = 1g: We have (1; 0); (0; 1) 2 U, but (1; 0) + (0; 1) = (1; 1) 2= U. Thus U is not a subspace of R2. Kyu-Hwan Lee Linear Algebra [18] • Spanning sets n Def. Assume that X1; X2; · · · ; Xk 2 F . An expression a1X1 + a2X2 + · · · + akXk is called a linear combination of X1; X2; · · · ; Xk (ai 2 F). The span of X1; X2; · · · ; Xk is the set of all linear combinations of X1; X2; · · · ; Xk. spanfX1; X2; · · · ; Xkg = fa1X1+a2X2+· · ·+akXkjai 2 Fg Kyu-Hwan Lee Linear Algebra [19] n Thm. Assume that X1; X2; · · · ; Xk 2 F . n 1. The spanfX1; X2; · · · ; Xkg is a subspace of F . 2. If W is a subspace containing X1; X2; · · · ; Xk, then spanfX1; X2; · · · ; Xkg ⊂ W : Proof. 1. Let U = spanfX1; X2; · · · ; Xkg. If Y = s1X1 + · · · + skXk; Z = t1X1 + · · · + tkXk 2 U; then Y + Z = (s1 + t1)X1 + · · · + (sk + tk)Xk 2 U and rY = rs1X1 + · · · + rskXk 2 U. 2. Clear! 2 Kyu-Hwan Lee Linear Algebra [20] The spanfX1; X2; · · · ; Xkg is the smallest subspace containing X1; · · · ; Xk. If U = spanfX1; X2; · · · ; Xkg, then fX1; X2; · · · ; Xkg 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. Kyu-Hwan Lee Linear Algebra [21] Assume A = C1 C2 · · · Cn : m × n matrix. Then imA = spanfC1; C2; · · · ; Cng: Proof. For X 2 Fn, x1 2 3 x2 AX = C1 C2 · · · Cn 6 . 7 = x1C1+x2C2+· · ·+xnCn 6 . 7 6 7 4xn5 n imA = fAXjX 2 F g = fx1C1 + x2C2 + · · · + xnCng = spanfC1; C2; · · · ; Cng 2 Kyu-Hwan Lee Linear Algebra [22] • Independence Def. fX1; X2; · · · ; Xkg : linearly independent if t1X1 + t2X2 + · · · + tkXk = 0 implies t1 = t2 = · · · = tk = 0. Thm. If fX1; X2; · · · ; Xkg is linearly independent, X 2 spanfX1; X2; · · · ; Xkg 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. 2 Kyu-Hwan Lee Linear Algebra [23] Eg. X1; X2; X1 + X2 2X1 + 2X2 = 2(X1 + X2) Eg. 1 0 1 1 0 1 0 2 3 2 3 2 3 2 3 2 3 2 3 2 3 0 1 1 0 1 1 0 6 7 ; 6 7 ; 6 7 r 6 7 + s 6 7 + t 6 7 = 6 7 617 617 617 617 617 617 607 6 7 6 7 6 7 6 7 6 7 6 7 6 7 425 425 435 425 425 435 405 1 0 1 0 1 0 0 2 3 r 2 3 2 3 0 1 1 2 3 0 0 1 0 6 7 s = 6 7 ) 6 7 ; r = s = t = 0 61 1 17 6 7 607 60 0 17 6 7 4t5 6 7 6 7 42 2 35 405 40 0 05 Kyu-Hwan Lee Linear Algebra [24] Eg. 1 1 −3 2 2 3 2 3 2 3 2 3 2 ; −2 ; 2 ; 0 6 7 6 7 6 7 6 7 4−15 4 1 5 4 1 5 405 1 1 −3 2 0 2 3 2 3 2 3 2 3 2 3 r1 2 + r2 −2 + r3 2 + r4 0 = 0 6 7 6 7 6 7 6 7 6 7 4−15 4 1 5 4−15 405 405 r1 1 1 −3 2 2 3 0 2 3 r2 2 3 2 −2 2 0 6 7 = 0 6 7 6r37 6 7 4−1 1 −1 05 6 7 405 4r45 Kyu-Hwan Lee Linear Algebra [25] 1 1 −3 2 3 2 3 2 3 2 ; −2 ; 2 6 7 6 7 6 7 4−15 4 1 5 4−15 1 1 −3 1 0 −1 2 3 2 3 2 −2 2 ) 0 1 −2 6 7 6 7 4−1 1 −15 40 0 0 5 Eg. fX; Y g : indep. ) f2X + 3Y; X − 5Y g: indep. r(2X + 3Y ) + s(X − 5Y ) = O (2r + s)X + (3r − 5s)Y = O 2r + s = 0; 3r − 5s = 0 r = s = 0 Kyu-Hwan Lee Linear Algebra [26] Eg. lin. dep. lin. indep. lin. dep. lin. indep. Kyu-Hwan Lee Linear Algebra [27] Thm. TFAE 1. A is invertible. 2. The columns of A are linearly independent. 3. The columns of A span Fn.
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