ENM 500 Linear Algebra Review  1. u = (2, -4, 3) and v = (-5, 1, 2). u + v = (2-5, -4+1, 3 +2) = (-3, -3, 5). 2u = (4, -8, 6); -v = (5, -1, -2); v u . v = ( 2*-5 + -4*1+ 3*2) = -8. ||u|| = is the length   d(u,v) = is distance 2. x(2, 3) = (y, 6) => 2x = y, 3x = 6 => x = 2, y = 4. u + v u 3u -2u 4/8/2019

Find k so that u and v and are orthogonal with = 0 => orthogonal u = (-1, k, -2) and v = (4, -2, 5). -4 –2k –10 = 0 => k = -7.  4. Let u = (k, (sqrt 3), 4). Find k so that ||u|| = 10. (k2 + 3 + 16)1/2 = 10 => k = 9.   Let u = (2, –5, 1) and v = (3, 0, 2). Find .   = 2*3 - 5*0 + 1*2 = 8. 6. Solve 2x + 3y = 6. Infinitely many solutions 4/8/2019

i) x = (4 – 3y)/2 substituted for x in 2nd equation yields Solve by i) substitution, ii) elimination, iii) Cramer’s Rule, iv) Gaussian reduction, v) elementary row operations. 2x + 3y = 4 5x + 4y = 3 i) x = (4 – 3y)/2 substituted for x in 2nd equation yields 10 – 7.5y + 4y = 3 => y = 2; x = -1.  ii) 2x + 3y = 4; multiply by 5  10x + 15y = 20 5x + 4y = 3; multiply by 2  10x + 8y = 6 Then subtract to get 7y = 14 => y = 2 and x = -1. 4/8/2019

= 7 / -7 = -1 = -14/-4 = 2 iii) Cramer's Rule 2x + 3y = 4 5x + 4y = 3 4/8/2019

iv) Matrix Inverse: Ax = b => A-1Ax = x = A- b Where x = (x, y) and b = (4, 3) (inverse #2A((2 3)(5 4)))  #2A((-4/7 3/7)(5/7 -2/7) (print-matrix *)  -4/7 3/7 5/7 -2/7 4/8/2019

v) Elementary Row Operations 2 3 4 1 1.5 2 1 1.5 2 1 0 -1 5 4 3 0 -3.5 -7 0 1 2 0 1 2 2x + 3y = 4 5x + 4y = 3 Find the inverse 2 3 1 0 1 3/2 ½ 0 1 3/2 ½ 0 1 0 -4/7 3/7 5 4 0 1 0 -7/2 -5/2 1 0 1 5/7 -2/7 0 1 5/7 -2/7 -4/7 3/7 5/7 -2/7 4/8/2019

Find the inverse of a 2 by 2 matrix. (inverse #2A((1 3)(4 -2)))  #2A((1/7 3/14)(2/7 -1/14)) 4/8/2019

Trace (A) = 3 + 2 = 5 = sum of main diagonal elements A matrix may be looked upon as a function. Consider a matrix A which maps all vectors in the plane. For example, A = A: (1, 3)  = Trace (A) = 3 + 2 = 5 = sum of main diagonal elements = a11 + a22 + … + ann 4/8/2019

Ax = tx => (tI – A)x = 0 => 9. Let A = and B = Find A2, AB, AT, (AB)T. Find A’s characteristic equation and show that A satisfies it. Ax = tx => (tI – A)x = 0 => A2 - 7I = O A2 = AB = = AT = (AB)T = 4/8/2019

A = and B = f(A) = 2A2 – 5A + 6I g(A) = A3-2A2 + A + 3I 10. Let f(t) = 2t2 –5t + 6 and g(t) = t3 –2t2 + t +3. Find f(A) and g(A) and f(B) and g(B) for matrices A = and B = f(A) = 2A2 – 5A + 6I g(A) = A3-2A2 + A + 3I A2 = A3 = (expt-mat #2A((2 -3)(5 1)) 2)  #2A((-11 -9)(15 -14)) (m-list-ops (list (expt-matrix #2A((2 -3)(5 1)) 2) (K*mat #2A((2 -3)(5 1)) -3) (K*mat (identity-matrix 2) 17)) #' M+) 4/8/2019

11. Find the inverse of the following matrices by forming the adjacency matrix, and then by fixing the identity matrix to it and converting the initial matrix to the identity matrix.   B adjB |B|I A = A-1 = B-1 = 1 4 1 0 1 4 1 0 1 0 -3/5 4/5 2 3 0 1 0 -5 -2 1 0 1 2/5 -1/5  4/8/2019

Matrix Inverse e.g., Cross out Row 3 and Column 1, evaluate the remaining determinant as 6, sum the indices 3 + 1 = 4, even => put 6 in transposed indices (1,3); if sum is odd put negated determinant in transposed indices. Do the (3, 2) entry -6. The determinant is -6 but the sum of the indices is 5, odd, thus a – (-6) = 6 is put in transposed indices (2, 3) . Do the (1, 2) entry -3. The determinant is -6 but the sum of the indices is 3, odd, thus a – (-6) = 6 is put in transposed indices (2, 1) . Do the (2, 3) entry 3. The determinant is 12 but the sum of the indices is 5, odd, thus a – 12 is put in transposed indices (3, 2). Continued and finally divide the transposed adjacently matrix by the determinant of B. B = Badj = 4/8/2019

Find the eigenvalues and corresponding eigenvectors of A = (tI – A)u = 0, where u= (x, y) = (t – 5)(t + 1) => t = 5, -1 = 0 => x = y or (1, 1); eigenvector = -2x –4y = 0 => x = -2y or (2, -1) let P = P-1AP is diagonal matrix with the eigenvalues. PAP-1 (m-list-ops (list (inverse #2A((1 2)(1 -1))) #2A((1 4)(2 3)) #2A((1 2)(1 -1))))  #2A((5 0)(0 -1)) (eigenvalues #2A((1 4)(2 3))) 4/8/2019

(eigenvalues (inverse am))  (1.0 0.5) 13. (setf am #2A((1 0)(1 2))) (eigenvalues am)  (2.0 1.0) (eigenvalues (inverse am))  (1.0 0.5) Observe inverse eigenvalues are reciprocals of original matrix's eigenvalues. 14. Show that AB and BA have same eigenvalues. (setf am #2A((1 0)(1 2)) bm #2A((4 7)(9 2))) (eigenvalues (mmult am bm))  (20.393796 -5.393797) (eigenvalues (mmult bm am))  (20.393796 -5.393797) 4/8/2019

Matrices 4/8/2019

Upper triangular and Det = product of main diagonal 5 * 1* 3* 2 15. Matrices Upper triangular and Det = product of main diagonal 5 * 1* 3* 2 Symmetric since A = AT Det (AB) = [(Det A)*(Det B)]; Note: AB  BA in general. 4/8/2019

A(BC) = (AB)C Associative A(B + C) = AB + AC Distributive 16. Matrix Properties A(BC) = (AB)C Associative A(B + C) = AB + AC Distributive (A + B)C = AC + BC Distributive (AB)T = BTAT Transpose AB  BA in general; unless (commutative) matrices. An = AAAA…AAAA Power of A n A's multiplied AmAn = Am+n (Am)n = Amn AI = IA = A Commutative with Identity matrix AB = 0 but A, B  0 Example: (setf ma #2a((0 1 0)(1 0 1)(0 1 0)) mb #2a((1 -2 1)(0 0 0)(-1 2 -1))) (Mmult ma mb)  #2a((0 0 0)(0 0 0)(0 0 0)) 4/8/2019

Determinants Compute the determinant of each matrix. A = B = |A| = (det #2A((1 2 3)(4 -2 3)(2 5 -1)))  79 |B| = (det #2A((2 0 1)(4 2 -3)(5 3 1)))  24 4/8/2019

|AB| = |A| |B| (setf am #2A((1 0)(1 2)) bm #2A((4 7)(9 2)) (det (mmult am bm))  -110 (* (det am) (det bm))  -110 (eigen (mmult A B)) = (eigen (mmult B A)) (mmult A B)  #2A((12 8 0)(6 4 0)(-7 -2 1)) (mmult B A)  #2A((12 6 1)(8 4 -2)(-2 -1 1)) AB  BA 4/8/2019

Eigenvalues: AB = BA (eigen (mmult am bm))  ((4.0 -8.0 -8.0) (#2A((4.0 -4 4)(-8 8.0 4)(0 0 12.0)) #2A((-8.0 -4 4)(-8 -4.0 4)(0 0 0.0)) #2A((-8.0 -4 4)(-8 -4.0 4)(0 0 0.0)))) (eigen (mmult bm am))  ((4.0 -8.0 -8.0) (#2A((10.0 -10 10)(-2 2.0 10)(0 0 12.0)) #2A((-2.0 -10 10)(-2 -10.0 10)(0 0 0.0)) #2A((-2.0 -10 10)(-2 -10.0 10)(0 0 0.0)))) 4/8/2019

A *(adj A)= (adj A)*A= |A|I (adj-mat am)  #2A((2 0)(-1 1) (adj-mat #2A((1 -3 3)(3 -5 3)(6 -6 4)))  #2A((-2 -6 6)(6 -14 6)(12 -12 4)) (det #2A((1 -3 3)(3 -5 3)(6 -6 4)))  16 (mmult #2A(( 1 -3 3)(3 -5 3)( 6 -6 4)) #2A((-2 -6 6)(6 -14 6)(12 -12 4)))  #2A((16 0 0)(0 16 0)(0 0 16)) 4/8/2019

Characteristic Equations (setf am #2A(( 1 -3 3)(3 -5 3)(6 -6 4)) bm #2A((-3 1 -1)(-7 5 -1)(-6 6 -2))) (char-e am)  1t3 + 0t2 -12t -16 (characteristic equation) (char-e bm)  1t3 + 0t2 -12t -16 Matrices am and bm have the same characteristic equations but am has 3 independent eigenvectors and bm has two; and thus the two matrices are not similar. (M+ (mmult am am am) (k*mat -12 am) (k*mat -16 (identity-matrix 3)))  #2A((0 0 0)(0 0 0)(0 0 0)); am satisfies C-E 4/8/2019

Powers of Square Matrix (setf A-mat #2A((1 2)(3 4))) (expt-matrix A-mat 5)  #2A((1069 1558)(2337 3406)) (m-list-ops (list-of 5 a-mat))  4/8/2019

AB = AC; but B  C (setf A #2a((4 2 0)(2 1 0)(-2 -1 1)) B #2a((2 3 1)(2 -2 -2)(-1 2 1)) C #2a((3 1 -3)(0 2 6)(-1 2 1))) (mmult a b)  #2A((12 8 0)(6 4 0)(-7 -2 1))) (mmult a c)  #2A((12 8 0)(6 4 0)(-7 -2 1))) Note that (det A)  0; A has no inverse. 4/8/2019

Induction Example Prove for n = 1; Assume true for n = k' prove true for n = k + 1 1 + 2 + … + n = n(n+1)/2 1 + 2 + … + n + (n + 1) = n(n+1)/2 + (n+1) = [n(n+1) + 2(n+1)]/2 = (n+1)(n+2)/2 4/8/2019