Math 636 Final - Quiz Component 1. The system of linear equations 2x2 x3 = 3 2x1 + x2 + 2x3 = 1 x1 + x2 + 3x3 = 2 (a) is consistent with a unique solution. (b) is consistent with infinitely many solutions. (c) is inconsistent. 2. Which of the sets of polynomials in P2 (R) is linearly dependent? (a) {1 + 2x + 2x2 , 1 + x + x2 , x2 + 4x + 4} (b) {2, 2 + 3x, 2 + 3x + 4x2 } (c) {3 + 2x 2x2 , 2 2x + 2x2 , 1 + 4x + 3x2 } (d) {1 + x + 3x2 , 2 + 4x2 , 6 3x 3x2 } 3. Let (a) (b) (c) (d) A be an invertible matrix. Which of the following is false? A must be n n. Row(A) = Null(AT ). det(A1 ) 6= det(A). The columns of A form a basis for Rn . 4. Which of the following statements is true? (a) If A~v = ~v , then ~v is an eigenvector of A. (b) If A is n n and rank(A) < n, then A has a row of zeroes. (c) If P is an orthogonal matrix, then P is invertible. \u0002 \u0003 (d) If {~v1 , . . . , ~vk } is a linearly independent set of vectors in Rn , then A = ~v1 , . . . , ~vk is invertible. 5. Let V be a finite dimensional vector space and let {~v1 , . . . , ~vk } be a linearly independent set in V. Which of the following statements is false? (a) dim V k. (b) Span{~v1 , . . . , ~vk } is a subspace of V. (c) Every basis for V contains the vectors ~v1 , . . . , ~vk . (d) If dim V = k, then Span{~v1 , . . . , ~vk } = V. 1 2 0 0 2 6. What is the geometric multiplicity of the eigenvalue = 2 of A = 1 2 1 ? 1 0 3 (a) g = 1 (b) g = 2 (c) g = 3 7. Which of the following is the vector ~x that is closest to being a solution for the system x1 + x2 = 2 x1 2x2 = 3 3x1 2x2 = 2 \u0014 \u0015 3/10 (a) 43/30 \u0014 \u0015 9/20 (b) 43/20 \u0014 \u0015 1/4 (c) 43/36 \u0014 \u0015 1/4 (d) 43/36 8. Which of the following is the best fitting equation of the form y = a + bx for the following data? x 1 0 1 y 5 2 2 (a) 5 3 72 x (b) 5 3 + 72 x (c) 2 3x (d) 11 6 9. In R4 using the inner product h~x, ~y i = x1 y1 + 2x2 y2 + 3x3 y3 + x4 y4 . the following four vectors are orthogonal: 5 0 1 1 1 3 1 1 ~v1 = 1 , ~v2 = 1 , ~v3 = 2 , ~v4 = 1 0 0 6 1 1 1 Let S = Span{~v1 , ~v2 , ~v3 }. The projection of ~x = 1 onto S is: 1 8/7 1 0 8/7 8/ 7 1 0 8/7 (b) (c) (a) (d) 1 2 6/ 7 6/7 1 7 1/7 1/ 7 10 x 3 3 \u0014 \u0015 \u0014 \u0015 \u0014 \u0015 1 1 1 2 2 1 10. Let A1 = , A2 = , A3 = M22 (R) and define S = Span{A1 , A2 , A3 }. 1 1 1 0 0 1 \u0014 \u0015 0 0 If D = , then projS (D) is: 0 2 \u0014 \u0015 \u0014 \u0015 \u0014 \u0015 \u0014 \u0015 0 0 6 2 7/6 1 2/3 0 (a) (b) (c) (d) 1 1 4 4 0 5/6 2/3 2/3 For questions 11 - 20, determine if the statement is True or False. You should make sure that you have a proof of each true statement and a counter example for each false statement. 11. If the reduced row echelon form of (A I) is I, then is not an eigenvalue of A. (a) True. (b) False. 12. If a system of 3 linear equations in 4 unknowns is such that the coefficient matrix has a rank of 2, then the system has infinitely many solutions. (a) True. (b) False. 13. If A is not diagonalizable, then A is not invertible. (a) True. (b) False. 14. If ~v1 , ~v2 R3 , then ~x = s~v1 + t~v2 , s, t R is a vector equation for a plane in R3 . (a) True. (b) False. 15. If A is a 3 3 matrix, and ~x, ~y , and ~z are vectors such that A~x = 2~x, A~y = 3~y , and A~z = 5~z, then det(A) = 30. (a) True. (b) False. 16. If A is diagonalizable, then AT is diagonalizable. (a) True. (b) False. 17. If A is invertible and A is similar to B, the B is invertible. (a) True. (b) False. 4 1 18. If ~x = 1 is a solution of a homogenous system of linear equations, then the system 1 has infinitely many solutions. (a) True. (b) False. 19. Let ~a, ~b, ~c Rn with ~a 6= ~0. If ~a ~b = 0 and ~a ~c = 0, then ~b = ~c. (a) True. (b) False. 20. Let ~b, ~v R3 . If ~b 6= ~0, then the set with vector equation ~x = ~b + t~v is not a subspace of R3 . (a) True. (b) False. Math 636 Final - Written Component 1. Find a and b to obtain the best fitting equation of the form y = a + bt2 for the given data. t 2 1 0 1 2 y 2 0 1 1 3 2. Assume that A is an m n matrix and B is an n m matrix where m 6= n. Prove that if AB = Im , then rank(B) = m. 3. Prove Corollary 6.3.4 which states: If A is an n n matrix with distinct eigenvalues 1 , . . . , k , then A is diagonalizable if and only if gi = ai for 1 i k. 4. Let W be a subspace of a finite dimensional inner product space V. Prove that projW (~v ) is independent of basis. That is, prove that if {~v1 , . . . , ~vk } and {w ~ 1, . . . , w ~ k } are both orthogonal bases for W, then h~v , ~vk i h~v , w ~ 1i h~v , w ~ ki h~v , ~v1 i ~v1 + + ~vk = w ~1 + + w ~k 2 2 2 k~v1 k k~vk k kw ~ 1k kw ~ k k2 NOTE: Notice that our definition of projW is dependent on the choice of orthogonal basis. So, you cannot assume that projW (~v ) = projW (~v ) if we are using different bases. This is what you are supposed to prove in this question. 1 MATH 636 Lecture 28 Slides (Module 5) (Last Updated: May 22, 2014) Linear Mappings Module 5: Linear Mappings and Diagonalization Linear Mappings Module 5: Linear Mappings and Diagonalization Linear Mappings One of the most fundamental concepts in mathematics is that of a function. Recall that a function f is a rule that assigns to every element x of an initial set V , called the domain of f , a unique value y in another set U , called the codomain of f . We say that f maps V to U and write f : V U . Additionally, we say that y is the image of x under f and write f (x) = y. We also recall that the range of a function is the subset of the codomain that contains all images of the elements of V under f . Mathematically, Range(f ) = {f (x) U | x V } You have seen and worked with functions f : R R in your earlier mathematics. However, in many real world applications, we require functions whose domain and/or codomain are other sets. For example, vector spaces! So, in this lecture, we will start looking at functions whose domain and codomain are vector spaces. Linear Mappings Module 5: Linear Mappings and Diagonalization Page 1 of 11 c University of Waterloo and others MATH 636 Lecture 28 Slides (Module 5) (Last Updated: May 22, 2014) Linear Mappings We want to make sure that our functions behave well with respect to vectors spaces. In particular, if we have two vector spaces V and U, and a function f : V U, then it would make sense to require that the range of f is a subspace of U. What conditions do we require on the function f to ensure that its range will be a subspace of U? We need to make sure that f preserves addition and scalar multiplication. That is, for any v 1, v 2 V and c R we must have f ( v 1 ) + f ( v 2 ) = f ( v1+ v 2) cf ( v ) = f (c v ) 1 1 In words, \"the sum of images is the image of the sum and the scalar multiple of an image is the image of the scalar multiple.\" Linear Mappings Module 5: Linear Mappings and Diagonalization Linear Mappings Theorem 8.1.1 If V and U are vector spaces and f : V U is a function that satisfies L1 f ( v ) + f ( v ) = f ( v + v ) 1 2 1 2 L2 cf ( v 1 ) = f (c v 1) for all v 1 , v 2 V and c R, then Range(f ) = {f ( x)U| x V} is a subspace of U. Proof: By definition, Range(f ) is a non-empty subset of U. So, let y 1, y 2 Range(f ). This means there exists x 1, x 2 V such that f ( x 1) = y1 f ( x 2) = y2 and To prove that Range(f ) is closed under addition, we need to show that y1+ y 2 Range(f ). So, we need to find a vector in V that maps to y1+ y 2. We see that x + x V and 1 2 f ( x1+ x 2 ) = f ( x 1 ) + f ( x 2 ), = y + y 1 by L1 2 as required. Thus, y1+ y 2 Range(f ) Linear Mappings Module 5: Linear Mappings and Diagonalization Page 2 of 11 c University of Waterloo and others MATH 636 Lecture 28 Slides (Module 5) (Last Updated: May 22, 2014) Linear Mappings Theorem 8.1.1 If V and U are vector spaces and f : V U is a function that satisfies L1 f ( v ) + f ( v ) = f ( v + v ) 1 2 1 2 L2 cf ( v 1 ) = f (c v 1) for all v 1 , v 2 V and c R, then Range(f ) = {f ( x)U| x V} is a subspace of U. Proof: Similarly, using L2 we get for any c R that f (c x 1 ) = cf ( x 1 ) = c y1 Hence, c y 1 Range(f ). Consequently, by the Subspace Test, Range(f ) is a subspace of U. Linear Mappings Module 5: Linear Mappings and Diagonalization Linear Mappings Definition Let V and U be vector spaces. A function L : V U is called a linear mapping if L(t x + s y ) = tL( x ) + sL( y) for all x , y V and s, t R. Remarks 1. Some textbooks will define linear mappings in terms of properties L1 and L2, while others will define it as L(t x + y ) = tL( x ) + L( y) The reason I chose L(t x + s y ) = tL( x ) + sL( y) is to stress the fact that the main property of linear mappings is that they preserve linear combinations. That is, L(c1 v 1 + + cn v n ) = c1 L( v 1 ) + + cn L( v n) This property, called the linearity property, is the one we will use most often. Linear Mappings Module 5: Linear Mappings and Diagonalization Page 3 of 11 c University of Waterloo and others MATH 636 Lecture 28 Slides (Module 5) (Last Updated: May 22, 2014) Linear Mappings Definition Let V and U be vector spaces. A function L : V U is called a linear mapping if L(t x + s y ) = tL( x ) + sL( y) for all x , y V and s, t R. Remarks 2. Linear mappings are some times called linear transformations. 3. A linear mapping L : V V is often called a linear operator. We will use linear operators a lot in the near future. 4. If f : Rn Rm , then we should write x1 y1 . . f .. = .. xn ym However, we will often write f (x1 , . . . , xn ) = (y1 , . . . , ym ) Linear Mappings Module 5: Linear Mappings and Diagonalization Linear Mapping Example Example Let f : R2 R2 be defined by f (x1 , x2 ) = (2x1 + x2 , 3x1 + 5x2 ). (a) Evaluate f (1, 2) and f (0, 0). Solution: We have f (1, 2) = (2(1) + 2, 3(1) + 5(2)) = (4, 7) f (0, 0) = (2(0) + 0, 3(0) + 5(0)) = (0, 0) Just a reminder that, although we write this for compactness, in your mind you should be seeing \u0012\u0014 \u0015\u0013 \u0014 \u0015 1 4 f = 2 7 \u0012\u0014 \u0015\u0013 \u0014 \u0015 0 0 f = 0 0 Linear Mappings Module 5: Linear Mappings and Diagonalization Page 4 of 11 c University of Waterloo and others MATH 636 Lecture 28 Slides (Module 5) (Last Updated: May 22, 2014) Linear Mapping Example Example Let f : R2 R2 be defined by f (x1 , x2 ) = (2x1 + x2 , 3x1 + 5x2 ). (b) Find a vector x R2 such that f ( x ) = (1, 1). \u0014 \u0015 x Solution: We need to find x = 1 such that x2 \u0014 \u0015 \u0012\u0014 \u0015\u0013 \u0014 \u0015 1 x1 2x1 + x2 =f = 1 x2 3x1 + 5x2 This gives us a system of linear equations 2x1 + x2 = 1 3x1 + 5x2 = 1 Row reducing the corresponding augmented matrix, we find that x1 = 5 x2 = 13 . \u0014 \u0015 4/13 Thus, x = . It is easy to verify that this vector works. 5/13 Linear Mappings 4 13 and Module 5: Linear Mappings and Diagonalization Linear Mapping Example Example Let f : R2 R2 be defined by f (x1 , x2 ) = (2x1 + x2 , 3x1 + 5x2 ). (c) Prove that f is linear. Solution: To prove that it is linear, we must show that it satisfies the linearity property. \u0014 \u0015 \u0014 \u0015 x y Let x = 1 , y = 1 R2 and let s, t R. x2 y2 Then, \u0001 f (s x + t y ) = f s(x1 , x2 ) + t(y1 , y2 ) = f (sx1 + ty1 , sx2 + ty2 ) \u0001 = 2(sx1 + ty1 ) + (sx2 + ty2 ), 3(sx1 + ty1 ) + 5(sx2 + ty2 ) = s(2x1 + x2 , 3x1 + 5x2 ) + t(2y1 + y2 , 3y1 + 5y2 ) = sf (x1 , x2 ) + tf (y1 , y2 ) = sf ( x ) + tf ( y) Hence, f is linear. Linear Mappings Module 5: Linear Mappings and Diagonalization Page 5 of 11 c University of Waterloo and others MATH 636 Lecture 28 Slides (Module 5) (Last Updated: May 22, 2014) Linear Mapping Example Example Let L : M22 (R) P2 (R) be defined by L (a) Evaluate L \u0012\u0014 1 1 \u0015\u0013 2 . 1 \u0012\u0014 a c b d \u0015\u0013 = (a + b + c)x + (a b d)x2 . Solution: \u0012\u0014 \u0015\u0013 1 2 L = [1 + 2 + (1)]x + (1 2 1)x2 = 2x 2x2 1 1 Linear Mappings Module 5: Linear Mappings and Diagonalization Linear Mapping Example Example \u0012\u0014 a Let L : M22 (R) P2 (R) be defined by L c b d \u0015\u0013 = (a + b + c)x + (a b d)x2 . (b) Find a matrix A such that L(A) = 2x + x2 . Solution: We need to find A = 2x + x2 = L \u0012\u0014 \u0014 a c a c \u0015 b such that d \u0015\u0013 b = (a + b + c)x + (a b d)x2 d Hence, we need a + b + c = 2 and a b d = 1. Solving this system, we see that one choice is a = 3/2, b = 1/2, c = 0, and d = 0. That is, \u0012\u0014 \u0015\u0013 3/2 1/2 L = 2x + x2 0 0 Linear Mappings Module 5: Linear Mappings and Diagonalization Page 6 of 11 c University of Waterloo and others MATH 636 Lecture 28 Slides (Module 5) (Last Updated: May 22, 2014) Linear Mapping Example Example Let L : M22 (R) P2 (R) be defined by L \u0012\u0014 a c b d \u0015\u0013 = (a + b + c)x + (a b d)x2 . (c) Prove that L is linear. Solution: \u0014 \u0015 \u0014 a b1 a Let 1 , 2 c1 d1 c2 Then, L s \u0014 a1 c1 \u0015 b2 M22 (R) and s, t R. d2 \u0015 \u0014 b1 a +t 2 d1 c2 \u0015 \u0012\u0014 b2 \u0001 sa1 + ta2 =L d2 sc1 + tc2 sb1 + tb2 sd1 + td2 \u0015\u0013 = ([sa1 + ta2 ] + [sb1 + tb2 ] + [sc1 + tc2 ])x + ([sa1 + ta2 ] [sb1 + tb2 ] [sd1 + td2 ])x 2 2 = s[(a1 + b1 + c1 )x + (a1 b1 d1 )x ] + t[(a2 + b2 + c2 )x + (a2 b2 d2 )x ] \u0012\u0014 \u0015\u0013 \u0012\u0014 \u0015\u0013 a1 b1 a2 b2 = sL + tL c1 d1 c2 d2 Hence, L is linear. Linear Mappings Module 5: Linear Mappings and Diagonalization Linear Mapping Example Example Determine if the mapping f : R3 R defined by f ( x ) = k x k is linear. Solution: Let x , y R3 and s, t R. We have that f (s x + t y ) = ks x + t yk Note that Theorem 1.3.3 tells us that ks x + t y k |s|k x k + |t|k y k. 1 0 Take x = 0 and y = 1. Then 0 0 1 f ( x + y ) = f (1, 1, 0) = 1 = 2 0 but, 1 0 f ( x ) + f ( y)= 0 + 1 =1+1=2 0 0 Thus, f ( x + y ) 6= f ( x ) + f ( y ), hence f is not linear. Linear Mappings Module 5: Linear Mappings and Diagonalization Page 7 of 11 c University of Waterloo and others 2 MATH 636 Lecture 28 Slides (Module 5) (Last Updated: May 22, 2014) Check-In Question 1 Determine if the mapping L : R2 P1 (R) defined by L(a1 , a2 ) = (a1 + a2 ) + (2a1 + 2a2 )x is linear. \u0014 \u0015 \u0014 \u0015 a b Solution: Let a = 1 , b = 1 R2 and s, t R. We have a2 b2 L(s a + t b ) = L(sa1 + tb1 , sa2 + tb2 ) = (sa1 + tb1 + sa2 + tb2 ) + (2[sa1 + tb1 ] + 2[sa2 + tb2 ])x = s[(a1 + a2 ) + (2a1 + 2a2 )x] + t[(b1 + b2 ) + (2b1 + 2b2 )x = sL( a ) + tL( b ) So, L is linear. Linear Mappings Module 5: Linear Mappings and Diagonalization Check-In Question 2 Determine if the mapping M : R2 P1 (R) defined by M (a1 , a2 ) = 1 + (a1 3a2 )x is linear. Solution: Let a R2 . Observe that \u0012 \u0013 M 0 a = M (0, 0) = 1 But, \u0012 \u0013 0M a = 0[1 + (a1 3a2 )x = 0 Therefore, property L2 does not always hold. Hence, M is not linear. Linear Mappings Module 5: Linear Mappings and Diagonalization Page 8 of 11 c University of Waterloo and others MATH 636 Lecture 28 Slides (Module 5) (Last Updated: May 22, 2014) Matrix Mapping Example 2 Let A = 4 5 mapping. 3 0 and define f (x) = A x . Such a function f is called a matrix 1 (a) Find the domain and codomain of f ? Solution: For A x to be defined, x must have 2 rows since A has 2 columns. Also, since A has 3 rows, the product A x will have 3 rows. Hence, the domain of f is R2 3 and the codomain of f is R . We write f : R2 R3 . \u0014 \u0015 2 (b) Calculate f ( v ) where v = . 1 Solution: We have 2 f ( v ) = A v = 4 5 Linear Mappings 1 3 \u0014 \u0015 2 0 = 8 1 1 9 Module 5: Linear Mappings and Diagonalization Matrix Mapping Example 2 Let A = 4 5 mapping. 3 0 and define f (x) = A x . Such a function f is called a matrix 1 (c) Determine f ( x ) for any x R2 . Solution: We have 2 x = 4 f ( x ) = A 5 Linear Mappings 3 \u0014 \u0015 2x1 + 3x2 x 0 1 = 4x1 x2 1 5x1 + x2 Module 5: Linear Mappings and Diagonalization Page 9 of 11 c University of Waterloo and others MATH 636 Lecture 28 Slides (Module 5) (Last Updated: May 22, 2014) Check-In Question 1 Prove that every matrix mapping L( x ) = A x is a linear mapping. Solution: Let A be any m n matrix, let x , y Rn , and let s, t R. We have L(s x + t y ) = A(s x + t y) = sA x + tA y, by Theorem 3.1.3 = sL( x ) + tL( y) Therefore, L is linear. Hence, every matrix mapping is a linear mapping. Linear Mappings Module 5: Linear Mappings and Diagonalization Operations on Linear Mappings Definition Let V and W be vector spaces and let L : V W and M : V W be linear mappings. We define L + M : V W by (L + M )( v ) = L( v ) + M ( v ), for all v V For any t R, we define tL by (tL)( v ) = tL( v ), for all v V Notice that we have defined L + M and tL for every vector v V, so the domain of these mappings is V. Also, since L( v ) and M ( v ) are in W and t R, then L( v ) + M ( v ) and tL( v ) are in W since W is closed under addition and scalar multiplication since it is a vector space. Theorem 8.1.2 If V and W are vector spaces and L : V W and M : V W are linear mappings, then L + M and tL are linear mappings for any t R. Linear Mappings Module 5: Linear Mappings and Diagonalization Page 10 of 11 c University of Waterloo and others MATH 636 Lecture 28 Slides (Module 5) (Last Updated: May 22, 2014) Check-In Question 1 Let V and W be vector spaces and let L : V W and M : V W be linear mappings. Prove that L + M is linear. Solution: Let x , y V and s, t R. We have that (L + M )(s x + t y ) = L(s x + t y ) + M (s x + t y ), by definition of L + M = sL( x ) + tL( y ) + sM ( x ) + tM ( y ), since L and M are linear = s[L( x ) + M ( x )] + t[L( y ) + M ( y )], by properties V3 and V9 = s[(L + M )( x )] + t[(L + M )( y )], by definition of L + M Therefore, L + M is linear. Linear Mappings Module 5: Linear Mappings and Diagonalization Composition of Linear Mappings Definition Let L : V W and M : W U be linear mappings, we define M L by (M L)( v ) = M (L( v )) for any v V Theorem 8.1.3 If L : V W and M : W U are linear mappings, then M L is a linear mapping from V to U. Linear Mappings Module 5: Linear Mappings and Diagonalization Page 11 of 11 c University of Waterloo and others MATH 636 Lecture 29 Slides (Module 5) (Last Updated: May 23, 2014) The Rank-Nullity Theorem Module 5: Linear Mappings and Diagonalization The Rank-Nullity Theorem Module 5: Linear Mappings and Diagonalization Rank-Nullity Theorem In the last lecture, we defined a linear mapping so that the range of a linear mapping is a subspace of its codomain. In this lecture, we will look more closely at the range of linear mapping and another special subspace related to linear mappings called the kernel. By looking at the dimensions of these special subspaces, we will be lead to the very important Rank-Nullity Theorem. The Rank-Nullity Theorem Module 5: Linear Mappings and Diagonalization Page 1 of 9 c University of Waterloo and others MATH 636 Lecture 29 Slides (Module 5) (Last Updated: May 23, 2014) Range of a Linear Mapping Definition Let L : V U be a linear mapping. We define the range of L by Range(L) = {L( v)U| v V} Theorem 8.2.1 Let V and U be vector spaces and let L : V U be a linear mapping. Then, L( 0 V ) = 0 U The Rank-Nullity Theorem Module 5: Linear Mappings and Diagonalization Kernel of a Linear Mapping Definition Let L : V U be a linear mapping. We define the kernel of L by ker(L) = { v V | L( v ) = 0 U} Notice that the kernel of L is a subset of the domain V. It is the special part of V such that every vector in this special part is mapped to 0 . V Ker(L) U L: V U v2 v3 v1 Theorem 8.2.2 If L : V W is a linear mapping, then ker(L) is a subspace of V. The Rank-Nullity Theorem Module 5: Linear Mappings and Diagonalization Page 2 of 9 c University of Waterloo and others 0U u1 MATH 636 Check-In Question 1 Prove that if L : V W is a linear mapping, then ker(L) is a subspace of V. Solution: We use the Subspace Test. ker(L) is a subset of V by definition and ker(L) is non-empty by Theorem 8.2.1. Let x , y ker(L). Then, we have that L( x ) = 0 and L( y)= 0. Since L is linear, we get that x + y ) = L( x ) + L( y)= 0 + 0 = 0 L( Therefore, x + y ker(L), so ker(L) is closed under addition. We also get that for any t R L(t x ) = tL( x ) = t( 0 ) = 0 Thus, t x ker(L), and so ker(L) is also closed under scalar multiplication. Consequently, ker(L) is a subspace of V by the Subspace Test. The Rank-Nullity Theorem Module 5: Linear Mappings and Diagonalization MATH 636 Lecture 30 Slides (Module 5) (Last Updated: May 22, 2014) Matrix of a Mapping and Special Subspaces Module 5: Linear Mappings and Diagonalization Matrix of a Mapping and Special Subspaces Module 5: Linear Mappings and Diagonalization Matrix of a Mapping and Special Subspaces In this lecture, we will use matrices as a tool to help us represent linear mappings from Rn to Rm . Matrix of a Mapping and Special Subspaces Module 5: Linear Mappings and Diagonalization Page 1 of 13 c University of Waterloo and others MATH 636 Lecture 30 Slides (Module 5) (Last Updated: May 22, 2014) Example Example Let L : R2 R3 be the linear mapping defined by \u0012\u0014 \u0015\u0013 2x1 + x2 2 1 x1 L = 3x1 + 5x2 = x1 3 + x2 5 , x2 x2 0 1 x1 , x2 R This tells us two things: 1 2 1. Range(L) = Span 3 , 5 0 1 \u0012\u0014 \u0015\u0013 2 1 \u0014 \u0015 x1 x 2. L = 3 5 1 x2 x2 0 1 That is, this linear mapping is a matrix-mapping. Matrix of a Mapping and Special Subspaces Module 5: Linear Mappings and Diagonalization Matrix of a Linear Mapping Theorem 8.3.1 If L : Rn Rm is a linear mapping, then L can be represented as a matrix mapping with the corresponding m by n matrix [L] given by \u0002 \u0003 e 1 ) L( e 2 ) L( e n) [L] = L( where e 1, . . . , e n are the standard basis vectors for Rn . Matrix of a Mapping and Special Subspaces Module 5: Linear Mappings and Diagonalization Page 2 of 13 c University of Waterloo and others MATH 636 Lecture 30 Slides (Module 5) (Last Updated: May 22, 2014) Check-In Question 1 Prove that if L : Rn Rm is a linear mapping, then L can be represented as a matrix mapping with the corresponding m by n matrix [L] given by \u0002 \u0003 e 1 ) L( e 2 ) L( e n) [L] = L( where e 1, . . . , e n are the standard basis vectors for Rn . Proof: Let x Rn . Then, x = x1 e 1 + x2 e 2 + + xn e n. Hence, L( x ) = L(x1 e 1 + x2 e 2 + + xn e n) = x1 L( e 1 ) + x2 L( e 2 ) + + xn L( e n) x1 \u0002 \u0003 . = L( e 1 ) L( e 2 ) L( e n ) .. xn = [L] x as required. Matrix of a Mapping and Special Subspaces Module 5: Linear Mappings and Diagonalization Standard Matrix Definition Let L : Rn Rm be a linear mapping. The matrix \u0002 e 1 ) L( e 2) [L] = L( \u0003 L( e n) is called the standard matrix of L and has the property that L( x ) = [L] x This is called the standard matrix of L since it is based on the standard basis vectors. We will look at the matrix of more general linear mappings with respect to any bases in a few lectures. Matrix of a Mapping and Special Subspaces Module 5: Linear Mappings and Diagonalization Page 3 of 13 c University of Waterloo and others MATH 636 Lecture 30 Slides (Module 5) (Last Updated: May 22, 2014) Example Example \u0014 \u0015 3 2 2 Let a = R2 . Define a linear mapping proj a : R R by 4 x a a proj a(x)= k a k2 \u0012\u0014 \u0015\u0013 1 Find the standard matrix [proj and use it to find proj . a a 2 Solution: By definition, the first column of [proj a ] is proj a ( e 1 ). \u0014 \u0015 \u0014 \u0015 e1 a 3 3 9/25 proj ( e ) = a = = 1 a 12/25 k a k2 25 4 The second column of [proj a ] is proj a ( e 2 ). \u0014 \u0015 \u0014 \u0015 e2 a 4 3 12/25 proj ( e ) = a = = 2 a 16/25 k a k2 25 4 Hence, the standard matrix of the linear mapping is \u0014 \u0015 9/25 12/25 [proj = a 12/25 16/25 Matrix of a Mapping and Special Subspaces Module 5: Linear Mappings and Diagonalization Example Example \u0014 \u0015 3 2 2 Let a = R2 . Define a linear mapping proj a : R R by 4 x a a proj a(x)= k a k2 Find the standard matrix [proj a ] and use it to find proj a \u0012\u0014 \u0015\u0013 1 . 2 Solution: By definition of the standard matrix, we get that \u0014 \u0015 \u0014 \u0015\u0014 \u0015 \u0014 \u0015 \u0012\u0014 \u0015\u0013 1 9/25 12/25 1 33/25 1 = [proj = = proj a a 2 12/25 16/25 2 44/25 2 Matrix of a Mapping and Special Subspaces Module 5: Linear Mappings and Diagonalization Page 4 of 13 c University of Waterloo and others MATH 636 Lecture 30 Slides (Module 5) (Last Updated: May 22, 2014) Example Example \u0014 \u0015 \u0014 \u0015 3 1 Let a = and x = 4 2 \u0012\u0014 \u0015\u0013 \u0014 \u0015 \u0014 1 1 9/25 Then proj = [proj = a ( x ) = proj a a 2 2 12/25 \u0015\u0014 \u0015 \u0014 \u0015 12/25 1 33/25 = 16/25 2 44/25 y a x proj a x x Matrix of a Mapping and Special Subspaces Module 5: Linear Mappings and Diagonalization Example Example Let G : R3 R2 be defined by G(x1 , x2 , x3 ) = (x1 , x2 + 2x3 ). Prove that G is linear and find the standard matrix of G. Solution: Let y , z R3 and s, t R. We get \u0001 G s y + t z ) = G s(y1 , y2 , y3 ) + t(z1 , z2 , z3 ) = G(sy1 + tz1 , sy2 + tz2 , sy3 + tz3 ) = sy1 + tz1 , sy2 + tz2 + 2(sy3 + tz3 ) = s(y1 , y2 + 2y3 ) + t(z1 , z2 + 2z3 ) = sG( y ) + tG( z) \u0001 Hence, G is linear. We have G(1, 0, 0) = (1, 0), G(0, 1, 0) = (0, 1), G(0, 0, 1) = (0, 2) Thus, we have \u0002 e 1) [G] = G( G( e 2) Matrix of a Mapping and Special Subspaces \u0014 \u0003 1 G( e 3) = 0 0 1 \u0015 0 2 Module 5: Linear Mappings and Diagonalization Page 5 of 13 c University of Waterloo and others MATH 636 Lecture 30 Slides (Module 5) (Last Updated: May 22, 2014) Standard Matrix In the first example, we saw that for the linear mapping L : R2 R3 defined by \u0012\u0014 \u0015\u0013 2x1 + x2 x1 L = 3x1 + 5x2 x2 x2 we had 2 [L] = 3 0 and 1 5 1 1 2 Range(L) = Span 3 , 5 0 1 Matrix of a Mapping and Special Subspaces Module 5: Linear Mappings and Diagonalization Standard Matrix Theorem 8.3.2 \u0002 v1 If [L] = \u0003 v n is the standard matrix of a linear mapping L : Rn Rm , then Range(L) = Span{ v 1, . . . , v n} Proof: Observe that for any x Rn we have x1 \u0002 \u0003 . v n . = x1 L( x ) = [L] x = v 1 v 1 + + xn v n, . xn x1 , . . . , xn R Thus, every vector in Range(L) is in Span{ v 1, . . . , v n } and every vector in Span{ v 1 , . . . , v n } is in Range(L) as required. Matrix of a Mapping and Special Subspaces Module 5: Linear Mappings and Diagonalization Page 6 of 13 c University of Waterloo and others MATH 636 Lecture 30 Slides (Module 5) (Last Updated: May 22, 2014) Columnspace Definition \u0002 a1 Let A = \u0003 a n be an m n matrix. The columnspace of A is Col(A) = Span{ a 1, . . . , a n } = {A x | x Rn } Observe that we have called the columnspace of a matrix a 'space' since Theorem 8.3.2 proves that the columnspace of any matrix A is the range of the matrix mapping L( x ) = A x and hence the columnspace is a subspace of Rm by Theorem 8.1.1. Moreover, we have that if L : Rn Rm is linear, then Range(L) = Col([L]). Theorem 8.3.3 Let L : Rn Rm be a linear mapping with standard matrix A = [L]. Then, x ker(L) if and only if A x = 0. Matrix of a Mapping and Special Subspaces Module 5: Linear Mappings and Diagonalization Nullspace Definition Let A be an m n matrix. The set of all x Rn such that A x = 0 is called the nullspace of A and is denoted Null(A) = { x Rn | A x = 0} Again, we see that ker(L) = Null([L]) and hence Null(A) is a subspace of Rn . As we will see, the columnspace and nullspace of a matrix are very important, so let's look at these a little more closely. Matrix of a Mapping and Special Subspaces Module 5: Linear Mappings and Diagonalization Page 7 of 13 c University of Waterloo and others MATH 636 Lecture 30 Slides (Module 5) (Last Updated: May 22, 2014) Example Example 4 1 Determine whether x = 5 is in the nullspace of A = 2 2 0 Solution: We have 1 A x = 2 0 1 1 1 Hence, A x 6= 0 , so x 6 Null(A). 1 1 1 1 3 . 5 1 4 1 3 5 = 3 5 2 5 Matrix of a Mapping and Special Subspaces Module 5: Linear Mappings and Diagonalization Example Example 4 1 Determine whether 0 is in the columnspace of A = 2 8 0 1 1 1 1 3 . 5 4 Solution: We need to determine if 0 is a linear combination of the columns of A. 8 That is, we need to determine if we can find c1 , c2 , c3 such that 4 1 1 1 c1 + c2 c3 0 = c1 2 + c2 1 + c3 3 = 2c1 + c2 + 3c3 8 0 1 5 c2 5c3 Comparing entries gives the system of linear equations c1 + c2 c3 = 4 2c1 + c2 + 3c3 = 0 c2 5c3 = 8 Matrix of a Mapping and Special Subspaces Module 5: Linear Mappings and Diagonalization Page 8 of 13 c University of Waterloo and others MATH 636 Lecture 30 Slides (Module 5) (Last Updated: May 22, 2014) Example Example 4 1 Determine whether 0 is in the columnspace of A = 2 8 0 Solution: Row reducing the augmented matrix, we get 1 1 1 4 1 0 4 2 1 3 0 0 1 5 0 1 5 8 0 0 0 1 1 1 1 3 . 5 4 8 0 4 Hence, since this has a solution, we have that 0 is in the columnspace of A. 8 We could find values of c1 , c2 , c3 by finding the general solution. Matrix of a Mapping and Special Subspaces Module 5: Linear Mappings and Diagonalization Example Example 1 0. Find a basis for the columnspace and nullspace of A. 2 2 1 1 Solution: By definition, Col(A) is spanned by 1 , 3 , 0 . 0 1 2 Consider 0 1 2 1 c1 + 2c2 + c3 0 = c1 1 + c2 3 + c3 0 = c1 + 3c2 0 0 1 2 c2 + 2c3 Comparing entries gives a homogeneous system of linear equations. Row reducing the coefficient matrix gives 1 2 1 1 0 0 1 3 0 0 1 0 0 1 2 0 0 1 2 1 1 Hence, the system has the unique solution c1 = c2 = c3 = 0. Therefore, 1 , 3 , 0 is 0 1 2 also linearly independent and hence is a basis for Col(A). 1 Let A = 1 0 2 3 1 Every vector x Null(A) sastifes 0 1 2 1 x1 + 2x2 + x3 0 = A x = x1 1 + x2 3 + x3 0 = x1 + 3x2 0 0 1 2 x2 + 2x3 This is the same system as before. Hence, the only solution is x1 = x2 = x3 = 0. Therefore, the only vector x Null(A) is x = 0 . Thus, we say a basis for Null(A) is the empty set. Matrix of a Mapping and Special Subspaces Module 5: Linear Mappings and Diagonalization Page 9 of 13 c University of Waterloo and others MATH 636 Lecture 30 Slides (Module 5) (Last Updated: May 22, 2014) Check-In Question 1 Let B = \u0014 1 2 Solution: By definition, 2 1 \u0015 3 . Find a basis for Col(B) and Null(B). 1 \u001a\u0014 \u0015 \u0014 \u0015 \u0014 \u0015\u001b 1 2 3 , , spans Col(B). However, observe that 2 1 1 \u0014 \u0015 \u0014 \u0015 \u0014 \u0015 1 2 3 + = 2 1 1 Hence, Col(B) = Span \u001a\u0014 \u0015 \u0014 \u0015\u001b 1 2 , 2 1 \u001a\u0014 \u0015 \u0014 \u0015\u001b 1 2 and , is clearly linearly independent, so it is a basis for Col(B). 2 1 Matrix of a Mapping and Special Subspaces Module 5: Linear Mappings and Diagonalization Check-In Question 1 Let B = \u0014 1 2 2 1 \u0015 3 . Find a basis for Col(B) and Null(B). 1 Solution Continued: Every vector x Null(B) sastifes \u0014 \u0015 \u0014 \u0015 \u0014 \u0015 \u0014 \u0015 \u0014 \u0015 0 1 2 3 x + 2x2 + 3x3 = B x = x1 + x2 + x3 = 1 0 2 1 1 2x1 x2 + x3 Row reducing the coefficient matrix of the corresponding system of linear equations gives \u0014 \u0015 \u0014 \u0015 1 2 3 1 0 1 2 1 1 0 1 1 Thus, x = x and x = x , so x has the form 1 3 2 3 x1 x3 1 x = x2 = x3 = x3 1 , x3 R x3 x3 1 1 1 Therefore, spans Null(B) and is clearly linearly independent, so it is a basis 1 for Col(B). Matrix of a Mapping and Special Subspaces Module 5: Linear Mappings and Diagonalization Page 10 of 13 c University of Waterloo and others MATH 636 Lecture 30 Slides (Module 5) (Last Updated: May 22, 2014) Rowspace Since we have considered the span of the columns of A, it makes sense to also consider the span of the rows of A. We again can use matrix-vector notation. But, since A x gives us a linear combination of columns, we need to turn the rows of A into columns by taking the transpose. That is, AT x gives a linear combination of the columns of AT , which is a linear combination of the columns of A. Definition Let A be an m n matrix with rows aT i for 1 i m. The span of the rows of A is called the rowspace of A and is denoted Row(A) = {AT x | x Rm } Observe that Row(A) = Col(AT ). Thus, to complete the set, we also look at the nullspace of AT . Matrix of a Mapping and Special Subspaces Module 5: Linear Mappings and Diagonalization Nullspace Definition Let A be an m n matrix. The nullspace of AT is called the left nullspace of A. Null(AT ) = { x Rm | AT x = 0} To find a basis for the rowspace and left nullspace of a matrix A, we could just find a basis for the columnspace and nullspace of AT . For an m n matrix A, the columnspace, nullspace, rowspace, and left nullspace are collectively called the four fundamental subspaces of A. Theorem 8.3.4 If A is an m n matrix, then Col(A) and Null(AT ) are subspaces of Rm , and Row(A) and Null(A) are subspace of Rn . Matrix of a Mapping and Special Subspaces Module 5: Linear Mappings and Diagonalization Page 11 of 13 c University of Waterloo and others MATH 636 Lecture 30 Slides (Module 5) (Last Updated: May 22, 2014) Example Example 1 2 Find a basis for the four fundamental subspaces of A = 0 1 2 1 1 Solution: 2 2 8 By definition, Col(A) = Span , , . 1 1 0 5 2 1 1 2 1 2 2 8 . 1 5 To determine if this spanning set is linearly independent, we consider 1 1 2 2 2 8 0 = x1 + x2 + x3 0 1 1 1 2 5 This gives a homogeneous system with corresponding coefficient matrix A. Row reducing, we get 1 1 2 1 0 3 2 8 0 1 1 2 0 1 1 0 0 0 0 0 0 1 2 5 1 1 2 2 Thus, the third vector is a linear combination of the first two vectors. Thus, B = , 0 1 1 2 spans Col(A) and is clearly linearly independent, so it is a basis for Col(A). Matrix of a Mapping and Special Subspaces Module 5: Linear Mappings and Diagonalization Example Continued Example 1 2 Find a basis for the four fundamental subspaces of A = 0 1 Solution Continued: 1 2 0 1 1 2 1 2 For Null(A), we need to solve A x = 0. 2 1 8 0 1 0 5 0 0 1 0 0 1 2 1 2 2 8 . 1 5 3 1 0 0 Using our row reduction above, we get that every x Null(A) has the form 3x3 3 x = x3 = x3 1 , x3 R x3 1 3 Thus, a basis for Null(A) is 1 . 1 Matrix of a Mapping and Special Subspaces Module 5: Linear Mappings and Diagonalization Page 12 of 13 c University of Waterloo and others MATH 636 Lecture 30 Slides (Module 5) (Last Updated: May 22, 2014) Example Continued Example 1 2 Find a basis for the four fundamental subspaces of A = 0 1 1 2 1 2 Solution Continued: 2 8 . 1 5 For Null(AT ), we need to solve AT x = 0. Row reducing the corresponding coefficient matrix AT , we get 1 2 0 1 1 0 1/2 1 2 1 2 0 1 1/4 2 8 1 5 0 0 0 1/2 3/4 0 So, we get that every x Null(AT ) has the form 1 1 1/2 1/2 2 x3 + 2 x4 14 x3 34 x4 1/4 3/4 x = = x3 1 + x4 0 , x3 0 1 x4 x3 , x4 R 1/2 1/2 3/4 1/4 Thus, a basis for Null(AT ) is , 0 since it is clearly linearly independent and 1 0 1 spans Null(AT ). Matrix of a Mapping and Special Subspaces Module 5: Linear Mappings and Diagonalization Example Continued Example 1 2 Find a basis for the four fundamental subspaces of A = 0 1 Solution Continued: 1 2 1 2 2 0 1 1 By definition, Row(A) = Span 1 , 2 , 1 , 2 . 2 8 1 5 2 8 . 1 5 We need to determine if the spanning set is linearly independent or not. Consider 1 2 0 1 0 = c1 1 + c2 2 + c3 1 + c4 2 8 1 5 2 This gives us a homogeneous system with coefficient matrix AT. Hence, using our row reducing 2 1 above, we find that the rowspace is spanned by C = 1 , 2 . 2 8 Since this set is now linearly independent and spans Row(A), it is a basis for Row(A). Matrix of a Mapping and Special Subspaces Module 5: Linear Mappings and Diagonalization Page 13 of 13 c University of Waterloo and others MATH 636 Lecture 33 Slides (Module 5) (Last Updated: May 23, 2014) Matrix of a Linear Operator Module 5: Linear Mappings and Diagonalization Matrix of a Linear Operator Module 5: Linear Mappings and Diagonalization Matrix of a Linear Operator \u0014 \u0015 3 We saw the standard matrix of the projection of a vector x R2 onto a = is 4 [proj a]= \u0014 9/25 12/25 \u0015 12/25 16/25 y a x proj a x x Matrix of a Linear Operator Module 5: Linear Mappings and Diagonalization Page 1 of 8 c University of Waterloo and others MATH 636 Lecture 33 Slides (Module 5) (Last Updated: May 23, 2014) Matrix of a Linear Operator It would make much more sense to find the matrix representation of the projection with respect to a basis that is more suited to\u0014the \u0015 geometry of the projection. In 3 . Of course, to make a basis for R2 , particular, a basis which contains the vector 4 we still need one more basis vector. \u0014 \u0015 4 We pick since it is nice to have the basis vectors \u0014 \u0015 orthogonal \u0014 \u0015 to each other and 3 3 3 because the projection of any vector orthogonal to onto will be 0 . 4 4 We \u0014 \u0015 need to figure out how to define \u001a\u0014the \u0015 matrix \u0014 \u0015\u001brepresentation of the projection onto 3 3 4 with respect to the basis B = , . 4 4 3 Matrix of a Linear Operator Module 5: Linear Mappings and Diagonalization Matrix of a Linear Operator Let L : Rn Rn be a linear operator. To define the matrix [L]B of L with respect to any basis B = { v 1, . . . , v n } of Rn , we want to mimic what we had for the standard matrix. In particular, we recall that the standard matrix [L] satisfies L( x ) = [L] x . Since we are now working with respect to the basis B, we want to put everything into B-coordinates. That is, we want [L( x )]B = [L]B [ x ]B For any x Rn , we can write x = b1 v 1 + + bn v n. b1 . Hence, we have [ x ]B = .. . bn Then, [L( x )] = [L(b v + + b v )] B 1 n 1 n B = [b1 L( v 1 ) + + bn L( v n )]B = b [L( v )] + + b [L( v )] 1 1 \u0002 v 1 )]B = [L( Matrix of a Linear Operator n B n B b1 \u0003 [L( v n )]B .. . bn Module 5: Linear Mappings and Diagonalization Page 2 of 8 c University of Waterloo and others MATH 636 Lecture 33 Slides (Module 5) (Last Updated: May 23, 2014) Matrix of a Linear Operator Definition If B = { v 1, . . . , v n } is a basis for Rn and L : Rn Rn is a linear operator, then the matrix of L with respect to the basis B is defined to be \u0002 \u0003 v 1 )]B [L( v n )]B [L]B = [L( It satisfies [L( x )]B = [L]B [ x ]B NOTE: For short, we may call the matrix of L with respect to the basis B the B-matrix of L. Matrix of a Linear Operator Module 5: Linear Mappings and Diagonalization Matrix of a Linear Operator Example Example Determine the matrix of the linear operator L : R3 R3 with respect to the basis B = { v 1, v 2, v 3 } where L( v 1 ) = 2 v 1 3 v 2, L( v 2 ) = 3 v 1 + 4 v2 v 3 , and L( v 3 ) = v 1 + 2 v 2 + 6 v 3. 5 Use the matrix to find L( x ) where [ x ]B = 3. 1 Solution: Thus, we have By definition, 2 [L( v 1 )]B = 3 , 0 \u0002 v )1 ]B [L]B = [L( 3 [L( v 2 )]B = 4 , 1 [L( v 2 )]B Matrix of a Linear Operator [L( v 3 )]B \u0003 1 [L( v 3 )]B = 2 6 2 = 3 0 3 4 1 1 2 6 Module 5: Linear Mappings and Diagonalization Page 3 of 8 c University of Waterloo and others MATH 636 Lecture 33 Slides (Module 5) (Last Updated: May 23, 2014) Matrix of a Linear Operator Example Example Determine the matrix of the linear operator L : R3 R3 with respect to the basis B = { v 1, v 2, v 3 } where L( v 1 ) = 2 v 1 3 v 2, L( v 2 ) = 3 v 1 + 4 v2 v 3 , and L( v 3 ) = v 1 + 2 v 2 + 6 v 3. 5 Use the matrix to find L( x ) where [ x ]B = 3. 1 Solution: By definition, [L]B \u0002 v )1 ]B = [L( [L( v 2 )]B Also, by definition, we get that 2 [L( x )]B = [L]B [ x ]B = 3 0 Thus, L( x ) = 0 v 1 25 v 2 + 9 v3 Matrix of a Linear Operator 2 \u0003 [L( v 3 )]B = 3 0 3 4 1 3 4 1 1 2 6 1 5 0 2 3 = 25 6 1 9 Module 5: Linear Mappings and Diagonalization Matrix of a Linear Operator Example Example \u0014 \u0015 \u001a\u0014 \u0015 \u0014 \u0015\u001b 3 3 4 Let a = and B = , . Determine the B-matrix of proj a. 4 4 3 Solution: We have \u0012\u0014 \u0015\u0013 \u0014 \u0015 \u0014 \u0015 \u0014 \u0015 3 3 3 4 = =1 +0 4 4 4 3 \u0012\u0014 \u0015\u0013 \u0014 \u0015 \u0014 \u0015 \u0014 \u0015 4 0 3 4 proj = =0 +0 a 3 0 4 3 proj a Thus, [proj a ]B = \u0012\u0014 \u0015\u0013\u0015 \u0014\u0014 3 proj a 4 B Matrix of a Linear Operator \u0014 proj a \u0012\u0014 \u0015\u0013\u0015 \u0015 \u0014 4 1 = 3 0 B \u0015 0 0 Module 5: Linear Mappings and Diagonalization Page 4 of 8 c University of Waterloo and others MATH 636 Lecture 33 Slides (Module 5) (Last Updated: May 23, 2014) Matrix of a Linear Operator Example Example 1 2 2 Let A = 2 1 2 and define L( x ) = A x. 2 2 1 1 1 1 Let B = 1 , 1 , 0 = { v 1, v 2, v 3 }. Find [L]B . 1 0 1 Solution: We have We have 5 1 1 1 L( v 1 ) = A v 1 = 5 = 5 1 + 0 1 + 0 0 5 1 0 1 1 1 1 1 L( v 2 ) = A v 2 = 1 = 0 1 1 1 + 0 0 0 1 0 1 1 1 1 1 L( v 3 ) = A v 3 = 0 = 0 1 + 0 1 1 0 1 1 1 0 \u0002 v 1 )]B [L]B = [L( [L( v 2 )]B [L( v 3 )]B \u0003 5 = 0 0 0 1 0 0 0 1 b1 By definition of the matrix of L with respect to B we get that if [ x ]B = b2 , then b3 5b1 Matrix of a Linear Operator Example Continued [(L( x )]B = [L]B [ x ]B = b2 b3 Matrix of a Linear Operator Module 5: Linear Mappings and Diagonalization This describes the action of the linear mapping. The linear mapping L takes a vector Example 1 1 2 it2by a factor of 1 and reflects it over the 5 1in the 11 direction, x and stretches Let A = 2 1 2 and B = 1 , 11 , 0 = { v 1, v 2, v 3 }. 2 2 1 1 0 1 1 1 1 vectors Define L( x )=and A x. 0 . 0 1 5 0 0 \u0002 \u0003 0 . Then [L]B = [L( v 1 )]B [L( v 2 )]B [L( v 3 )]B = 0 1 0 0 1 By definition of the matrix of L with respect to B, b1 5b1 we get that if [ x ]B = b2 , then [(L( x )]B = [L]B [ x ]B = b2 . b3 b3 This describes the action of the linear mapping. 1 The linear mapping L takes a vector x and stretches it by a factor of 5 in the 1 1 1 1 direction and reflects it over the vectors 1 and 0 . 0 1 Matrix of a Linear Operator Module 5: Linear Mappings and Diagonalization Page 5 of 8 c University of Waterloo and others MATH 636 Lecture 33 Slides (Module 5) (Last Updated: May 23, 2014) Diagonal Matrix Definition An n n matrix D is said to be a diagonal matrix if dij = 0 for all i 6= j. We denote a diagonal matrix by D = diag(d11 , d22 , . . . , dnn ) NOTE: We could also define a diagonal matrix to be a matrix that is both upper and lower triangular. Example \u0015 0 . 5 1 0 The diagonal matrix A = diag(1, 2, 3, 0) is A = 0 0 The diagonal matrix D = diag(4, 5) is D = \u0014 4 0 0 2 0 0 0 0 3 0 0 0 . 0 0 Thus, for any linear operator L : Rn Rn , we want to determine if there exists a basis B for Rn such that the matrix of L with respect to B is diagonal. Moreover, if such a basis does exist, we would like to have a method for finding it. Matrix of a Linear Operator Module 5: Linear Mappings and Diagonalization Similar Matrices Let L : Rn Rn be a linear operator with standard matrix A = [L] and let B = { v 1, . . . , v n } be a basis for Rn . By definition, we have \u0002 v 1 ]B [L]B = [A [A v n ]B \u0003 (1) We know \u0002that the change \u0003 of coordinates matrix from B-coordinates to S-coordinates v 1 vn . is P = Hence, we have [ x ] = P 1 x for any x Rn . B Since A v i Rn for 1 i n, we can apply this to (1) to get \u0002 \u0003 v 1 ]B [A v n ]B [L]B = [A \u0002 \u0003 v 1 ) P 1 (A v n) = P 1 (A \u0002 \u0003 v 1 (P 1 A) vn = (P 1 A) \u0002 \u0003 v 1 vn = P 1 A = P 1 AP = P 1 [L]P Matrix of a Linear Operator Module 5: Linear Mappings and Diagonalization Page 6 of 8 c University of Waterloo and others MATH 636 Lecture 33 Slides (Module 5) (Last Updated: May 23, 2014) Similar Matrices Example \u0015 3 be the standard matrix of a linear mapping L : R2 R2 , 1 \u001a\u0014 \u0015 \u0014 \u0015\u001b 1 2 and let B = , . 3 7 Let A = \u0014 1 2 Find [L]B . Solution: \u0014 1 Let P = 3 \u0015 2 , then we know that P 1 = 7 1 1 [L]B = P 1 AP = \u0014 Hence, Matrix of a Linear Operator \u0014 7 3 \u0015 2 . 1 72 31 \u0015 167 72 Module 5: Linear Mappings and Diagonalization Similar Matrices Theorem 6.1.1 If there exists an invertible matrix P such that P 1 AP = B, then (1) rank A = rank B (2) det A = det B (3) tr A = tr B where tr A is defined by tr A = n P aii and is called the trace of a i=1 matrix. Definition Let A and B be n n matrices. If there exists an invertible matrix P such that P 1 AP = B, then A and B are said to be similar. NOTE: If P 1 AP = B, then we can take Q = P 1 to get that Q1 BQ = A. So, the similarity property is symmetric. Matrix of a Linear Operator Module 5: Linear Mappings and Diagonalization Page 7 of 8 c University of Waterloo and others MATH 636 Lecture 33 Slides (Module 5) (Last Updated: May 23, 2014) Conclusion Our earlier work shows us that the B-matrix of a linear operator L is similar to the standard matrix of L. So, we can now reword our main question. Given a linear operator L on Rn , can we find a basis B for Rn such that [L]B is similar to a diagonal matrix, and, if we can, how do we find the basis B? Matrix of a Linear Operator Module 5: Linear Mappings and Diagonalization Page 8 of 8 c University of Waterloo and others MATH 636 Lecture 34 Slides (Module 5) (Last Updated: May 22, 2014) Eigenvalues and Eigenvectors Module 5: Linear Mappings and Diagonalization Eigenvalues and Eigenvectors Module 5: Linear Mappings and Diagonalization Eigenvalues and Eigenvectors In the last lecture, we looked at how to find the matrix of a linear operator L with respect to a basis B. At the end of the lecture we stated our main goal: to determine if a linear operator L : Rn Rn has a basis B such that [L]B is diagonal, and, if it does, how to find B. In this lecture, we will figure out how to do this. Eigenvalues and Eigenvectors Module 5: Linear Mappings and Diagonalization Page 1 of 8 c University of Waterloo and others MATH 636 Lecture 34 Slides (Module 5) (Last Updated: May 22, 2014) Eigenvalues and Eigenvectors Let L : Rn Rn be a linear operator. v 1, . . . , v n } such that the matrix of L with respect to B is Assume that there is a basis B = { diagonal. In particular, assume that [L]B = diag(1 , . . . , n ). In the last lecture, we found that [L]B was similar to the A \u0002standard matrix \u0003 = [L] of L. In v1 v n we got particular, using the change of coordinates matrix P = P 1 AP = diag(1 , . . . , n ) PP 1 AP = P diag(1 , . . . , n ) AP = P diag(1 , . . . , n ) \u0002 \u0002 \u0003 \u0003 v1 vn = v1 v n diag(1 , . . . , n ) A \u0002 \u0003 \u0002 \u0003 v1 n vn A v 1 A v n = 1 By comparing columns, we see that we must have A v i = i v i for 1 i n. Moreover, since P is invertible, the columns of P must be linearly independent. In particular, v i 6= 0 for 1 i n. On the other hand, if A v i = i v i for 1 i n, then we have [L( v i )]B = [i v i ]B = i e i, 1in Hence, \u0002 e1 [L]B = 1 \u0003 n e n = diag(1 , . . . , n ) So, [L]B is diagonal. Eigenvalues and Eigenvectors Module 5: Linear Mappings and Diagonalization Eigenvalues and Eigenvectors Definition Let A be an n n matrix. If there is a non-zero vector v such that A v = v for some scalar , then is called an eigenvalue of A and v is called an eigenvector of A corresponding to . We call (, v ) an eigenpair. Definition Let L be a linear operator on Rn . If there is a non-zero vector v such that L( v ) = v for some scalar , then is called an eigenvalue of L and v is called an eigenvector of L corresponding to . Eigenvalues and Eigenvectors Module 5: Linear Mappings and Diagonalization Page 2 of 8 c University of Waterloo and others MATH 636 Lecture 34 Slides (Module 5) (Last Updated: May 22, 2014) Eigenvalues and Eigenvectors Example Example Let a R2 , a 6= 0 . What are the eigenvectors and corresponding eigenvalues of 2 2 proj a : R R defined by x a a proj a(x)= k a k2 Solution: Recall that, geometrically, the projection finds that amount of the vector x in the direction of a . That is, the image of x under the projection is a scalar multiple of a. y x proj a x a x Eigenvalues and Eigenvectors Module 5: Linear Mappings and Diagonalization Eigenvalues and Eigenvectors Example Continued Example Let a R2 , a 6= 0 . What are the eigenvectors and corresponding eigenvalues of 2 2 proj a : R R defined by x a proj a a(x)= k a k2 Solution Continued: y x proj a x =x a x We want to find all vectors x in R2 such that proj a(x)=x. (t a) a proj a = t a = (1)(t a) a (t a ) = k a k2 Thus, t a for t 6= 0 is an eigenvector with eigenvalue 1. Eigenvalues and Eigenvectors Module 5: Linear Mappings and Diagonalization Page 3 of 8 c University of Waterloo and others MATH 636 Lecture 34 Slides (Module 5) (Last Updated: May 22, 2014) Eigenvalues and Eigenvectors Example Continued Example Let a R2 , a 6= 0 . What are the eigenvectors and corresponding eigenvalues of 2 2 proj a : R R defined by x a proj a a(x)= k a k2 Solution Continued: y a v x We observe that if v is any vector orthogonal to a , then we have v a proj a = 0 = 0 a a(v)= k a k2 Hence, all non-zero vectors orthogonal to a are eigenvectors with corresponding eigenvalue 0. Eigenvalues and Eigenvectors Module 5: Linear Mappings and Diagonalization Eigenvalues and Eigenvectors Example Example Thinking geometrically, what are the eigenvectors and corresponding eigenvalues of the linear mapping R : R2 R2 that rotates a vector x by an angle where 0 < < 2 radians? y R ( x) R ( x) x x R ( x) Solution: We first realize that if = radians, then R rotates x around to x . Hence, every non-zero vector x is an eigenvector of R with corresponding eigenvalue 1. On the other hand, if we are rotating x by an angle with 0 < < 2 and 6= , then the result will not be a scalar multiple of x . So, in this case, R will not have any real eigenvalues. Eigenvalues and Eigenvectors Module 5: Linear Mappings and Diagonalization Page 4 of 8 c University of Waterloo and others MATH 636 Lecture 34 Slides (Module 5) (Last Updated: May 22, 2014) Eigenvectors Example Example 3 Consider A = 3 5 of A. 1 (a) v 1 = 2 1 Solution: We have 6 3 6 7 7. Determine which of the following vectors are eigenvectors 5 3 A v 1 = 3 5 6 3 6 2 7 1 7 2 = 4 = 2 v1 2 5 1 Thus, v 1 is an eigenvector of A with corresponding eigenvalue = 2. Eigenvalues and Eigenvectors Module 5: Linear Mappings and Diagonalization Eigenvectors Example Example 3 Consider A = 3 5 of A. 1 (b) v 2 = 1 1 Solution: We have 6 3 6 7 7. Determine which of the following vectors are eigenvectors 5 3 A v 2 = 3 5 6 3 6 7 1 16 7 1 = 13 5 1 16 v 2 is not an eigenvector of A. which is not a scalar multiple of v 2 . Thus, Eigenvalues and Eigenvectors Module 5: Linear Mappings and Diagonalization Page 5 of 8 c University of Waterloo and others MATH 636 Lecture 34 Slides (Module 5) (Last Updated: May 22, 2014) Eigenvectors Example Example 3 Consider A = 3 5 of A. 0 (c) v 3 = 0 0 6 3 6 7 7. Determine which of the following vectors are eigenvectors 5 Solution: We have A v 3 = 0 = 1 v 3 , so . . . v 3 is NOT an eigenvector of A. Because, by definition, the zero vector is not allowed to be an eigenvector! Eigenvalues and Eigenvectors Module 5: Linear Mappings and Diagonalization Check-In Question 1 3 5 5 Consider A = 7 9 5. 7 7 3 Determine which of the following vectors are eigenvectors of A. 1 1 2 a x = 1 b y = 0 c z = 3 1 1 1 5 1 3 1 5 1 = 3 = 3 1 3 1 3 1 3 5 5 1 2 1 7 9 5 0 = 2 6= t 0 7 7 3 1 4 1 3 5 5 2 4 2 7 9 5 3 = 8 6= t 3 7 7 3 1 4 1 1 Therefore, x = 1 is the only eigenvector. 1 Solution: We have 3 7 7 5 9 7 Eigenvalues and Eigenvectors Module 5: Linear Mappings and Diagonalization Page 6 of 8 c University of Waterloo and others MATH 636 Lecture 34 Slides (Module 5) (Last Updated: May 22, 2014) Eigenvalues Example Example 3 Consider A = 3 5 7 7. Is = 1 an eigenvalue of A? 5 6 3 6 Solution: We need to determine if For this, we would need 3 3 5 This gives there is a non-zero vector v such that A v = 1 v. 6 7 v1 v1 3 7 v2 = v2 6 5 v3 v3 3v1 + 6v2 + 7v3 = v1 3v1 + 3v2 + 7v3 = v2 5v1 + 6v2 + 5v3 = v3 Moving the variables on the right side to the left side gives the homogeneous system 2v1 + 6v2 + 7v3 = 0 3v1 + 2v2 + 7v3 = 0 5v1 + 6v2 + 4v3 = 0 Eigenvalues and Eigenvectors Module 5: Linear Mappings and Diagonalization Eigenvalues Example Example 3 Consider A = 3 5 6 3 6 Solution Continued: 7 7. Is = 1 an eigenvalue of A? 5 2v1 + 6v2 + 7v3 = 0 3v1 + 2v2 + 7v3 = 0 5v1 + 6v2 + 4v3 = 0 Solving this system, we find that the only solution is the trivial solution. Hence, the only vector that satisfies A v = 1 v is the zero vector. Hence, = 1 is not an eigenvalue of A. It is instructive to notice that the coefficient matrix of the homogeneous system was 2 6 7 3 6 7 3 2 7 = 3 3 7 = A I 5 6 4 5 6 5 Eigenvalues and Eigenvectors Module 5: Linear Mappings and Diagonalization Page 7 of 8 c University of Waterloo and others MATH 636 Lecture 34 Slides (Module 5) (Last Updated: May 22, 2014) Check-In Question 1 Use the method in the previous example to determine if = 4 is an eigenvalue of 3 5 5 A = 7 9 5. 7 7 3 Solution: We need to determine For this we would need 3 7 7 This gives if there is a non-zero vector v such that A v = v. 5 9 7 5 v1 v1 5 v2 = 4 v2 3 v3 v3 3v1 + 5v2 5v3 = 4v1 7v1 + 9v2 5v3 = 4v2 7v1 + 7v2 3v3 = 4v3 Moving the variables on the right side to the left side gives the homogeneous system 7v1 + 5v2 5v3 = 0 7v1 + 5v2 5v3 = 0 7v1 + 7v2 7v3 = 0 We can see that this homogeneous system will have at least one free variable, and hence will have infinitely many solutions. Consequently, = 4 is an eigenvalue of A. Eigenvalues and Eigenvectors Module 5: Linear Mappings and Diagonalization Conclusion Our real goal was to determine for which matrices A, where we are thinking of A as the standard matrix of a linear mapping L : Rn Rn , is there a basis B of eigenvectors so that [L]B is diagonal. Thus, we actually want a quick way of finding all eigenvalues and corresponding eigenvectors to a matrix. Eigenvalues and Eigenvectors Module 5: Linear Mappings and Diagonalization Page 8 of 8 c University of Waterloo and others MATH 636 Lecture 35 Slides (Module 5) (Last Updated: May 22, 2014) Eigenvalues and Eigenvectors Continued Module 5: Linear Mappings and Diagonalization Eigenvalues and Eigenvectors Continued Module 5: Linear Mappings and Diagonalization Eigenvalues and Eigenvectors Continued In the last lecture, in the pursuit of trying to find a basis B for a linear mapping L : Rn Rn such that [L]B was diagonal, we defined eigenvalues and eigenvectors. Recall that if A v = v where v 6= 0 , then is called an eigenvalue of A and v is called an eigenvector of A corresponding to . In this lecture, we will form a general procedure for finding all the eigenvalues and corresponding eigenvectors for a matrix and look at the theory of eigenvalues and eigenvectors. Eigenvalues and Eigenvectors Continued Module 5: Linear Mappings and Diagonalization Page 1 of 10 c University of Waterloo and others MATH 636 Lecture 35 Slides (Module 5) (Last Updated: May 22, 2014) Finding Eigenvalues Let A be an n n matrix. We want to find all such that A v = v Av v = 0 v 6= 0 (A I) v = 0 Hence, is an eigenvalue of A if and only if det(A I) = 0. Moreover, if is an eigenvalue, then all corresponding eigenvectors are the non-trivial solutions of the homogeneous system. Eigenvalues and Eigenvectors Continued Module 5: Linear Mappings and Diagonalization Finding Eigenvalues and Eigenvectors Example Example \u0014 2 Find the eigenvalues and corresponding eigenvectors of A = 1 \u0015 12 . 5 Solution: We have \u0012\u0014 \u0015 \u0014 \u0015\u0013 2 12 1 0 1 5 0 1 \u0012\u0014 \u0015\u0013 2 12 = det 1 5 det(A I) = det = (2 )(5 ) (12)(1) = 2 + 3 + 2 = ( + 2)( + 1) So, det(A I) = 0 if and only if = 2 or = 1. Hence, the eigenvalues of A are 1 = 2 and 2 = 1. Eigenvalues and Eigenvectors Continued Module 5: Linear Mappings and Diagonalization Page 2 of 10 c University of Waterloo and others MATH 636 Lecture 35 Slides (Module 5) (Last Updated: May 22, 2014) Finding Eigenvalues and Eigenvectors Example Continued Example Find the eigenvalues and corresponding eigenvectors of A = Solution Continued: For 1 = 2, we have A 1 I = \u0014 4 1 \u0015 \u0014 12 1 3 0 \u0014 2 1 \u0015 12 . 5 \u0015 3 0 We get that a vector equation for the solution set is \u0014 \u0015 3 x =t , tR 1 Therefore, all eigenvectors corresponding to 1 = 2 are \u0014 \u0015 3 x =t , t R, t 6= 0 1 Eigenvalues and Eigenvectors Continued Module 5: Linear Mappings and Diagonalization Finding Eigenvalues and Eigenvectors Example Continued Example \u0014 2 Find the eigenvalues and corresponding eigenvectors of A = 1 Solution Continued: Similarly, for 2 = 1, we have A 2 I = \u0014 3 1 \u0015 \u0014 12 1 4 0 \u0015 12 . 5 \u0015 4 0 Hence, a vector equation for the solution set is \u0014 \u0015 4 x =t , tR 1 Therefore, all eigenvectors corresponding to 2 = 1 are \u0014 \u0015 4 x =t , t R, t 6= 0 1 Eigenvalues and Eigenvectors Continued Module 5: Linear Mappings and Diagonalization Page 3 of 10 c University of Waterloo and others MATH 636 Lecture 35 Slides (Module 5) (Last Updated: May 22, 2014) Finding Eigenvalues and Eigenvectors Example Continued Check: We have and \u0014 2 1 \u0015\u0014 \u0015 \u0014 \u0015 \u0014 \u0015 12 3t 6t 3t = = (2) 5 t 2t t \u0014 2 1 \u0015\u0014 \u0015 \u0014 \u0015 \u0014 \u0015 4t 4t 12 4t = = (1) 5 t t t Eigenvalues and Eigenvectors Continued Module 5: Linear Mappings and Diagonalization Finding Eigenvalues and Eigenvectors Example Example 1 Find the eigenvalues and corresponding eigenvectors of B = 2 6 Solution: We have det(B I) = = = = = 2 5 6 2 2. 3 1 2 2 1 0 0 det 2 5 2 0 1 0 6 6 3 0 0 1 1 2 2 2 5 2 6 6 3 1 2 0 2 5 3 6 6 3 1 2 0 4 1 0 6 6 3 1 2 2 2 (3 ) = ( 3)( 9) = ( 3) ( + 3) 4 1 So, det(B I) = 0 if and only if = 3 or = 3. Hence, the eigenvalues of A are 1 = 3 and 2 = 3. Eigenvalues and Eigenvectors Continued Module 5: Linear Mappings and Diagonalization Page 4 of 10 c University of Waterloo and others MATH 636 Lecture 35 Slides (Module 5) (Last Updated: May 22, 2014) Finding Eigenvalues and Eigenvectors Example Continued Example 1 Find the eigenvalues and corresponding eigenvectors of B = 2 6 Solution Continued: For 1 = 3, we have 2 B 1 I = 2 6 2 2 6 2 1 2 0 6 0 Hence, a vector equation for the solution set is 1 1 x = s 1 + t 0 , 0 1 2 5 6 2 2. 3 1 0 0 1 0 0 s, t R Therefore, all eigenvectors corresponding to 1 = 3 are 1 1 x = s 1 + t 0 0 1 for any s, t R with s and t not both equal to 0. Eigenvalues and Eigenvectors Continued Module 5: Linear Mappings and Diagonalization Finding Eigenvalues and Eigenvectors Example Continued Example 1 Find the eigenvalues and corresponding eigenvectors of B = 2 6 Solution Continued: Similarly, for 2 = 3, we have 4 A 2 I = 2 6 2 8 6 2 1 2 0 0 0 Hence, a vector equation for the solution set is 1 x = t 1 , 3 0 1 0 2 5 6 2 2. 3 1/3 1/3 0 tR Therefore, all eigenvectors corresponding to 2 = 3 are 1 x = t 1 , t R, t 6= 0 3 Eigenvalues and Eigenvectors Continued Module 5: Linear Mappings and Diagonalization Page 5 of 10 c University of Waterloo and others MATH 636 Lecture 35 Slides (Module 5) (Last Updated: May 22, 2014) Check-In Question 1 Find the eigenvalues and corresponding eigenvectors of A = \u0014 1 1 \u0015 1 . 1 Solution: We have \u0015 \u0014 \u0015\u0013 \u0012\u0014 1 1 1 0 1 1 0 1 \u0012\u0014 \u0015\u0013 1 1 = det 1 1 det(A I) = det = (1 )(1 ) 1(1) = 2 2 = ( 2) So, det(A I) = 0 if and only if = 0 or = 2. Hence, the eigenvalues of A are 1 = 0 and 2 = 2. Eigenvalues and Eigenvectors Continued Module 5: Linear Mappings and Diagonalization Check-In Question 1 Find the eigenvalues and corresponding eigenvectors of A = Solution Continued: For 1 = 0, we have A 1 I = \u0014 1 1 \u0015 \u0014 1 1 1 0 \u0014 1 1 \u0015 1 . 1 \u0015 1 0 \u0014 \u0015 1 , tR Hence, a vector equation for the solution set is x =t 1 \u0014 \u0015 1 Therefore, all eigenvectors corresponding to 1 = 0 are x =t , t R, t 6= 0 1 \u0015 \u0014 \u0015 1 1 1 1 0 0 \u0014 \u0015 1 Hence, a vector equation for the solution set is x =t , tR 1 \u0014 \u0015 1 Therefore, all eigenvectors corresponding to 2 = 2 are x =t , t R, t 6= 0 1 Similarly, for 2 = 2, we have A 2 I = Eigenvalues and Eigenvectors Continued \u0014 1 1 Module 5: Linear Mappings and Diagonalization Page 6 of 10 c University of Waterloo and others MATH 636 Lecture 35 Slides (Module 5) (Last Updated: May 22, 2014) Characteristic Polynomial Definition If A is an n n matrix, then we call the n-th degree polynomial given by C() = det(A I) the characteristic polynomial of A. Theorem 6.2.1 A scalar is an eigenvalue of a square matrix A if and only if C() = 0. Eigenvalues and Eigenvectors Continued Module 5: Linear Mappings and Diagonalization Algebraic Multiplicity Definition The algebraic multiplicity, a , of an eigenvalue is the number of times appears as a root of the characteristic polynomial C(). Example 1 We saw that the characteristic polynomial of B = 2 6 C() = ( 3)2 ( + 3). 2 5 6 2 2 was 3 Thus, a1 = 2 and a2 = 1. Eigenvalues and Eigenvectors Continued Module 5: Linear Mappings and Diagonalization Page 7 of 10 c University of Waterloo and others MATH 636 Lecture 35 Slides (Module 5) (Last Updated: May 22, 2014) Eigenspace and Geometric Multiplicity Definition If is an eigenvalue of A, then Null(A I) is called the eigenspace of and is denoted E . Definition The geometric multiplicity g of an eigenvalue is the dimension of its eigenspace. That is, g = dim Null(A I). Eigenvalues and Eigenvectors Continued Module 5: Linear Mappings and Diagonalization Eigenspace and Geometric Multiplicity Example Example 1 For B = 2 6 2 5 6 2 2, 3 1 1 we found that a basis for the eigenspace of 1 = 3 was 1 , 0 . 0 1 Thus, g1 = dim E1 = 2. 1 We also found that a basis for the eigenspace of 2 = 3 was 1 . 3 Hence, g2 = dim E2 = 1. Eigenvalues and Eigenvectors Continued Module 5: Linear Mappings and Diagonalization Page 8 of 10 c University of Waterloo and others MATH 636 Lecture 35 Slides (Module 5) (Last Updated: May 22, 2014) Eigenspace and Geometric Multiplicity Example Example Find the algebraicand geometric multiplicities of all eigenvalues of 1 2 3 A = 2 5 3. 4 4 5 Solution: We have 1 C() = det(A I) = 2 4 3 = 3 0 3 = 0 0 2 5 4 2 5 4 2 3 4 3 3 5 3 3 5 3 0 5 = ( 5)( 3)2 Hence, the eigenvalues of A are 1 = 5 and 2 = 3. We have a1 = 1 and a2 = 2. Eigenvalues and Eigenvectors Continued Module 5: Linear Mappings and Diagonalization Eigenspace and Geometric Multiplicity Example Example Find the algebraicand geometric multiplicities of all eigenvalues of 1 2 3 A = 2 5 3. 4 4 5 Solution: For 1 = 5, we get 4 A 1 I = 2 4 3 1 0 3/2 3 0 1 3/2 0 0 0 0 3 Thus, a basis for the eigenspace of 1 = 5 is 3 and hence g1 = 1. 2 For 2 = 3, we get 2 2 3 1 1 0 0 1 A 2 I = 2 2 3 0 4 4 2 0 0 0 1 Thus, a basis for the eigenspace of 2 = 3 is 1 and hence g2 = 1. 0 2 0 4 Eigenvalues and Eigenvectors Continued Module 5: Linear Mappings and Diagonalization Page 9 of 10 c University of Waterloo and others MATH 636 Lecture 35 Slides (Module 5) (Last Updated: May 22, 2014) Check-In Question 1 2 Find the algebraic and geometric multiplicities of all eigenvalues of A = 0 0 Solution: We have 2 C() = det(A I) = 0 0 1 2 0 = ( 2)3 0 1 2 Hence, the only eigenvalue of A is 1 = 2. We have a1 = 3. For 1 = 2, we get 0 1 0 A 2 I = 0 0 1 0 0 0 1 Thus, a basis for the eigenspace of 1 = 2 is 0 and hence g1 = 1. 0 Eigenvalues and Eigenvectors Continued Module 5: Linear Mappings and Diagonalization Theorems Lemma 6.2.2 If A and B are similar matrices, then A and B have the same characteristic polynomial, and hence the same eigenvalues. Theorem 6.2.3 If is an eigenvalue of A, then 1 g a . Note that an eigenvalue such that g < a is said to be deficient. Theorem 6.2.4 If 1 , . . . , n are all eigenvalues of an n n matrix A, then det A = 1 n tr A = 1 + + n Eigenvalues and Eigenvectors Continued Module 5: Linear Mappings and Diagonalization Page 10 of 10 c University of Waterloo and others 1 2 0 0 1. 2 MATH 636 Lecture 36 Slides (Module 5) (Last Updated: May 22, 2014) Diagonalization Theory Module 5: Linear Mappings and Diagonalization Diagonalization Theory Module 5: Linear Mappings and Diagonalization Diagonalization Theory Definition An n n matrix A is said to be diagonalizable if there exists an invertible matrix P such that P 1 AP = D is diagonal. We say that P diagonalizes A. Recall that we previously showed that if B = { v 1, . . . , v n } is a basis for Rn of eigenvectors of a matrix A, and 1 , . . . , n \u0002are eigenvalues of\u0003 A corresponding to v 1 v n gives v 1, . . . , v n respectively, then taking P = P 1 AP = diag(1 , . . . , n ) Diagonalization Theory Module 5: Linear Mappings and Diagonalization Page 1 of 5 c University of Waterloo and others MATH 636 Lecture 36 Slides (Module 5) (Last Updated: May 22, 2014) Diagonalization Theory Lemma 6