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Page of 8 Solutions for Spring Semester MATH 32/52 (Barsamian) Homework 8 (Due Wednesday, April 9, 24) (Three problems similar to suggested exercise 5-#3(a),(b),(c)) For each matrix A M n n (F) (i)

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Page of 8 Solutions for Spring Semester MATH 32/52 (Barsamian) Homework 8 (Due Wednesday, April 9, 24) (Three problems similar to suggested exercise 5-#3(a),(b),(c)) For each matrix A M n n (F) (i) Determine all the eigenalues of A. Show all details of the calculation. (ii) For each eigenalue λ of A, find the eigenectors corresponding to λ. (iii) If possible, find a basis for F n consisting of eigenectors of A. (i) If successful in finding such a basis, determine an inertible matrix Q and a diagonal matrix D such that Q AQ = D. 3 [] (2 points) Let A = ( 2 2 ) M 3 3 (R) 2 2 (i) Determine all the eigenalues of A. Show all details of the calculation. Solution: We start by finding the characteristic polynomial f(λ). 3 λ f(λ) = det(a λi 3 ) = det ( 2 2 λ ) = do determinant along 3 rd column = 2 2 λ 2 (2 λ) (3 λ) λ) = ( )det ( ) ( )det ( ) + ( λ)det (( (2 λ) ) = ( )((2)(2) (2)(2 λ)) + ((3 λ)(2) (2)()) (λ)((3 λ)(2 λ) (2)()) = = λ 3 + 5λ 2 8λ + 4 (A check with Wolfram Alpha confirms that this is correct.) The eigenalues of A are the roots of the characteristic polynomial. We find them by setting f(λ) equal to zero and soling for λ. = f(λ) = λ 3 + 5λ 2 8λ + 4 = (λ )(λ 2) 2 We see that the eigenalues will be λ = and λ 2 = 2. (Wolfram Alpha confirms this.) (ii) For each eigenalue λ of A, find the eigenectors corresponding to λ. Solution: Find the eigenector corresponding to eigenalue λ =. Corresponding to eigenalue λ = will be an eigenector that we can denote = ( ). 3 The ector must satisfy the matrix equation (A λ I 3 ) =. That is 3 λ ( 2 2 λ ) ( 2 ) = ( ) 2 2 λ 3 3 λ This corresponds to the augmented matrix (A λ I 3 ) = ( 2 2 λ ) 2 2 λ 2 Using the alue λ =, this augmented matrix becomes (A λ I 3 ) = ( 2 ) 2 2 Add (-)R to R2. Add (-)R to R3. Add R3 to R. Multiply R by (/2). Multiply R3 by (-). Interchange R2 and R3. ( /2) We obtain the reduced row echelon form ( ) (Wolfram Alpha confirms this.) 2 Page 2 of 8 (/2) 3 = This corresponds to the system { 2 = = Clearly 2 must be zero. But can be any real number, as long as 3 = 2. So the solution set is K H = {(s,,2s), s R}. A basis for this solution set is β = {(,,2)}. We can let the basis ector be. That is, we can let = ( ) () + () (2) We check by multiplying A = ( 2 2 ) ( ) = ( 2() + 2() (2) ) = ( ) = () + 2() + (2) 2 This confirms that is in fact an eigenector with eigenalue. Now find the eigenector corresponding to eigenalue λ 2 = 2. Corresponding to eigenalue λ 2 = will be an eigenector that we can denote 2 = ( ). 32 The ector x must satisfy the matrix equation (A λ 2 I 3 ) 2 =. That is 3 λ 2 2 ( 2 2 λ 2 ) ( 22 ) = ( ) 2 2 λ λ 2 This corresponds to the augmented matrix (A λ 2 I 3 ) = ( 2 2 λ 2 ) 2 2 λ 2 Using the alue λ 2 = 2, this augmented matrix becomes (A λ 2 I 3 ) = ( 2 ) Add (-2)R to R2. Add (-2)R to R3. Multiply R2 by (-/2). Add (-2)R2 to R. ( /2) 2 (/2) 32 = We obtain the echelon form ( ( /2) ) corresponding to system { 22 (/2) 32 = = 32 can be any real number, as long as 2 = (/2) 32 and 22 = (/2) 32. So the solution set is K H = {((/2)s, (/2)s, s), s R}. A basis for this set is β = {(,,2)}. We can let the basis ector be 2. That is, we can let 2 = ( ) () + () (2) 2 We check by multiplying A = ( 2 2 ) ( ) = ( 2() + 2() (2) ) = ( 2) = () + 2() + (2) 4 This confirms that 2 is in fact an eigenector with eigenalue 2. (iii) If possible, find a basis for F n consisting of eigenectors of A. Solution: Since we only found two eigenectors for A, we conclude that it is not possible to find a basis for F n consisting of eigenectors of A. (i) If successful in finding such a basis, determine an inertible matrix Q and a diagonal matrix D such that Q AQ = D. Solution: Since it is not possible to find a basis for F n consisting of eigenectors of A, we conclude that A is not diagonalizable. [2] (2 points) Let A = ( 2 2 ) M 2 2(C) (i) Determine all the eigenalues of A. Show all details of the calculation. 2 22 Page 3 of 8 Solution: We start by finding the characteristic polynomial f(λ). f(λ) = det(a λi 2 ) = det ( λ 2 2 λ ) = ( λ)( λ) (2)( 2) = λ2 2λ + 5 (Wolfram Alpha confirms this.) The eigenalues of A are the roots of the characteristic polynomial. We find them by setting f(λ) equal to zero and soling for λ. = f(λ) = λ 2 2λ + 5 It is not obious how to factor this polynomial, so we use the quadratic formula. λ = ( 2) ± ( 2)2 4()(5) = 2 ± 6 = 2 ± 4 = 2 ± 4i = ± 2i 2() We see that the eigenalues will be λ = + 2i and λ 2 = 2i. (Wolfram Alpha confirms this.) Remark: This tells us that the characteristic polynomial factors oer the field C in the following way: f(λ) = λ 2 2λ + 5 = (λ λ )(λ λ 2 ) = (λ ( + 2i))(λ ( 2i)) (ii) For each eigenalue λ of A, find the eigenectors corresponding to λ. Solution: Find the eigenector corresponding to eigenalue λ = + 2i. Corresponding to eigenalue λ = + 2i will be an eigenector that we can denote = ( ). 2 The ector must satisfy the matrix equation (A λ I 2 ) =. That is ( λ 2 ) ( 2 λ ) = ( 2 ) This corresponds to the augmented matrix (A λ I 3 ) = ( λ 2 2 λ ) Using the alue λ = + 2i, this augmented matrix becomes ( + 2i) (A λ I 3 ) = ( 2 2 ( + 2i) 2i 2 ) = ( 2 2i ) Multiply R by (i/2). Add (-2)R to R2. We obtain the echelon form ( i ) corresponding to the system { i 2 = = 2 can be any real number, as long as = i 2. So the solution set is K H = {(is, s), s C}. A basis for this solution set is β = {(i, )}. We can let the basis ector be. That is, we can let = ( i ). Check by multiplying A = ( 2 2 ) (i 2() + i ) = ((i) ) = ( 2 2(i) + () + 2i ) = ( + 2i) (i ) = λ This confirms that = ( i ) is in fact an eigenector with eigenalue λ = + 2i. Find the eigenector corresponding to eigenalue λ 2 = 2i. Corresponding to eigenalue λ 2 = 2i will be an eigenector that we can denote 2 = ( 2 ). 22 The ector 2 must satisfy the matrix equation (A λ 2 I 2 ) =. That is ( λ 2 2 ) ( 2 λ 2 ) = ( 2 ) This corresponds to the augmented matrix (A λ 2 I 3 ) = ( λ λ 2 ) Using the alue λ = + 2i, this augmented matrix becomes ( 2i) (A λ 2 I 3 ) = ( 2 2 ( 2i) 2i 2 ) = ( 2 2i ) Multiply R by (-i/2). Add (-2)R to R2. We obtain the echelon form ( i ) corresponding to the system { 2 + i 22 = = Page 4 of 8 22 can be any real number, as long as 2 = i 22. So the solution set is K H = {( is, s), s C}. A basis for this solution set is β = {( i, )}. We can let the basis ector be 2. That is, we can let 2 = ( i ). Check by multiplying A = ( 2 2 ) ( i 2() i ) = (( i) ) = ( 2 ) = ( 2i) ( i 2( i) + () 2i ) = λ 2 2 This confirms that 2 = ( i ) is in fact an eigenector with eigenalue λ 2 = 2i. (iii) If possible, find a basis for F n consisting of eigenectors of A. Solution: We found two eigenectors, = ( i ) and 2 = ( i ). Because they are linearly independent ectors in C 2, we can use them as a basis of C 2. So our basis is β = {, 2 } = {( i ), ( i )} (i) If successful in finding such a basis, determine an inertible matrix Q and a diagonal matrix D such that Q AQ = D. Solution: The inertible matrix Q can be made by using the eigenectors = ( i ) and 2 = ( i ) as columns. That is Q = ( i i 2 ) = ( ). We know that Q is inertible because it has rank 2. The diagonal matrix D can be made by using the corresponding eigenalues on the diagonal. That is D = ( λ ( + 2i) ) = ( λ 2 ( 2i) ). [3] (2 points) Let A = ( 2 2 ) M 2 2(R) (i) Determine all the eigenalues of A. Show all details of the calculation. Solution: This is the same matrix that was used in problem [2], but now the field is R instead of C. In problem [2], we found that the characteristic polynomial was f(λ) = λ 2 2λ + 5. We used the quadratic formula to find roots of this polynomial, and the result was that we found two complex roots: λ = + 2i and λ 2 = 2i. Those are the eigenalues when the field is C. This told us that the characteristic polynomial factors oer the field C in the following way: f(λ) = λ 2 2λ + 5 = (λ λ )(λ λ 2 ) = (λ ( + 2i))(λ ( 2i)) But now in that we are using the field R instead of C, we must say that there are no real eigenalues! Remark: This tells us that the polynomial f(λ) = λ 2 2λ + 5 does not factor oer the field R. (ii),(iii) Since there are no real eigenalues, there are no eigenectors, either. (i) Since there are no real eigenalues, the matrix A is not diagonalizable oer the field R. [4] (2 points) Let V = R 2 and define T: V V by T(a, b) = (a + b, 4a + b). Find the eigenalues of T and find an ordered basis β for V such that [T] β is diagonal. Solution: First we get a matrix representation for T in the standard basis. Let α = {e, e 2 } = {(,), (,)} be the standard basis for R 2. [T] α will be a 2 2 matrix [T] α = ( A A 2 )with entries obtained from the two equations: A 2 A 2 { T(e ) = A e + A 2 e 2 T(e 2 ) = A 2 e + A 22 e 2 So we inestigate what happens when these basis ectors are used as input to T. { T(e ) = T(,) = ( +,4() + ) = (,4) = (,) + 4(,) = ()e + (4)e 2 T(e 2 ) = T(,) = ( +,4() + ) = (,) = (,) + (,) = ()e + ()e 2 Matching the coefficients in the two sets of equations, we find [T] α = ( 4 ) Now we find the characteristic polynomial and the eigenalues of the matrix [T] γ. Page 5 of 8 f(λ) = det([t] α λi 2 ) = det (( 4 ) (λ λ) )) = det (( λ 4 ( λ) ) = ( λ)( λ) (4)() = λ 2 2λ 3 = (λ + )(λ 3) We conclude that the eigenalues of the matrix [T] α = ( 4 ) are λ = and λ 2 = 3. (Wolfram Alpha confirms this.) Now we find the corresponding eigenectors. Find the eigenector corresponding to eigenalue λ =. Corresponding to eigenalue λ = will be an eigenector that we can denote = ( ). 2 The ector must satisfy the matrix equation ([T] α λ I 2 ) =. That is, ( ( λ ) 4 ( λ ) ) ( ) = ( 2 ) This corresponds to the augmented matrix ([T] α λ I 2 ) = ( ( λ ) 4 ( λ ) ) Using the alue λ =, this augmented matrix becomes ([T] α λ I 2 ) = ( ) Add (-2)R to R2. Multiply R by (/2). We obtain the echelon form ( (/2) ) corresponding to the system { + (/2) 2 = = can be any real number, as long as 2 = 2. So the solution set is K H = {(s, 2s), s R}, and a basis for the solution set is β = {(, 2)}. We can let the basis ector be. That is, let = ( 2 ). Check by multiplying [T] α = ( 4 ) ( + ( 2) ) = (() 2 4() + ( 2) ) = ( 2 ) = ( ) ( 2 ) = λ This confirms that = ( 2 ) is in fact an eigenector with eigenalue λ =. Find the eigenector corresponding to eigenalue λ 2 = 3. Corresponding to eigenalue λ 2 = 3 will be an eigenector that we can denote 2 = ( 2 ). 22 The ector 2 must satisfy the matrix equation ([T] α λ 2 I 2 ) 2 =. That is, ( ( λ 2) 4 ( λ 2 ) ) ( ) = ( 2 ) This corresponds to the augmented matrix ([T] α λ 2 I 2 ) = ( ( λ 2) 4 ( λ 2 ) ) Using the alue λ 2 = 3, this augmented matrix becomes ([T] α λ 2 I 2 ) = ( ) Add (2)R to R2. Multiply R by (-/2). We obtain the echelon form ( ( /2) ) corresponding to the system { 2 + ( /2) 22 = = 2 can be any real number, as long as 22 = 2 2. So the solution set is K H = {(s, 2s), s R}, and a basis for the solution set is β = {(,2)}. We can let the basis ector be 2. That is, we can let 2 = ( 2 ). Check by multiplying [T] α = ( 4 ) ( + (2) ) = (() 2 4() + (2) ) = (3 6 ) = (3) ( 2 ) = λ 2 2 This confirms that 2 = ( 2 ) is in fact an eigenector with eigenalue λ 2 = 3. Conclusion Switching from column form to ordered-pair form for the eigenectors, we hae the following: The ector = (, 2) is an eigenector with eigenalue λ =. The ector 2 = (,2) is an eigenector with eigenalue λ 2 = 3. Page 6 of 8 Introduce ordered basis β = {, 2 } = {(, 2), (,2)}. In this basis, the matrix representation of T will be the diagonal matrix [T] β = ( λ ) = ( λ 2 3 ) [5] (2 points) Let V = P 2 (R) and define T: V V by T(f(x)) = xf (x) + f(2)x + f(3). (a) Let α = {, 2, 3 } = {, x, x 2 }, the standard basis for P 2 (R). Find [T] α. Solution: Obsere that the formula for T is gien with a general function f(x) as input. But we know that our inputs will be elements of P 2 (R), that is, functions of the form f(x) = a + bx + cx 2. So it will be helpful to hae a formula for T(a + bx + cx 2 ). For that, we get some of the parts first. If f(x) = a + bx + cx 2, then The deriatie of f will be f (x) = b + 2cx. The alue of f at x = 2 will be the number f(2) = a + b(2) + c(2) 2 = a + 2b + 4c. The alue of f at x = 3 will be the number f(3) = a + b(3) + c(3) 2 = a + 3b + 9c. We use those parts to get the formula for T(a + bx + cx 2 ). T(a + bx + cx 2 ) = x(b + 2cx) + (a + 2b + 4c)x + (a + 3b + 9c) = (a + 3b + 9c) + (a + 3b + 4c)x + (2c)x 2 A A 2 A 3 [T] γ will be a 3 3 matrix [T] α = ( A 2 A 2 A 23 ) with entries obtained from the three equations: A 3 A 32 A 33 T( ) = A + A A 3 3 { T( 2 ) = A 2 + A A 32 3 T( 3 ) = A 3 + A A 33 3 So we inestigate what happens when the basis ectors are used as input to T. T( ) = T( + x + x 2 ) = () + ()x + ()x 2 = () + () 2 + () 3 { T( 2 ) = T( + x + x 2 ) = (3) + (3)x + ()x 2 = (3) + (3) 2 + () 3 T( 3 ) = T( + x + x 2 ) = (9) + (4)x + (2)x 2 = (9) + (4) 2 + (2) Matching the coefficients in the two sets of equations, we find [T] α = ( 3 4) 2 (b) Find the eigenalues of the matrix [T] α and the corresponding eigenectors of [T] α. These eigenectors should be column ectors representing elements of R 3. Solution: We start by finding the characteristic polynomial and the eigenalues. The characteristic polynomial is ( λ) 3 9 ( λ) 3 f(λ) = det([t] α λi 3 ) = det ( (3 λ) 4 ) = (2 λ)det ( (3 λ) ) (2 λ) = (2 λ)[( λ)(3 λ) ()(3)] = (2 λ)[λ 2 4λ] = (2 λ)(λ)(λ 4) We conclude that the eigenalues are λ = and λ 2 = 2 and λ 3 = 4. (Wolfram Alpha confirms this.) Now we find the corresponding eigenectors. Find the eigenector corresponding to eigenalue λ =. Corresponding to eigenalue λ = will be an eigenector that we can denote = ( ). 3 The ector must satisfy the matrix equation ([T] α λ I 3 ) =. That is, ( λ ) 3 9 ( (3 λ ) 4 ) ( 2 ) = ( ) (2 λ ) 3 2 Page 7 of 8 ( λ ) 3 9 This corresponds to the augmented matrix ([T] α λ I 3 ) = ( (3 λ ) 4 ) (2 λ ) 3 9 Using the alue λ =, this augmented matrix becomes ([T] α λ I 2 ) = ( 3 4 ) 2 Add (-)R to R2. Multiply R2 by (-/5). Add (-9)R2 to R. Add (-2)R2 to R = We obtain the echelon form ( ) corresponding to the system { 3 = = Clearly, we must hae 3 =. We see that 2 can be any real number, as long as = 2. So the solution set is K H = {(s, s, ), s R}, and a basis for this solution set is β = {(,,)}. We can let the basis ector be. That is, we can let = ( ). 3 9 Check by multiplying[t] α = ( 3 4) ( ) = = ( ) = () = λ 2 This confirms that = ( ) is in fact an eigenector with eigenalue λ =. Find the eigenector corresponding to eigenalue λ 2 = 2. Corresponding to eigenalue λ 2 = 2 will be an eigenector that we can denote 2 = ( ). 32 The ector 2 must satisfy the matrix equation ([T] α λ 2 I 3 ) 2 =. That is, ( λ 2 ) ( (3 λ 2 ) 4 ) ( 22 ) = ( ) (2 λ 2 ) 32 ( λ 2 ) 3 9 This corresponds to the augmented matrix ([T] α λ 2 I 3 ) = ( (3 λ 2 ) 4 ) (2 λ 2 ) 3 9 Using the alue λ 2 = 2, this augmented matrix becomes ([T] α λ 2 I 2 ) = ( 4 ) Add R to R2. Multiply R2 by (/4). Add (-3)R2 to R. Multiply R by (-). (3/4) 2 + (3/4) 32 = We obtain the echelon form ( (3/4) ) corresponding to the system { 22 + (3/4) 32 = = 32 can be any real number, as long as 2 = (/4) 32 and 22 = (/4) 32. So the solution set is K H = {((/4)s, (/4)s, s), s R}. A basis for the solution set is β = {(,,4)}. We can let the basis ector be 2. That is, we can let 2 = ( ) Page 8 of Check by multiplying [T] α = ( 3 4) ( ) = = ( 26) = ( 2) 2 = λ This confirms that 2 = ( ) is in fact an eigenector with eigenalue λ 2 = 2. 4 Find the eigenector corresponding to eigenalue λ 3 = 4. Corresponding to eigenalue λ 3 = 4 will be an eigenector that we can denote 3 = ( ). 33 The ector 3 must satisfy the matrix equation ([T] α λ 3 I 3 ) 3 =. That is, ( λ 3 ) ( (3 λ 3 ) 4 ) ( 23 ) = ( ) (2 λ 3 ) 33 ( λ 3 ) 3 9 This corresponds to the augmented matrix ([T] α λ 3 I 3 ) = ( (3 λ 3 ) 4 ) (2 λ 3 ) 3 9 Using the alue λ 3 = 4, this augmented matrix becomes ([T] α λ 3 I 2 ) = ( 4 ) 2 Multiply R by (-/3). Add (-)R to R2. Add (3)R3 to R2. Add (3)R2 to R. Add (2)R2 to R = We obtain the echelon form ( ) corresponding to the system { 33 = = Clearly we must hae 33 =. We see that 3 can be any real number, as long as 23 = 3. So the solution set is K H = {(s, s, ), s R}. A basis for the solution set is β = {(,,)}. We can let the basis ector be 3. That is, we can let 3 = ( ) Check by multiplying [T] α = ( 3 4) ( ) = = ( 4) = (4) 3 = λ This confirms that 3 = ( ) is in fact an eigenector with eigenalue λ 3 = 4. (c) Using your result from (b), find a corresponding ordered basis β forp 2 (R) such that [T] β is a diagonal matrix. (The basis ectors in β should be polynomials, elements of P 2 (R).) Solution: Conerting our column ectors to functions, we hae The eigenector = ( ) and 3 = ( ) and 3 = ( ) correspond to the functions = + x and 2 = 3x + 4x 2 and 3 = + x. So our basis of eigenectors is β = {, 2, 3 } = { + x, 3 3x + 4x 2, + x}. In this basis, the matrix representation of T will be [T] α = ( 2 )

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