Unformatted text preview: Math 330 Homework 5.5 (Pages 341,342)
(2) Find the eigenvalues and a basis for each eigenspace in C2 of A = SOLUTION: det(A  xI) = 5  x 5 = (5  x)(1  x)  (5)(1) = x2  6x + 10 = 0 1 1x 6 62  4(10) 2 = 6 2 4 = 6 2i =3i 2 5 5 1 1 x= x=3+i A  (3 + i)I = 2i 5 1 2  i Now assuming we have indeed calculated the eigenvalues correctly, row 1 and row 2 are multiples of each other and hence all the information is contained in the first row. Thus with x1 eigenvector x = we have that x2 (2  i)x1  5x2 = 0 (2  i)x1 = 5x2 x2 = 2i x1 5 So that we can think of x1 as a free variable which we set to 1, and then the pivot variable 1 x2 = 2i . Therefore, x = 2i is our eigenvector. We can also multiply by 5 and thus for 5
5 eigenvalue x = 3 + i the eigenspace is the span of We can immediately conjugate to conclude that for the eigenvalue x = 3  i the eigenspace is 5 the span of . 2+i (4) Find the eigenvalues and a basis for each eigenspace in C2 of A = SOLUTION: We streamline the same approach used in problem 2: det(A  xI) = 5  x 2 = (5  x)(3  x)  (2)(1) = x2  8x + 17 = 0 1 3x 8 82  4(17) 2 = 8 2 4 = 8 2i =4i 2 5 2 1 3 5 2i . x= 1 For x = 4 + i the first row of A  (4 + i)I is [1  i  2] so (1  i)x1  2x2 = 0 or x2 = 1i x1. 2 2 We let x1 = 2 and hence for x = 4 + i our eigenspace is the span of . For x = 4  i 1i 2 our eigenspace is the span of . 1+i (10) The transformation x Ax is the composition of a rotation and a scaling. Give the angle 5 5 of the rotation, where  < , and give the scale factor r. A = . 5 5 SOLUTION: Notice that since 5 5 1 5 = we see that a point on the xaxis, namely (1, 0), 5 5 0 5 is rotated into the middle of the second quadrant and increased in length:1 This an angle of 3 4 and a scale factor of 52 + 52 = 50 = 5 2.2 a b b a so that A = 5 2 1 3 = (16) Find an invertible matrix P and a matrix C of the form P CP 1 . SOLUTION: We note from problem (4) that with eigenvalue x = 4  i we have eigenvector
Notice that the point (0, 1) on the yaxis is rotated to (5, 5) which is the same angle of 3 . 4 Notice that we do not have to calculate the eigenvalues and eigenvectors of this matrix but that this analysis a b only applies to matrices of the form as we have here. It is the form of the C in theorem 9. b a
1 2 2 2 1+i Therefore, P = 2 0 1 1 = 2 1 +i 0 1 and C = 4 1 1 4 Notice that you can do this calculation with the eigenvalue x = 4 + i instead of 4  i as we did. The key to making it work though is realizing that the imaginary part of the eigenvalue must equal the (1, 2) entry of C. (22) Let A be a complex (or real) n n matrix , and let x in Cn be an eigenvector corresponding to an eigenvalue in C. Show that for each nonzero complex scalar , the vector x is an eigenvector of A. SOLUTION: If Ax = x then A(x) = (Ax) = (x). This makes sense since eigenvectors come from eigenspaces so they have to be closed under scalar multiplication. Of course the reason we require that = 0 is because we do not consider the zero vector as an eigenvector. If we did, then every scalar would be an eigenvalue of every matrix. 3 ...
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 Spring '08
 Johnson,J
 Math, Linear Algebra, Algebra, Eigenvalue, eigenvector and eigenspace, Eigenspaces

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