H55 - Math 330 Homework 5.5 (Pages 341,342) (2) Find the...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 1-x 6 62 - 4(10) 2 = 6 2 -4 = 6 2i =3i 2 5 -5 1 1 x= x=3+i A - (3 + i)I = 2-i -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 = 2-i 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 = 2-i . Therefore, x = 2-i 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 3-x 8 82 - 4(17) 2 = 8 2 -4 = 8 2i =4i 2 5 -2 1 3 5 2-i . 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 = 1-i x1. 2 2 We let x1 = 2 and hence for x = 4 + i our eigenspace is the span of . For x = 4 - i 1-i 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 x-axis, 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 y-axis 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 ...
View Full Document

Ask a homework question - tutors are online