Math54 quiz 8- - s is an eigenvalue of A(I did not require this part for full credit but it’s great if you realized that it should be checked

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Math 54 Quiz 8 Solutions 1. (3 pts) Consider the linear transformation T from P 2 ( x ) to P 2 ( x ) given by T ( p ( x )) = x p (0) + p (2 x ) . Using the standard basis B = { 1 , x, x 2 } on P 2 ( x ), find the matrix representation [ T ] B of T . The formula for [ T ] B gives [ T ] B = ± [ T (1)] B [ T ( x )] B [ T ( x 2 )] B ² . Plugging 1 , x, x 2 into T we get T (1) = x + 1, T ( x ) = 0 + 2 x , and T ( x 2 ) = 0 + 4 x 2 . Thus, [ T ] B = 1 0 0 1 2 0 0 0 4 2. (4 points) Find all the eigenvalues of A , and for each eigenvalue, find a basis of the associated eigenspace. A = ³ 4 - 3 2 - 1 ´ . By computing that det( A - λI ) = (4 - λ )( - 1 - λ ) + 6 = λ 2 - 3 λ + 2 , one finds that the eigenvalues are 2 and 1. Therefore the eigenspaces are the nullspaces of A - 2 I = ³ 2 - 3 2 - 3 ´ ³ 1 - 3 / 2 0 0 ´ and A - I = ³ 3 - 3 2 - 2 ´ ³ 1 - 1 0 0 ´ and so we can take as bases ³ 3 2 ´ and ³ 1 1 ´ .
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3. (3 points) Let A be an n × n matrix. Show that every eigenvector of A is also an eigenvector of A 2 , and describe how their eigenvalues are related. Suppose that ~v is an eigenvector of A with eigenvalue λ . Then A 2 ~v = A ( A~v ) = A ( λ~v ) = λA ( ~v ) = λ 2 ~v So this shows that if λ is an eigenvalue of A , then λ 2 is an eigenvalue of A 2 . Let’s show the reverse: if s 2 is an eigenvalue of A 2 , let’s show that
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Unformatted text preview: s is an eigenvalue of A . (I did not require this part for full credit, but it’s great if you realized that it should be checked!) This time we should use the ”other version” of eigenvalue. Since s 2 is an eigenvalue of A 2 , we know that A 2-s 2 I is not invertible. But, ( A + sI )( A-sI ) = A 2 + sIA-sAI-s 2 I 2 = A 2 + sA-sA-s 2 I = A 2-s 2 I So if that matrix on the far right is noninvertible, at least one of the two matrices in the factorization on the left is noninvertible as well (because otherwise they would both be invertible and then so would their product). One of those matrices being noninvertible shows that one of ± s is an eigenvalue of A . But, ± s are the two possible squareroots of s 2 , so we again see that the squares of the eigenvalues of A are eigenvalues of A 2 ....
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This note was uploaded on 09/17/2011 for the course MATH 54 taught by Professor Chorin during the Spring '08 term at University of California, Berkeley.

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Math54 quiz 8- - s is an eigenvalue of A(I did not require this part for full credit but it’s great if you realized that it should be checked

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