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Unformatted text preview: Winter 2011 Math 67 Linear Algebra Homework 6 Problems 5.9, 5.14 and 5.19 from Axler. Solution. (5.9) Suppose that T : V V is a linear map with rank k . If v is an eigenvector with a non-zero eigenvector then T ( v/ ) = v . This implies that v is in the range of T . There are at most k distinct non-zero eigenvalues of T , since each such eigenvector is in the range (which is k-dimensional), and eigenvectors with distinct eigenvalues are linearly independent. Taking into account that 0 is a possible eigenvalues, one sees that there are at most k + 1 distinct eigenvalues of T . Solution. (5.14) We only need to check that this for p ( z ) = z m . In this case p ( STS- 1 ) = ( STS- 1 )( STS- 1 )( STS- 1 ) ... ( STS- 1 )( STS- 1 ) Rearranging the parenthesis we get the pairs S 1 S cancel, except for the S on the left and the S- 1 on the right. We get p ( STS- 1 ) = STTT ...TS- 1 = ST m S- 1 ....
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