Homework 9 Solutions - Math 110 Spring 2008 Solutions to...

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Math 110 Spring 2008 Solutions to HW9 Section 5.1 20. Let f ( t ) = det( A - tI ) = ( - 1) n t n + ··· + a 1 t + a 0 . Then a 0 = f (0) = det( A - 0 I ) = det( A ). ± 21. a. Following the suggestion in the book, we prove this by induction. We may as well take the base case to be n = 2, in which case the proposition is easily verified. Now assume it is true for n - 1, and take a matrix A M n × n ( F ). Computing det( A - tI ) along the first row yields f ( t ) = ( A 11 - t ) det( ˜ A 11 - tI n - 1 ) + p ( t ) where p ( t ) is the sum of the (1 ,j )-cofactors for j 2. Clearly p ( t ) will be a polynomial of degree at most n - 2, as in any of these cofactors you have a product of at most ( n - 2) factors of t . (We could make this a little more rigorous with another proof by induction, but I believe this is plausible enough.) Now by induction det( ˜ A 11 - tI n - 1 ) = Q n i =2 ( A ii - t ) + r ( t ), where r ( t ) is a polynomial of degree at most n - 3. Plugging this back into the expression above yields f ( t ) = n Y i =1 ( A ii - t ) + ( A 11 - t ) r ( t ) + p ( t ) = n Y i =1 ( A ii - t ) + q ( t ) where
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This note was uploaded on 06/15/2009 for the course MATH 110 taught by Professor Gurevitch during the Spring '08 term at University of California, Berkeley.

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Homework 9 Solutions - Math 110 Spring 2008 Solutions to...

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