assig3_solutions

# assig3_solutions - MATH235 Fall 2007 Assignment 3...

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MATH235 Fall 2007 Assignment 3. Solutions. (32 marks total) 1. Let A be an nxn matrix. Suppose that λ 1 ,...,λ k are the (distinct) eigenvalues of A and that their multiplicites are m λ 1 ,...,m λ k respectively. Then, prove that det ( A ) = k Y i =1 λ m λ i i Solution. (4 marks) The characteristic polynomial of A is given by: p ( x ) = det ( A - xI ) = ( λ 1 - x ) m λ 1 ( λ 2 - x ) m λ 2 ··· ( λ k - x ) m λ k By letting x = 0 we get det ( A ) = λ m λ 1 1 λ m λ 2 2 λ m λ k k 2. Let A be a 2x2 matrix. Prove that the characteristic polynomial of A is λ 2 - trace ( A ) λ + det ( A ) Solution. (4 marks) Let A = ± a 11 a 12 a 21 a 22 . Then the characteristic polynomial of A is given by p ( λ ) = det ( A - λI ) = ( a 11 - λ )( a 22 - λ ) - a 21 a 12 = λ 2 - ( a 11 + a 22 ) λ + ( a 11 a 22 - a 21 a 12 ) = λ 2 - trace ( A ) λ + det ( A ) 3. Let A be a 5x5 matrix. Suppose we are given the following facts about A : trace ( A ) = 10, 2 is an eigenvalue of A with multiplicity 3, and A is not invertible. Prove that this information completely determines the characteristic polynomial of A , and state the characteristic polynomial of A . Solution. (4 marks) A is not invertible if and only if it has determinant zero. It follows from question 1 that det ( A ) = 0 if and only if at least one of the eigenvalues is zero. Thus we already know four out ﬁve eigenvalues of A : λ 1 = 2 2 = 2 3 = 2 4 = 0 1

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So we need to ﬁnd the last one. Using the fact that the trace of A is equal to the sum of the eigenvalues of A (counted with multiplicities), we get λ 1 + λ 2 + λ 3 + λ 4 + λ 5 = 10
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assig3_solutions - MATH235 Fall 2007 Assignment 3...

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