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20105ee205A_1_2010ee205A_HW8_sol

# 20105ee205A_1_2010ee205A_HW8_sol - Chapter 10 Canonical...

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Chapter 10 Canonical Forms 1. Show that if a triangular matrix is normal, then it must be diagonal. Answer 10.1 We prove this by induction on n . Let T C n × n be normal and, without loss of generality, assume it is upper triangular. For n =1 , the matrix T is simply a complex number, so it is diagonal. Suppose T is diagonal for 1 n < k . Consider the case n = k . Since T is upper triangular, it has the form ± α v H 0 S ² for some α C , v C n - 1 , and upper triangular S C ( n - 1) × ( n - 1) . Since T is normal, we have that TT H = ± α v H 0 S ²± α 0 vS H ² = ± | α | 2 + v H vv H S H Sv SS H ² = T H T = ± α 0 H α v H 0 S ² = ± | α | 2 α v H α v vv H + S H S ² . Thus | α | 2 + v H v = | α | 2 , which is true only if v =0 . Hence H = S H S , i.e., S is normal. But S is also upper triangular, so by the induc- tion hypothesis, S is diagonal. Therefore, T = ± α 0 0 S ² is diagonal. 43

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44 CHAPTER 10. CANONICAL FORMS Similarly, if T is lower triangular and normal, then it is diagonal. Therefore for all positive integers n , if T C n × n is triangular and normal, then T is diagonal. Note: This result can also be obtained directly by comparing diagonal elements of TT H and T H T . 2. Prove that if A IR n × n is normal, then N ( A )= N ( A T ). Answer 10.2 If x N ( A ) , then Ax =0 . Hence, A T Ax and x T A T Ax . Since A is normal, we then have x T AA T x . But this implies that ± A T x ± and hence A T x , i.e., x N ( A T ) .
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20105ee205A_1_2010ee205A_HW8_sol - Chapter 10 Canonical...

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