ps3_solution_key_revised - TA: Yankun Wang Economics 617...

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TA: Yankun Wang Economics 617 Problem Set 3 Solution Key 1. [Matrix Multiplication and Transpose] Suppose A and B are symmetric n × n matrices. (a) Under what conditions on A and B is the matrix AB also symmetric? Explain. So lu t ion : Wha tw eknowi s A = A 0 ,B = B 0 , andwewan tasu cient condition such that ( AB ) 0 = AB. Notice that ( AB ) 0 = B 0 A 0 = BA, thus as long as we have BA = AB, AB is symmetric. Notice that this problem is just asking for a su cient condition, so you might come up with di f erent conditions: that’s OK. (b) Under the conditions identi f ed by you in (i), is the matrix BA also symmetric? Explain. Solution: Yes. Because ( BA ) 0 = A 0 B 0 = AB = BA, where the last step follows from our condition. 2. [Non-Singular and Invertible Matrices] (a) If A is an n × n non-singular matrix, we know that there exists an n × n matrix B such that BA = AB = I. Show that this B is unique. Proof: Suppose there exists a matrix C such that CA = AC = I, and C is not the same as B. Then: B = BI = B ( AC )=( BA ) C = IC = C. Thus we get contradiction. (b) Suppose A is an n × n non-singular matrix, and B is an n × n matrix satisfying AB = I. Does it follow that BA = I ?Exp la in . Proof: Yes. Prof. Mitra actually proved this in class. Or you can look at Page 15 and 16 in the Lectures Notes. Notice that why we can f nd a X such at AX = I, ( X is like B in our problem), and why we can f nd a Y such at YA = I, and f nally, how we prove that X = Y. (c) Suppose A is an n × n non-singular matrix, and B is an n
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Solution: Yes. The reason is: AB = A × £ B 1 B 2 ...B n ¤ =[00 ... 0] , so we must have AB i =0 ,i =1 , ..., n. Since all the column of A are linearly independent, B i =0 ,i =1 , ..., n. Thus B is the null matrix. 3. [Proof of the Rank Theorem]
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ps3_solution_key_revised - TA: Yankun Wang Economics 617...

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