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Midterm1_practice_sols

Midterm1_practice_sols - Solutions to Practice Midterm 1 By...

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Solutions to Practice Midterm 1 By Chu-Wee Lim Author’s note: I’ve attempted to be as complete as possible in my solution, hence the wordiness. In the midterm, you can probably miss one or two small steps. (Question 1): True or False (with justification): If A and B are n -by- n matrices with entries from F , then AB = 0 if and only BA = 0. Solution. False. Counterexample: if A = ( 0 0 0 1 ) and B = ( 0 1 0 0 ), then AB = ( 0 0 0 0 ) but BA = ( 0 1 0 0 ). (Question 2): True or False (with justification): If A and B are n -by- n matrices with entries from R , then AB = 7 I n if and only BA = 7 I n . Solution. True. First, a general statement. We claim that a linear map T : R n R n is an isomorphism iff it is one-to-one. Now, is obvious. For the converse, suppose T is one-to-one. Then n = dim( R n ) = dim N ( T ) + dim R ( T ). Since T is one-to-one, dim N ( T ) = 0, and so dim R ( T ) = n . Since R ( T ) R n , this implies R ( T ) = R n . Now look at L A and L B as linear maps R n R n . By symmetry, it suﬃces to prove = . Since L A L B = 7 · 1 V is one-to-one, L A is also one-to-one. By the above paragraph, L A is an isomorphism and thus has an inverse U , i.e. UL A = L A U = 1 V . So, UL A L B = L B = 7 U = L B = U = 1 7 L B . Thus, 1 7 L B is the inverse of L A and 1 7 L B L A = 1 V = L B L A = 7 · 1 V . (Question 3): True or False (with justification): If x, y V and a, b F , then ax + by = 0 if and only if x is a scalar multiple of y , or y is a scalar multiple of x .

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Midterm1_practice_sols - Solutions to Practice Midterm 1 By...

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