Assignment1Solution

Assignment1Solution - MATH 223, Linear Algebra Fall 2010...

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Unformatted text preview: MATH 223, Linear Algebra Fall 2010 Solutions to Assignment 1 1. Let z =- 6- 2 i and w = 5 + 4 i . Find ¯ z , ¯ w , z + w , z- w , z · w and z w (all in the form a + bi with a and b real numbers). Find the absolute values of each of these 6 numbers. Solution: ¯ z =- 6 + 2 i , ¯ w = 5- 4 i , z + w =- 1 + 2 i , z- w =- 11- 6 i , z · w = (- 6- 2 i )(5 + 4 i ) =- 30- 10 i- 24 i- 8 i 2 =- 22- 34 i and z w =- 6- 2 i 5 + 4 i = (- 6- 2 i )(5- 4 i ) (5 + 4 i )(5- 4 i ) =- 30- 10 i + 24 i + 8 i 2 5 2 + 4 2 =- 38 41 + 14 41 i. | ¯ z | = p (- 6) 2 + (- 2) 2 = √ 36 + 4 = √ 40. This answer is perfectly fine, but you should know that it is also 2 √ 10. | ¯ w | = √ 5 2 + 4 2 = √ 41. | z + w | = √ 5. | z- w | = √ 157. | z · w | = √ 1640. | z w | = r (- 38 41 ) 2 + ( 14 41 ) 2 = 1 41 √ 1640 . 2. Prove that for any complex numbers z 1 , z 2 and z 3 , z 1 · ( z 2 + z 3 ) = ( z 1 · z 2 )+ ( z 1 · z 3 ). You may use any properties about multiplication and addition of real numbers. Solution: Say z j = a j + ib j for j = 1 , 2 , 3 and each a j and b j is real. The left hand side of the equation to be shown is ( a 1 + b 1 i ) · [( a 2 + b 2 i ) + ( a 3 + b 3 i )] = ( a 1 + b 1 i )[( a 2 + a 3 ) + ( b 2 + b 3 ) i ] = [ a 1 ( a 2 + a 3 )- b 1 ( b 2 + b 3 )] + [ b 1 ( a 2 + a 3 ) + a 1 ( b 2 + b 3 )] i. The right-hand side is ( a 1 + b 1 i )( a 2 + b 2 i )+( a 1 + b 1 i )( a 3 + b 3 i ) = [( a 1 a 2- b 1 b 2 )+( a 1 b 2 + b 1 a 2 ) i ]+[( a 1 a 3- b 1 b 3 )+( a 1 b 3 + b 1 a 3 ) i ] = [ a 1 a 2- b 1 b 2 + a 1 a 3- b 1 b 3 ] + [ a 1 b 2 + b 1 a 2 + a 1 b 3 + b 1 a 3 ] i. Using distributivity and commutativity and associativity of addition in the real numbers, we see that a 1 ( a 2 + a 3 )- b 1 ( b 2 + b 3 ) = a 1 a 2- b 1 b 2 + a 1 a 3- b 1 b 3 and b 1 ( a 2 + a 3 ) + a 1 ( b 2 + b 3 ) = a 1 b 2 + b 1 a 2 + a 1 b 3 + b 1 a 3 , so indeed the two sides are equal. 3. If A is a matrix over the complex numbers, we let ¯ A be the most obvi- ous thing — it is obtained from A by replacing each of its entries by its conjugate. Supposing that A · B is defined, show that A · B = ¯ A · ¯ B. Solution: First notice that for any complex numbers z 1 and z 2 , we have z 1 + z 2 = ¯ z 1 + ¯ z 2 and z 1 z 2 = ¯ z 1 ¯ z 2 . (The first is pretty clear. For the 1 second, say z 1 = a 1 + b 1 i and z 2 = a 2 + b 2 i with each a j and b j real, so z 1 z 2 = ( a 1 a 2- b 1 b 2 ) + ( a 1 b 2 + b 1 a 2 ) i and its conjugate is ( a 1 a 2- b 1 b 2 )- ( a 1 b 2 + b 1 a 2 ) i . Now multiply ( a 1- b 1 i ) by ( a 2- b 2 i ) and see that it’s the same. Do it — don’t just take my word for it.) Say A is ‘ × m and B is m × n . Then all three of AB , AB and ¯ A · ¯ B are ‘ × n . For any j , k with 1 ≤ j ≤ ‘ and 1 ≤ k ≤ n , the ( j,k )-entry of AB is ∑ m p =1 a j,p b p,k , where a j,p is the ( j,p )-entry of A and b p,k is the ( p,k )-entry of B . So the ( j,k )-entry of AB is the conjugate of this; using the first paragraph repeatedly, we see that this is...
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This note was uploaded on 10/03/2010 for the course MATH 223 taught by Professor Loveys during the Spring '07 term at McGill.

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Assignment1Solution - MATH 223, Linear Algebra Fall 2010...

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