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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 righthand 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.
 Spring '07
 Loveys
 Linear Algebra, Algebra, Real Numbers

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