ass3 - MATH 223, Linear Algebra Fall, 2007 Solutions to...

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MATH 223, Linear Algebra Fall, 2007 Solutions to Assignment 3 1. Suppose that A and B are invertible matrices of the same size. (a) Show that B ( A - 1 + B - 1 ) = ( A + B ) A - 1 . Solution: Easy. By distributivity, the left-hand side is BA - 1 + BB - 1 = BA - 1 + I and the right-hand side is AA - 1 + BA - 1 = I + BA - 1 . By commutativity of +, the sides are equal. (b) Suppose that A + B is also invertible. Show that A - 1 + B - 1 is invertible, and its inverse is A ( A + B ) - 1 B . Solution: Being a product of invertible matrices, A ( A + B ) - 1 B is invertible, and its inverse is B - 1 (( A + B ) - 1 ) - 1 A - 1 . This equals B - 1 ( A + B ) A - 1 = ( B - 1 A + I ) A - 1 = B - 1 AA - 1 + IA - 1 = B - 1 + A - 1 = A - 1 + B - 1 . So A - 1 + B - 1 is invertible and its inverse is A ( A + B ) - 1 B . 2. (a) Let i = 0 1 0 0 - 1 0 0 0 0 0 0 1 0 0 - 1 0 , a matrix over the reals. Show that i 2 = - I , that the set C = { aI + bi : a,b ∈ R} is a subspace of M 4 ( R ) (regarded as a vector space over R ), C is closed under matrix multi- plication, and that every nonzero matrix in C has an inverse, which is also in C . (We can identify C with the field of complex numbers in a pretty obvious manner. Reminder: M 4 ( R )) is the collection of all 4 × 4 matrices with real entries.) Solution: I’ll skip the direct and easy calculation that i 2 = - I . The zero matrix is in C since we can take a = b = 0. Now if we have two elements of C , say they are a 1 I + b 1 i and a 2 I + b 2 i (with a 1 , b 1 , a 2 and b 2 real); their sum is ( a 1 + a 2 ) I + ( b 1 + b 2 ) i by basic properties of matrix algebra, and this is in C . So C is closed under addition. If α ∈ R and aI + bi C , then α ( aI + bi ) = ( αa ) I + ( αb ) i , also an element of C , so C is closed under scalar multiplication. So it’s a subspace of M 4 ( R ). Next, suppose that a 1 I + b 1 i and a 2 I + b 2 i are elements of C . Using the basic rules, we get ( a 1 I + b 1 i )( a 2 I + b 2 i ) = ( a 1 I + b 1 i )( a 2 I ) + ( a 1 I + b 1 i )( b 2 i ) = ( a 1 I )( a 2 I )+( b 1 i )( a 2 I )+( a 1 I )( b 2 i )+( b 1 i )( b 2 i ) = a 1 a 2 I + b 1 a 2 i + a 1 b 2 i + b 1 b 2 i 2 = ( a 1 a 2 - b 1 b 2 ) I + ( b 1 a 2 + a 1 b 2 ) i, 1
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just as in the complex numbers; we use i 2 = - I here. Finally, if aI + bi C is not zero, then of course at least one of a and b
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ass3 - MATH 223, Linear Algebra Fall, 2007 Solutions to...

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