2.4b - 2.4.15 : Suppose that T ( ) is a basis for W . Then...

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1 Math 110 Homework 4 Partial Solutions If you have any questions about these solutions, or about any problem not solved, please ask via email or in office hours, etc. 2.4.4 ( AB )( B - 1 A - 1 ) = A ( BB - 1 ) A - 1 = AA - 1 = I n . Additionally, ( B - 1 A - 1 ) AB = BB - 1 = I n . Thus AB is invertible, and ( AB ) - 1 = B - 1 A - 1 . 2.4.5 A t ( A - 1 ) t = ( A - 1 A ) t = I t n = I n . Also, ( A - 1 ) t A t = ( AA - 1 ) t = ( I n ) t = I n . Thus A t is invertible with inverse ( A - 1 ) t . 2.4.6 Suppose that AB = 0. Since A is invertible, we have A - 1 . Multiplying both sides of the equation by A - 1 gives B = A - 1 AB = A - 1 0 = 0.
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Unformatted text preview: 2.4.15 : Suppose that T ( ) is a basis for W . Then R ( T ) = span( T ( )) = W , so T is onto. Since dim( V ) = dim( W ), T is also one-to-one and thus an isomorphism. : Suppose that T is an isomorphism. Then, since T is one-to-one, we have proved (previous hw) that T ( ) is linearly independent. Since T ( ) has dim( W ) elements, it must be a basis for W ....
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