hw4sol - HOMEWORK 4 - SOLUTIONS 1. Easy, just apply the...

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HOMEWORK 4 - SOLUTIONS 1. Easy, just apply the techniques from class, namely cofactor expansion along any row, cofactor expansion down any column, and row reduction. The answer is - 48 - 34 i . 2. (GRADED - MEDIUM) (a) tr( CAC - 1 ) = tr( C - 1 CA ) = tr( A ). (b) A simple computation: det( CAC - 1 ) = det( C 1 )det( A )det( C ) = det( A )det( C )det( C - 1 ) = det( A )det( CC - 1 ) = det( A )det( I ) = det( A ) (c) First observe: Ax = 0 AC - 1 ( Cx ) = 0 CAC - 1 ( Cx ) = 0 AC - 1 ( Cx ) = C - 1 0 Ax = 0 Thus x ker( L A ) iff Cx ker( L CAC - 1 ). Since C is invertible, L C is an isomor- phism, and so it’s not hard to see that L C ± ker( L A ) : ker( L A ) ker( L CAC - 1 ) is an isomorphism. Thus unravelling the definitions gives us: null( A ) = null( L A ) = dim(ker( L A )) = dim(ker( L CAC - 1 )) = null( L CAC - 1 ) = null( CAC - 1 ) (d) This follows from part (c): rank( A ) = rank( L A ) = n - null( L A ) = n - null( L CAC - 1 ) = rank( L CAC - 1 ) = rank( CAC - 1 )
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3. (a) No. Let T = id R and U = - id R . Then clearly T and U are isomorphisms R R but T + U is the zero map, which is clearly not an isomorphism. (b) Yes. ( αT )( x ) = 0 α ( T ( x )) = 0 T ( x ) = 0 x = 0 where the first impli- cation follows by definition of the scalar multiple of a transformation, the second by the fact that α 6 = 0, and the last by the fact that T is an isomorphism. Since T is an isomorphism, it’s a map between vector spaces of the same dimension, hence αT is a map between spaces of the same dimension. Since the only vector it maps to zero is zero, αT is an isomorphism. 4. (a) It’s easy to see that UT has an inverse, in fact the formula for it is given to us: ( UT )( T - 1 U - 1 ) = UTT - 1 U - 1 = U id W U - 1 = UU - 1 = id X ( T - 1 U - 1 )( UT ) = T - 1 U - 1 UT = T - 1 id W T = T - 1 T = id V (b) Do a similar computation to the one above. (c)
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hw4sol - HOMEWORK 4 - SOLUTIONS 1. Easy, just apply the...

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