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Unformatted text preview: Math 235: Linear Algebra HW 6 Problem 3.2.3 Problem 3.2.3 Let T be the linear operator on R 3 defined by T ( x 1 ,x 2 ,x 3 ) = (3 x 1 ,x 1 x 2 , 2 x 1 + x 2 + x 3 ) Is T invertible? If so, find a rule for T 1 like the one which defines T . Yes it is invertible, and the inverse is given by T 1 ( x 1 ,x 2 ,x 3 ) = (1 / 3 x 1 , 1 / 3 x 1 x 2 , x 1 ) Problem 3.2.4 For the linear operator T of Excercise 3, prove that ( T 2 I ) ( T 3 I ) = 0 We start by computing ( T 3 I )( x,y,z ): ( T 3 I )( x,y,z ) = (3 x,x y, 2 x + y + z ) (3 x, 3 y, 3 z ) = (0 ,x 4 y, 2 x + y 2 z ) We now plug this into ( T 2 I ) : ( T 2 I ) (0 ,x 4 y, 2 x + y 2 z ) = T ( T (0 ,x 4 y, 2 x + y 2 z )) (0 ,x 4 y, 2 x + y 2 z ) = T (0 , x + 4 y, 3 x 3 y 2 z ) (0 ,x 4 y, 2 x + y 2 z ) = (0 ,x 4 y, 2 x + y 2 z ) (0 ,x 4 y, 2 x + y 2 z ) = (0 , , 0). Problem 3.2.8 Let V be a vectors space over a field F and T a linear operator on V . If T 2 = 0, whac can you say about the relation of the range of T to the null space of T ? Give an example of a linear operator T on R 2 such that T 2 = 0 but T 6 = 0. Since T 2 = 0 we know that T ( T ( v )) = 0 for each v V . So this means that T ( v ) is in the null space of V . Clearly T ( v ) is also in the range of V . This shows us that the range of V is contained in the null space of V . Let T be the linear operator on R 2 defined b T ( x,y ) = ( y, 0). T 2 ( a,b ) = T ( T ( a,b )) = T ( b, 0) = (0 , 0). So T 2 = 0. Also T (1 , 1) = (1 , 0) so T 6 = 0. Page 1 of 5 Math 235: Linear Algebra HW 6 Problem 3.2.11 Problem 3.2.11 Let V be a finitedimensional vector space and let T be a linear operator on V . Suppose that rank ( T 2 ) =rank( T )....
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 Spring '10
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