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554-HW-3q - 3 Let V be an n dimensional vectorspace Let T V...

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3) Let V be an n dimensional vectorspace. Let T : V V be a linear tranformation such that T ( V ) = null T . Show n is even. 3) Let V be a vectorspace and T : V V a linear transformation. Show that “ T ( V ) null T = { 0 } ⇐⇒ T ( ) = 0 = 0”. 3) Let V be a vectorspace and T a linear operator on V . If T 2 = 0 but T 6 = 0 what can we say about the relationship of the range of T to the nullspace of T ? 3) Let V be a vectorspace with dim V = n and let T be a linear operator on V such that rank T = rank T 2 . Show that T ( V ) null T = { 0 } . 3) Let T L ( F n , F n ), let A be the matrix of T in the standard ordered basis for F n , and let W be the subspace spanned by the column vectors of A . 3) Find a basis for the range and the null space of A = 1 2 1 0 1 1 - 1 3 4 3) T ( x, y ) = ( - y, x ) (a) The matrix for T in the standard ordered basis for R 2 is given by Te 1 = (0 , 1) = 0 e 1 + 1 e 2 ; Te 2 = ( - 1 , 0) = ( - 1) e 1 + 0 e 2 , i.e.[ T ] B = 0 1 - 1 0 (b) Using the new ordered basis B 0 = { (1 , 2); (1 , - 1) } we have Te 0 1 = (2 , 1) = 1 e 0 1 + 1 e 0 2 ; Te 0 2 = ( - 1 , 1) = 0 e 0 1 + ( - 1) e 0 2 , i.e.[ T ] B 0 = 1 1 0 - 1 (c) Prove that for every real number c the linear operator ( T - cI ) is invertible. ( T - cI )( x, y ) = T ( x, y ) - c ( x, y ) = ( - y, x ) - ( cx, cy ) = ( - cx - y, x - cy ) To compute [( T - cI )] B , ( T - cI ) e 1 = ( - c, 1) = ( - c ) e 1 + 1 e 2 ; ( T - cI ) e 2 = ( - 1 , - c ) = ( - 1) e 1 + ( - c ) e 2 , i.e.[( T - cI )] B = - c 1 - 1 - c . But the matrix M = 1 c 2 +1 - c - 1 1 - c can be seen to be the inverse of [( T - cI )] B by multiplying them out, so the linear operator corresponding to M must be the inverse of I . We know there exists a linear operator corresponding the the matrix M by theorem 11 from chapter 3. Since M is a left and right inverse to [( T - cI )] B we see that the corresponding
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