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Unformatted text preview: 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 = { } ” ⇐⇒ “ T ( Tα ) = ⇒ Tα = 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 = { } . 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 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 = 1 1 0 (b) Using the new ordered basis B = { (1 , 2);(1 , 1) } we have Te 1 = (2 , 1) = 1 e 1 + 1 e 2 ; Te 2 = ( 1 , 1) = 0 e 1 + ( 1) e 2 , i.e.[ T ] B = 1 1 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...
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This document was uploaded on 01/25/2012.
 Spring '09
 Vectors

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