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Homework 4 - Solution to 2.3, problem 16

Homework 4 - Solution to 2.3, problem 16 - Section 2.3#16...

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Section 2.3, #16: Let V be a finite-dimensional vector space, and let T : V V be linear. (a) If rank( T ) = rank( T 2 ), prove that R ( T ) N ( T ) = { 0 } . Deduce that V = R ( T ) N ( T ). Proof. Assume that rank( T ) = rank( T 2 ), i.e., that dim( R ( T )) = dim( R ( T 2 )). We want to show first that R ( T ) N ( T ) = { 0 } . The proof takes a cou- ple of steps. First, we note the following two facts: Claim 1 . R ( T 2 ) R ( T ). Proof of claim 1. Let v R ( T 2 ). Then v = T 2 ( x ) for some x V , so v = T ( T ( x )), so v R ( T ). Thus R ( T 2 ) R ( T ). Claim 2 . N ( T ) N ( T 2 ). Proof of claim 2. Let v N ( T ). Then T ( v ) = 0, so T 2 ( v ) = T ( T ( v )) = T (0) = 0, so v N ( T 2 ). Thus N ( T ) N ( T 2 ). Now recall (from Theorem 1.11) that if W is a subspace of V , then dim( W ) dim( V ), and if dim( W ) = dim( V ) then W = V . Applying this to the subspace R ( T 2 ) of R ( T ), since we are assuming dim( R ( T )) = dim( R ( T 2 )), we have that R ( T ) = R ( T 2 ). But also, by the Rank- Nullity Theorem, we have rank( T )+nullity( T ) = dim( V ). We can also apply this to the linear map T 2 , so we have rank( T 2 ) + nullity( T 2 ) = dim( V ). Thus rank( T ) + nullity( T ) = rank( T 2 ) + nullity( T 2 ) . Since rank( T ) = rank( T 2 ), we must also have nullity( T ) = nullity( T 2 ). Combining this fact with claim 2 and Theorem 1.11, since N ( T ) N ( T 2 ) and dim( N ( T )) = dim( N ( T 2 )), we must have N ( T ) = N ( T 2 ) . (*)
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