MA353_HW6_sol

MA353_HW6_sol - U is also a nonzero linear transformation,...

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Purdue University MA 353: Linear Algebra II with Applications Homework 6, due Feb. 26, Some Solutions Sec. 2.2 #13: Let V and W be vector spaces, and let T and U be nonzero linear transformations from V to W . If R ( T ) R ( U ) = { 0 } , prove that { T,U } is a linearly independent subset of L ( V,W ). Proof. Assume aT + bU = 0 for a,b F , where 0 is the zero transformation. We must show that a = b = 0. Now aT + bU = 0 gives aT ( v )+ bU ( v ) = 0 W for all v V . For the sake of obtaining a contradiction, let us assume that a 6 = 0. Then T ( v ) = - b a U ( v ) . Then since R ( U ) is a subspace and so in particular closed under scalar multiplication, we have - b a U ( v ) R ( U ). So T ( v ) R ( U ). But of course, T ( v ) R ( T ), and hence T ( v ) R ( T ) R ( U ). But R ( T ) R ( U ) = { 0 W } and so T ( v ) = 0 W . Since v V is arbitrary, this shows T = 0, which is a contradiction because T is a nonzero linear transformation. So a = 0. Then aT + bU = 0 gives bU = 0. But since
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Unformatted text preview: U is also a nonzero linear transformation, we must have b = 0. Sec. 2.3 #11: Let V be a vector space, and let T : V V be linear. Prove that T 2 = T if and only if R ( T ) N ( T ). Proof. = Assume T 2 = T . We must show R ( T ) N ( T ). So let x R ( T ). Then x = T ( y ) for some y V . Then T ( x ) = T ( T ( y )) = T 2 ( y ) = T ( y ) = 0 , which shows that x N ( T ). = Conversely, assume R ( T ) N ( T ). Let v V be arbitrary. We must show T 2 ( v ) = 0. But T 2 ( v ) = T ( T ( v )), and T ( v ) R ( T ). So R ( T ) N ( T ) implies T ( v ) N ( T ). So T ( T ( v )) = 0, i.e. T 2 ( v ) = 0. 1...
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