{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

MA353_HW6_sol

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

This preview shows page 1. Sign up to view the full content.

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
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online