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section3 - 9/9/11 EE221A Section 3 1 1.1 Functions, linear...

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EE221A Section 3 9/9/11 1 Functions, linear maps 1.1 Solutions to linear equations Theorem. (range and nullspace of linear operators) [LN3 p. 4] Consider A : U V with ( U,F ) , ( V,F ) linear spaces. Let b V . Then: a) A ( u ) = b has at least one solution ⇐⇒ b ∈ R ( A ) b) If b ∈ R ( A ) then i) A ( u ) = b has a unique solution ⇐⇒ N ( A ) = { θ U } ii) Let x 0 be such that A ( x 0 ) = b . Then A ( x ) = b ⇐⇒ x - x 0 ∈ N ( A ) Proof. a) = : Let x be one solution. Then b = A ( x ) for some x U , thus b ∈ R ( A ) by the definition of R ( A ) . =: There exists an x U such that A ( x ) = b . Therefore there is at least one solution x . b) i) = : Call the unique solution x 0 . Now suppose that there exists a nonzero vector x n in N ( A ) , and let z = x 0 + x n 6 = x 0 . Then by the linearity of A , A ( z ) = A ( x 0 + x n ) = A ( x 0 ) + A ( x n ) = b + 0 = b . But this says that there is another solution z distinct from x 0 which is unique. →← (thus there are no nonzero vectors in
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section3 - 9/9/11 EE221A Section 3 1 1.1 Functions, linear...

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