m235notes8 - Notes 8: Kernel, Image, Subspace Fix a matrix...

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Fix a matrix M of size n × m. Question. For what b R n does the system of equations Fx = b have a solution? We can ask this same question in another way. The matrix M induces a linear map: M : R m R n . By definition the image of F is the set of b R n so that the equation F ( x ) = b has a solution. This set of all such b has structure. Theorem 1 . Let F be a linear map R m -→ R n . 1. Let y 1 ,y 2 be elements in the image of F . Then y 1 + y 2 is also in the image of F . 2. Let y be an element in the image of F and λ R . Then λy is also in the image of F . Proof. Since each of y 1 ,y 2 is in the image of F , there are elements x 1 ,x 2 in the domain of F , that is, they are elements of R m , so that F ( x 1 ) = y 1 , F ( x 2 ) = y 2 . Now F ( x 1 + x 2 ) = ( our function is linear ) F ( x 1 ) + F ( x 2 ) = y 1 + y 2 . This proves the first statement. Since y is in the image of F , there is an element x in the domain of F so that F ( x ) = y . Then F ( λx ) = ( our function is linear ) λF ( x ) = λy. This says that λy is in the image of F . Definition 1. Let S be a subset of R n . We say that S is a subspace provided If u,v are both elements of S , then so is u + v. If u S and if λ is any real number, then λ · u S also. The above theorem says that the image of a linear map is a subspace. Indeed it is a fact that any subspace of R n is the image of a linear map. Subspaces can appear in many other ways besides being images of linear maps. 1
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This note was uploaded on 11/22/2011 for the course MATH 235 taught by Professor Markman during the Fall '08 term at UMass (Amherst).

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m235notes8 - Notes 8: Kernel, Image, Subspace Fix a matrix...

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