Finding Bases and Dimensions of Images and Kernels

Finding Bases and Dimensions of Images and Kernels -...

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Finding Bases and Dimensions of Images and Kernels Let T : R 5 R 4 be the linear transformation defined by T ( ~x ) = A~x , where A = 1 5 4 3 2 1 6 6 6 6 1 7 8 10 12 1 6 6 7 8 . We want to find the basis and dimension for both the image and kernel of T . First we compute the reduced row–echelon form of A . rref ( A ) = 1 0 - 6 0 6 0 1 2 0 - 2 0 0 0 1 2 0 0 0 0 0 To find a basis for the kernel, we want to find all vectors ~x satisfying A~x = ~ 0. This amounts to solving the system x 1 - 6 x 3 +6 x 5 = 0 x 2 +2 x 3 - 2 x 5 = 0 x 4 +2 x 5 = 0 If we rename the free variables x 3 = s,x 5 = t , the solutions are x 1 x 2 x 3 x 4 x 5 = 6 s - 6 t - 2 s + 2 t s - 2 t t = s 6 - 2 1 0 0 + t - 6 2 0 - 2 1 So, the kernel of A is Span 6
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This test prep was uploaded on 09/26/2007 for the course MATH 2940 taught by Professor Hui during the Fall '05 term at Cornell.

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Finding Bases and Dimensions of Images and Kernels -...

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