5 5 let t x 1 x 2 x 3 x 4 x 1 x 2 x 3 x 4 x 1 x 2 x 3

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5. Let T ( x 1 , x 2 , x 3 , x 4 ) = ( x 1 + x 2 - x 3 - x 4 , x 1 - x 2 - x 3 + x 4 ). Show that T is an onto linear transformation from R 4 to R 2 . Is T one to one? Solution. Observe that T is given by the matrix transformation T ( ~ X ) = A ~ X , where A = 1 1 - 1 - 1 1 - 1 - 1 1 Thus T is obviously a linear transformation. We compute the row reduced echelon from of A . 1 1 - 1 - 1 1 - 1 - 1 1 - 1 1 1 1 - 1 - 1 0 - 2 0 2 - 1 - 2 - 2 0 2 0 0 - 2 0 2 - 1 2 - 1 2 1 0 - 1 0 0 1 0 - 1 The last entry of the above tableau is the row reduced echelon form of A . The first two columns of the row reduced echelon form from the 2 by 2 identity matrix and are thus obviously linearly independent. Therefor the first two columns of A are linearly independent. Consequently T is onto. We also see that T ( x 1 , x 2 , x 3 , x 4 ) = (0 , 0) if and only if x 1 - x 3 = 0 and x 2 - x 4 = 0. There are several solutions to this system of equations for example ( x 1 , x 2 , x 3 , x 4 ) = (0 , 0 , 0 , 0) and ( x 1 , x 2 , x 3 , x 4 ) = (1 , 1 , 1 , 1). Consequently T is not one to one. 6
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