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Unformatted text preview: SOLUTION TO ASSIGNMENT 5  MATH 251, WINTER 2007 1. (A) Let W be a kdimensional subspace of F n . Prove that there are n k linear equations such that W is the solutions to that homogenous system. (B) Let W 1 = Span( { (1 , 1 , , 0) , (0 , 1 , 1 , 1) } ) and W 2 = Span( { (3 , 2 , 1 , 0) , (4 , 4 , 2 , 1) } ). Find a system of homogeneous linear equations such that W 1 is their solutions. The same for W 2 . Find then a basis for W 1 W 2 , and a basis for W 1 + W 2 . (Note that you are not allowed to make any assumption on the field over which these equations are given.) Proof. (1) Choose a basis { w 1 ,...,w k } for W and complete it to a basis { w 1 ,...,w n } for F n . There is a linear transformation T : F n F n k such that T ( w i ) = 0 for i = 1 ,...,k and T ( w k + i ) = e i for i = 1 ,...,n k . The kernel of T is then precisely W . Let A be the matrix representing T w.r.t. the standard basis. Then A defines a system of linear equations whose solution space is W . There are n k rows in A and so n k equations. The image of T is F n k and so rank c ( A ) = rank r ( A ) = n k and we even find that those equations are linearly independent. (2) Consider the two systems of linear equations: ( x 1 x 2 + x 3 = 0 x 3 + x 4 = 0 , ( x 1 x 2 x 3 2 x 4 = 0 x 1 3 x 3 2 x 4 = 0 . Note that W 1 (resp. W 2 ) is contained in the solutions to the first (resp. second) system of equations. On the other hand, the corresponding matrices 1 1 1 0 1 1 , 1 1 1 2 1 3 2 have both rank 2. Hence, the dimensions of the solutions to the homogeneous systems are 4 2 = 2. It follows that W 1 (resp. W 2 ) is equal to the solutions to the first (resp. second) system of equations. We may compute the intersection W 1 W 2 as the solutions to the common system of equations x 1 x 2 + x 3 = 0 x 3 + x 4 = 0 x 1 x 2 x 3 2 x 4 = 0 x 1 3 x 3 2 x 4 = 0 . One may solve it using the usual technique of row reduction. We find that the solutions are the span of ( 1 , 2 , 1 , 1). The dimension of W 1 + W 2 is dim( W 1 )+dim( W 2 ) dim( W 1 W 2 ) = 3. Therefore, it is enough to find 3 independent vectors in W 1 + W 2 . We take (1 , 1 , , 0) , (0 , 1 , 1 , 1) , (3 , 2 , 1 , 0). One can check that the vectors are independent either by noting the 3 3 subdeterminant consisting of the 3 last columns is nonzero, or by performing row reduction: 1 1 0 0 1 1 1 3 2 1 1 0 1 1 0 1 1 1 3 2 1 1 0 1 1 0 1 1 1 0 0 2 1 Whether 2 = 0 in the field or not, this matrix has rank 3....
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 Spring '08
 Goren
 Linear Equations, Equations

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