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# prac04s-1 - THE UNIVERSITY OF SYDNEY PURE MATHEMATICS...

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Unformatted text preview: THE UNIVERSITY OF SYDNEY PURE MATHEMATICS Linear Mathematics 2012 Practice session 4 — Solutions 1. Find a basis of the subspace V = braceleftBigparenleftBig w x y z parenrightBig ∈ R 4 vextendsingle vextendsingle vextendsingle 2 x- y = z- 3 w bracerightBig of R 4 . Solution A vector parenleftBig w x y z parenrightBig ∈ R 4 belongs to V if and only if 2 x- y = z- 3 w or, equivalently, if y = 2 x- z + 3 w . Thus, an arbitrary element parenleftBig w x y z parenrightBig of V can be written in the form parenleftBig w x y z parenrightBig = parenleftBig w x 2 x- z +3 w z parenrightBig = w parenleftBig 1 3 parenrightBig + x parenleftBig 1 2 parenrightBig + z parenleftBig- 1 1 parenrightBig . Therefore, the vectors parenleftBig 1 3 parenrightBig , parenleftBig 1 2 parenrightBig and parenleftBig- 1 1 parenrightBig span V . We now test this set for linear indepen- dence. Suppose that a parenleftBig 1 3 parenrightBig + b parenleftBig 1 2 parenrightBig + c parenleftBig- 1 1 parenrightBig = parenleftBig parenrightBig , for some a, b, c ∈ R . The corresponding augmented matrix, and a row echelon form for this matrix, is 1 1 3 2- 1 1 Row reduce-----→ 1 1 1 . Hence, a = b = c = 0 so these three vectors are linearly independent. (Note that there are no free variables. The last line of the row echelon form says that this system of equations is consistent.) Therefore, V is a subspace of R 4 with basis braceleftBigparenleftBig 1 3 parenrightBig , parenleftBig 1 2 parenrightBig , parenleftBig- 1 1 parenrightBigbracerightBig . 2. Let X = braceleftBigparenleftBig 1 1 1 parenrightBig , parenleftBig 1 1 parenrightBig , parenleftBig 1 parenrightBig , parenleftBig 3 2 parenrightBigbracerightBig . a ) Show that X spans R 3 . b ) Explain why X is not a basis for R 3 . c ) Find a subset of X which is a basis for R 3 . Solution a ) X spans R 3 if there is a solution, for every parenleftBig x y z parenrightBig ∈ R 3 , to the equation parenleftBig x y z parenrightBig = a parenleftBig 1 1 1 parenrightBig + b parenleftBig 1 1 parenrightBig + c parenleftBig 1 parenrightBig + d parenleftBig 3 2 parenrightBig = parenleftBig 1 1 1 3 1 1 0 2 1 0 0 0 parenrightBigparenleftBig...
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