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# tut02s - THE UNIVERSITY OF SYDNEY PURE MATHEMATICS Linear...

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Unformatted text preview: THE UNIVERSITY OF SYDNEY PURE MATHEMATICS Linear Mathematics 2012 Tutorial 2 — Solutions 1. Prove that the set S = braceleftBigparenleftBig x y parenrightBig ∈ R 2 | y = 4 x bracerightBig is a subspace of R 2 . Solution S is clearly non-empty, since parenleftBig parenrightBig satisfies the equation y = 4 x and is therefore in S . Now suppose that parenleftBig a 1 a 2 parenrightBig and parenleftBig b 1 b 2 parenrightBig both belong to S . That is, a 2 = 4 a 1 and b 2 = 4 b 1 . Then a 2 + b 2 = 4( a 1 + b 1 ) , and so parenleftBig a 1 + b 1 a 2 + b 2 parenrightBig ∈ S , and S is closed under addition. If k ∈ R , then ka 2 = k (4 a 1 ) = 4( ka 1 ) , and so parenleftBig ka 1 ka 2 parenrightBig ∈ S , and S is closed under scalar multiplication. 2. For each of the sets of vectors X ⊂ R 3 below, explicitly describe all of the vectors in the subspace Span( X ) of R 3 . a ) X = { } . b ) X = braceleftBigparenleftBig 1 1 1 parenrightBigbracerightBig . c ) X = braceleftBigparenleftBig 1 1 1 parenrightBig , parenleftBig 2 2 2 parenrightBigbracerightBig . d ) X = braceleftBigparenleftBig 1 1 1 parenrightBig , parenleftBig 1 parenrightBigbracerightBig . e ) X = braceleftBigparenleftBig 1 1 1 parenrightBig , parenleftBig 1 parenrightBig , parenleftBig 1 1 parenrightBigbracerightBig . Solution Recall that if X = { v 1 , . . . , v k } then Span( X ) = { λ 1 v 1 + ··· + λ k v k | λ 1 , . . . , λ k ∈ R } . a ) Span( parenleftBig parenrightBig ) = braceleftBigparenleftBig parenrightBigbracerightBig . b ) Span parenleftBigparenleftBig 1 1 1 parenrightBigparenrightBig = braceleftBigparenleftBig r r r parenrightBigvextendsingle vextendsingle vextendsingle r ∈ R bracerightBig . c ) As parenleftBig 2 2 2 parenrightBig = 2 parenleftBig 1 1 1 parenrightBig , or by part (b), Span parenleftBigparenleftBig 1 1 1 parenrightBig , parenleftBig 2 2 2 parenrightBigparenrightBig = braceleftBigparenleftBig r r r parenrightBigvextendsingle vextendsingle vextendsingle r ∈ R bracerightBig . d ) Span parenleftBigparenleftBig 1 1 1 parenrightBig , parenleftBig 1 parenrightBigparenrightBig = braceleftBigparenleftBig r + s r r parenrightBigvextendsingle vextendsingle vextendsingle r, s ∈ R bracerightBig . We can also describe Span( X ) = braceleftBigparenleftBig x y z parenrightBigvextendsingle vextendsingle vextendsingle y = z bracerightBig geometrically as the plane in R 3 with equation y − z = 0 . To see this observe that a vector parenleftBig x y z parenrightBig ∈ Span( X ) if and only if parenleftBig x y z parenrightBig = r parenleftBig 1 1 1 parenrightBig + s parenleftBig 1 parenrightBig , for some r, s ∈ R . Using Gaussian elimination we can simplify this to parenleftBigg 1 1 x 1 y 1 z parenrightBigg R 2 := R 2- R 1 R 3 := R 3- R 1 −−−−−−−→ parenleftBigg 1 1 x − 1 y − x − 1 z − x parenrightBigg R 2 := R 2 ×- 1 R 3 := R 3 + R 2 −−−−−−−→ parenleftBigg 1 1 x 1 x −...
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tut02s - THE UNIVERSITY OF SYDNEY PURE MATHEMATICS Linear...

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