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# sol1 - ISyE 7682 FALL 2010 1st Homework Problem 0.1 Assume...

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ISyE 7682 FALL 2010 – 1st Homework Problem 0.1 Assume that L is a subspace and A ⊆ R n is an affine manifold. Show that: (a) for any x 0 ∈ ℜ n , the set L + x 0 is an affine manifold; (b) for any x 0 ∈ A , the set A − x 0 is a subspace which does not depend on x 0 . (c) A is a subspace if and only if 0 ∈ A . Solution . a) Let x, ˜ x ∈ L + x 0 and λ ∈ ℜ be given. Then, x x 0 and ˜ x x 0 are in L , and since L is a subspace, it follows that (1 λ )( x x 0 ) + λ x x 0 ) ∈ L , or equivalently (1 λ ) x + λ ˜ x ∈ L + x 0 . Hence, L + x 0 is an affine manifold. b) Recall that A is an affine manifold if and only if it contains every affine combination of its elements, that is, for every integer k 1: λ 1 , . . . , λ k R λ 1 + . . . + λ k = 1 S 1 ⊆ A , . . . , S k ⊆ A = λ 1 S 1 + . . . + λ k S k ⊆ A . (1) Using this implication, we conclude that A + x 1 x 0 ⊆ A for every x 1 ∈ A . This is equivalent to A − x 0 ⊆ A − x 1 for every x 1 ∈ A . Interchanging the role of x 0 and x 1 in this last inclusion, we see that the reverse inclusion also holds. Hence, A − x 0 = A − x 1 for every x 1 ∈ A , showing that the set A − x 0 does not depend on the choice of the point x 0 ∈ A . To show that A − x 0 is a subspace, we have to show that α [ A − x 0 ] + β [ A − x 0 ] ⊆ A − x 0 for any scalars α and β . Indeed, using implication (1) several times, we obtain α [ A − x 0 ] + β [ A − x 0 ] = bracketleftbig α A + (1 α ) x 0 bracketrightbig + bracketleftbig β A + (1 β ) x 0 bracketrightbig 2 x 0 A + A − 2 x 0 ⊆ A − x 0 . We have thus shown that A − x 0 is a subspace and the result follows. c) If A is a subspace then clearly 0 ∈ A . Now if 0 ∈ A then by (b), it follows that A = A − 0 is a subspace. Problem 0.2 Let A : n → ℜ m be a given map. Show that the following properties are equivalent: (a) there exists T ∈ ℜ m × n and b ∈ ℜ m such that A ( x ) = Tx + b for every x ∈ ℜ n ; (b) A ( λx + (1 λ ) y ) = λA ( x ) + (1 λ ) A ( y ) for every x, y ∈ ℜ n and λ ∈ ℜ . 1

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Solution . Suppose first that a) holds and let x, y ∈ ℜ n and λ ∈ ℜ be given. Then, A ( λx + (1 λ ) y ) = T ( λx + (1 λ ) y )) + b = λTx + (1 λ ) Ty + b = λ ( Tx + b ) + (1 λ )( Ty + b ) = λA ( x ) + (1 λ ) A ( y ) , showing that b) holds.
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