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hw1_sol

# hw1_sol - Department of Electrical and Computer Engineering...

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Department of Electrical and Computer Engineering University of Maryland, College Park Fall 2011 ENEE 660 Homework Set 1 Solution Nuno C. Martins 1. Exercise 1.2 Solution. U = { x R 5 | [x] 1 = 3[x] 2 , [x] 3 = 7[x] 4 } Choose b 1 = 3 1 0 0 0 , b 2 = 0 0 7 1 0 , b 3 = 0 0 0 0 1 . First, observe that { b 1 , b 2 , b 3 } are linearly independent. Claim. span { b 1 , b 2 , b 3 } = U U span { b 1 , b 2 , b 3 } x U, [ x ] 1 = 3[ x ] 2 , [ x ] 3 = 7[ x ] 4 , thus, x = 3 [ x ] 2 [ x ] 2 7 [ x ] 4 [ x ] 4 [ x ] 5 = [ x ] 2 b 1 + [ x ] 4 b 2 + [ x ] 5 b 3 span { b 1 , b 2 , b 3 } span { b 1 , b 2 , b 3 } ⊆ U x span { b 1 , b 2 , b 3 } , α 1 , α 2 , α 3 s.t. x = α 1 b 1 + α 2 b 2 + α 3 b 3 = x = 3 α 1 α 1 7 α 2 α 2 α 3 U From above, span { b 1 , b 2 , b 3 } = U , and { b 1 , b 2 , b 3 } are linearly independent. Therefore, { b 1 , b 2 , b 3 } are one basis for U . (Ming Tse P. Laiu, 2011) 2. Exercise 1.3 Solution. By dimension theorem, dim ( U + W ) = dim ( U ) + dim ( W ) - dim ( U W ) dim ( U W ) = 0 Since U , W are subspaces, { 0 } ∈ U , { 0 } ∈ W ⇒ { 0 } ∈ U W U W = { 0 } (Ming Tse P. Laiu, 2011)

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3. Exercise 1.4 Solution. The statement is NOT true. Consider this counterexample below: U 1 , U 2 , U 3 R 2 U 1 = span { (0 , 1) } , U 2 = span { (1 , 0) } , U 3 = span { (1 , 1) } (Ming Tse P. Laiu, 2011) 4. Exercise 1.5 Solution. Denote ~x = ( x 1 , x 2 ) , ~ y = ( y 1 , y 2 ). From definition of triangle function, sin θ 1 = x 2 p x 2 1 + x 2 2 cos θ 1 = x 1 p x 2 1 + x 2 2 sin θ 2 = y 2 p y 2 1 + y 2 2 cos θ 2 = y 2 p y 2 1 + y 2 2 Therefore, cos θ = cos( θ 2 - θ 1 ) = cos θ 2 cos θ 1 + sin θ 2 sin θ 1 = x 2 y 2 + x 1 y 1 p x 2 1 + x 2 2 · p y 2 1 + y 2 2 But, < x, y > = x 1 y 1 + x 2 y 2 k x k = q x 2 1 + x 2 2 k y k = q y 2 1 + y 2 2 Substitute the corresponding term, we could easily write cos θ = < x, y > k x kk y k which is exactly < x, y > = k x kk y k cos θ (Xiangnan Weng, 2011)
5. Exercise 1.6

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hw1_sol - Department of Electrical and Computer Engineering...

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