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275A-Solutions-2

# 275A-Solutions-2 - ECE 275A Homework 2 Solutions 1 ECE 174...

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ECE 275A Homework # 2 Solutions 1. ECE 174 Midterm solutions are in a separate file. 2. Let V and W be independent (disjoint) subspaces of a vector space X , such that X = V + W . 1 Consider any two vectors x = x 1 + x 2 and y = y 1 + y 2 in X , where x 1 , y 1 ∈ V and x 2 , y 2 ∈ W give the unique decomposition of x and y along the companion subspaces V and W respectively. Let P = P V|W denote the projection operator which projects X onto V along W . Then P x = x 1 and P y = y 1 . For any two scalars α and β we have αx + βy = α ( x 1 + x 2 ) + β ( y 1 + y 2 ) = ( αx 1 + βy 1 ) + ( αx 2 + βy 2 ) ∈ X , with ( αx 1 + βy 1 ) ∈ V and ( αx 2 + βy 2 ) ∈ W since V and W are vector subspaces. Therefore P ( αx + βy ) = αx 1 + βy 1 = αPx + βPy , showing that a (possibly non-orthogonal) projection operator is linear. 3. Recall the the columns of [ V W ] are a basis for X iff they form a spanning set of vectors for X which are also linearly independent. Also recall that a linear operator P is a projection operator iff it is idempotent, P = P 2 . (a) The complementary subspace condition X = V ⊕W and the fact that the columns of V and and the columns of W each respectively forms a basis for V and W implies that for each x ∈ X there exists a v = V α ∈ V and a w = ∈ W such that x = v + w = V α + = [ V W ] ( α β ) showing that the columns of [ V W ] span X . The complementary subspace condition X = V ⊕ W implies that V and W are disjoint, which is true iff V ∩ W = { 0 } . Now suppose that the columns of V and W taken together are not linearly independent. Then there exists α ̸ = 0 and β ̸ = 0 such that 0 = V α + . (Recall that the columns of V are a linearly independent set, as are the columns of W .) This yields V α = ̸ = 0. But V α ∈ V and ∈ W , implying that 0 ̸ = V α ∈ V ∩ W which contradicts the assumption that V and W are disjoint, V ∩ W = { 0 } . Therefore the assumption that the columns of V and W taken together are not linearly independent must be incorrect. 1 I.e., let V and W be companion subspaces of X . 1

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(b) Note that to show that a linear operator is a projection operator onto V we need to show both that it is idempotent (i.e., that it is indeed a projector ) and that its range is V (i.e., that it is a projector onto V ). Let n be the dimension of X . With the columns M = [ V W ] linearly independent, M is an n × n invertible matrix. Consider the matrix P = M ( I 0 0 0 ) M - 1 = [ V 0] M - 1 .
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