275A-Solutions-3

# 275A-Solutions-3 - ECE 275A Homework#3 Solutions 1 Proof...

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Unformatted text preview: ECE 275A Homework #3 Solutions 1. Proof of (a). Obviously Ax = 0 ⇒ ⟨ y,Ax ⟩ = 0 for all y . To show suﬃciency, note that if ⟨ y,Ax ⟩ = 0 for all y , then it must certainly be true for the particular value of y = Ax . This yields ∥ Ax ∥ 2 = 0 ⇒ Ax = 0. Proof of (b). Let A be m × n (i.e., be a mapping from an n-dimensional space to an m-dimensional space) and r ( A ) = dim R ( A ). Recall that r ( A ) = r ( A ∗ ). We have A is onto ⇔ r ( A ) = r ( A ∗ ) = m ⇔ dim N ( A ) = m − r ( A ∗ ) = 0 ⇔ A ∗ is one-to-one. Applying this result, we have that A ∗ is onto if and only if A ∗∗ = A is one-to-one. Proof of (c). We have, y ∈ N ( A ∗ ) ⇔ A ∗ y = 0 ⇔ ⟨ A ∗ y,x ⟩ = 0 , ∀ x (from part (a)) ⇔ ⟨ y,Ax ⟩ = 0 , ∀ x ⇔ y ∈ R ( A ) ⊥ . Proof of (d). Let A be m × n . Then A ∗ is n × m , A ∗ A is n × n , and AA ∗ is m × m . First note that it is obvious that N ( A ) ⊂ N ( A ∗ A ) as x ∈ N ( A ) ⇒ Ax = 0 ⇒ A ∗ Ax = 0 ⇒ x ∈ N ( A ∗ A ) . On the other hand, x ∈ N ( A ∗ A ) ⇔ A ∗ Ax = 0 ⇔ ⟨ ξ,A ∗ Ax ⟩ = 0 , ∀ ξ (from part (a)) ⇔ ⟨ Aξ,Ax ⟩ = 0 , ∀ ξ ⇒ ⟨ Ax,Ax ⟩ = ∥ Ax ∥ 2 = 0 (in particular take ξ = x ) ⇔ Ax = 0 ⇔ x ∈ N ( A ) showing that N ( A ∗ A ) ⊂ N ( A ). It is therefore the case that indeed N ( A ∗ A ) = N ( A ) thereby ensuring that ν ( A ) = ν ( A ∗ A ). Thus A ∗ A is invertible (has rank n ) if and only 1 if A is one-to-one (has rank n ). (And which is true if and only if A ∗ is onto from part (b) above.) If we apply the above result to the n × m operator M = A ∗ , we obtain N ( AA ∗ ) = N ( A ∗ ). Thus AA ∗ is invertible if and only if A ∗ is one-to-one which is true if and only if A is onto, from part (b) above. Proof of (e). Most generally, let us assume that the space, X , is a vector space (not necessarily a normed or inner product space). Consider two (complete) disjoint subspaces, V ∩ W = { } such that X = V + W . We call two such subspaces V and W complementary subspaces. It is the case that every vector, x , in X can be uniquely represented as x = v + w where v ∈ V ⊂ X and w ∈ W ⊂ X . 1 Furthermore, the ‘component’ vectors v and w are linearly independent. Note that v and w remain well defined vectors in X . The projection of X onto V along a complementary space W , is defined by the mapping, x = v + w 7→ v , for all x ∈ X . Because the decomposition of x into components along V and W is unique for every x in X , this mapping is obviously unique. Regardless of what we call this mapping, say P or Q , the mapping always has the same domain, codomain, and value v = Px = Qx for all x in the domain X . By the definition of a function, operator, or mapping, if any two mappings P and Q have the same domain and codomain and Px = Qx for all x then it must be the case that P = Q ....
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## This note was uploaded on 01/14/2011 for the course ECE 210a taught by Professor Chandrasekara during the Fall '08 term at UCSB.

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275A-Solutions-3 - ECE 275A Homework#3 Solutions 1 Proof...

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