svd - ⊥ Applying this result to A t gives Range A t...

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The Singular Value Decomposition Let A R m × n . Then there exist orthogonal matrices U R m × m , V R n × n , and a diagonal matrix of singular values Σ = diag( σ 1 , σ 2 , . . . , σ p ), where p = min ( m, n ) and σ 1 σ 2 ≥ ··· σ p 0, such that A = U Σ V t . So what? Recall the two fundamental subspaces associated with any matrix (or linear transformation) A : The range of A is the subspace of R m defined as Range( A ) = { y R m : y = Ax, for some x R n } , and the nullspace of A is the subspace of R n defined as Nullsp( A ) = { x R n : Ax = 0 } . The rank of a matrix A is the dimension of the range of A , and the nullity of A is the dimension of the nullspace of A . One of the fundamental properties of an m × n matrix A is rank( A ) + nullity( A ) = n. In an inner product space, this result should be seen as a corollary to another fundamental result which says that the range of A is the orthogonal complement of the nullspace of A t : Range( A ) = [Nullsp( A t )]
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Unformatted text preview: ⊥ . Applying this result to A t gives Range( A t ) = [Nullsp( A )] ⊥ . Back to the SVD: If r = rank( A ), then σ r > 0 and σ r +1 = 0. If we write U = [ U 1 , U 2 ], and V = [ V 1 , V 2 ], where U 1 ∈ R m × r and V 1 ∈ R n × r , then (the columns of) U 1 form an orthonormal basis (O.B.) for Range( A ), U 2 an O.B. for Nullsp( A t ), V 1 an O.B. for Range( A t ), and V 2 an O.B. for Nullsp( A ). It’s all there in the SVD. And more. A matrix of rank s which best approximates A in the 2-norm is A s ≡ s X j =1 σ j u j v t j . This implies that the singular values tell us about how close A is to matrices of a given rank (e.g. how close to singular is this square matrix?), and helps us to quantify the uncertainties in our data....
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This note was uploaded on 12/18/2010 for the course PHYS 5073 taught by Professor Mark during the Fall '10 term at Arkansas.

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