Unformatted text preview: E t ( b + te ) and solve for ˙ x (0): ˙ x (0) = ( A t A )1 [ A t ( eEx LS ) + E t ( bAx LS )] . Now deﬁne κ 2 ( A ) = k A kk ( A t A )1 A t k . (generalizing the condition number for Ax = b to rectangular matrices). Then with r ≡ ( bAx LS ), k ˙ x (0) k k x LS k ≤ t n k ( A t A )1 A t ( eE x LS ) k k x LS k + k ( A t A )1 E t r k k x LS k o + O( t 2 ) ≤ κ 2 ( A )( k te k k A kk x LS k + k tE k k A k ) + k ( A t A )1 kk A kk tE kk r k k A kk x LS k + O( t 2 ) Let θ be the angle between b and ColSp( A )). Since r ⊥ Ax LS , k Ax LS k = p k b k 2 k r k 2 = k b k cos ( θ ) ≤ k A kk x LS k . Thus, the condition number of (1) is roughly κ ( A ) ± 1 + 1  cos ( θ )  κ ( A ) k r k k b k ² . This signals trouble when κ ( A ) is large, and becomes quadratic in κ ( A ) if b points away from ColSp( A )....
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 Fall '10
 MARK
 Matrices, Rank, Invertible matrix, Linear least squares

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