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Let A be a real m x n matrix and let A¹ = VEUT, where A = UEVT where U is m × m, V is nxn, both unitary, Σ is m x n and Σ is n x m have the form «N with σ₁ 20₂2 Show: and St = 20₂ > 0. A¹b is a least-squares solution to A = b. Previously we used â = (ATA)-¹ATb for our least-squares solution, but we had the restriction that the columns of the "data" matrix A were independent, this guarantees that NS(A) = NS(ATA): {0}. It is not hard to see that A¹ = (ATA)-¹AT if A has linear independent columns. =

Let A be a real m x n matrix and let A¹ = VEUT, where A = UEVT where U is m × m, V is nxn, both unitary, Σ is m x n and Σ is n x m have the form «N with σ₁ 20₂2 Show: and St = 20₂ > 0. A¹b is a least-squares solution to A = b. Previously we used â = (ATA)-¹ATb for our least-squares solution, but we had the restriction that the columns of the "data" matrix A were independent, this guarantees that NS(A) = NS(ATA): {0}. It is not hard to see that A¹ = (ATA)-¹AT if A has linear independent columns. =

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Let A be a real m x n matrix and let A = VEUT, where A = UEVT where U is m × m, V is nxn, both unitary, Σ is m x n and Σ is nx m have the form Σ = σ1 or 0 and ΣΤ – = with σ₁ ≥ 0₂ ≥ ... >σ, > 0. Show: â = A¹b is a least-squares solution to Ax = b. Previously we used â = (ATA)-¹ ATb for our least-squares solution, but we had the restriction that the columns of the "data" matrix A were independent, this guarantees that NS(A) = NS(ATA) = {0}. It is not hard to see that A¹ = (ATA)-¹AT if A has linear independent columns.
Transcribed Image Text:Let A be a real m x n matrix and let A = VEUT, where A = UEVT where U is m × m, V is nxn, both unitary, Σ is m x n and Σ is nx m have the form Σ = σ1 or 0 and ΣΤ – = with σ₁ ≥ 0₂ ≥ ... >σ, > 0. Show: â = A¹b is a least-squares solution to Ax = b. Previously we used â = (ATA)-¹ ATb for our least-squares solution, but we had the restriction that the columns of the "data" matrix A were independent, this guarantees that NS(A) = NS(ATA) = {0}. It is not hard to see that A¹ = (ATA)-¹AT if A has linear independent columns.
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