# Practice Problems 6 Solutions

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Information about Practice Problems 6 Solutions

Published on March 6, 2014

Author: anhtuantran509

Source: slideshare.net

Practice Problems 6: Solutions 1. (a) Let A be an m × n matrix (m ≥ n) with linearly independent columns: Full QR factorization: A = QR • Q is an m × m orthogonal matrix • R is an m × n upper triangular matrix ˜˜ Reduced QR factorization: A = QR ˜ ˜ ˜ • Q is an m × n matrix with QT Q = I ˜ • R is an n × n upper triangular matrix (b) H = A AT A −1 ˜˜ ˜˜ ˜˜ AT = QR (QR)T (QR) ˜˜ ˜ ˜ ˜˜ = QR RT QT QR −1 −1 ˜˜ (QR)T ˜ ˜ RT QT −1 ˜˜ ˜ ˜ ˜ ˜ = QR RT R RT QT ˜ ˜ ˜˜˜ ˜ = QRR−1 R−T RT QT T ˜˜ = QQ =I ˜ ˜ Only two matrices are inverted: R and RT . 2. Consider the matrix Q = cos θ sin θ − sin θ . cos θ An eigenvector x must be in the same direction as Qx, that is, Qx = λx. However, multiplying Q by a real vector x rotates the vector in the xy-plane through the angle θ, so x cannot be in the same direction as Qx. 3. (a) Let X be an m × n matrix: Full singular value factorization X = U ΣV T • U is an m × m orthogonal matrix • Σ is an m × n diagonal matrix • V is an n × n orthogonal matrix ˜ β = XT X −1 XT y = U ΣV T T U ΣV T T −1 −1 U ΣV T = V ΣT U T U ΣV V ΣT U T y T T −1 T T VΣ U y = V Σ ΣV −1 T T =V Σ Σ V V ΣU T y −1 T =V Σ Σ ΣU T y T y

Note: While it may be tempting to say that because Σ is a diagonal matrix Σ = ΣT , keep in mind that Σ is an m × n matrix. If m = n, then Σ is not square, so it can’t be symmetric. Furthermore, −1 −1 when Σ isn’t square, it cannot be invertible, so ΣT Σ = Σ−1 ΣT . ˜˜˜ (b) Reduced singular value factorization X = U ΣV T ˜ ˜ ˜ • U is an m × n matrix with U T U = I ˜ • Σ is an n × n diagonal matrix ˜ • V is an n × n orthogonal matrix ˜ β = XT X −1 XT y = ˜˜˜ U ΣV T T −1 ˜˜˜ U ΣV T ˜˜˜ ˜˜˜ = V ΣU T U ΣV T −1 ˜˜˜ U ΣV T T y ˜˜˜ V ΣU T y −1 ˜˜ ˜ ˜˜˜ = V Σ2 V T V ΣU T y ˜˜ ˜ ˜˜˜ = V Σ−2 V T V ΣU T y ˜˜ ˜ = V Σ−1 U T y ˜ ˜ ˜ Note: Σ−1 is actually Σ† , the pseudoinverse of Σ. This matrix is created by replacing all the ˜ and then transposing the resulting matrix. If Σ has full rank, then ˜ non-zero diagonal entries of Σ ˜ ˜ † = Σ−1 . In addition, the expression for β contains the pseudoinverse of ˜ it is invertible and Σ −1 ˜˜ ˜ matrix X: X † = X T X X T = V Σ† U T . 4. ˆ Σ= 1 ˜T ˜ m−1 X X = 1 ˜˜ T ˜˜ m−1 (QR) (QR) d2 = xT Σ−1 xi = xT ˜u ˆ ˜ ˜i i 1 ˜T ˜ m−1 R R = −1 xi ˜ 1 ˜T ˜T ˜ ˜ m−1 R Q QR = 1 ˜T ˜ m−1 R R

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