Practice Problems 6 Solutions
Upcoming SlideShare
Loading in...5
×

Like this? Share it with your network

Share

Practice Problems 6 Solutions

  • 57 views
Uploaded on

 

  • Full Name Full Name Comment goes here.
    Are you sure you want to
    Your message goes here
    Be the first to comment
    Be the first to like this
No Downloads

Views

Total Views
57
On Slideshare
55
From Embeds
2
Number of Embeds
1

Actions

Shares
Downloads
0
Comments
0
Likes
0

Embeds 2

http://www.slideee.com 2

Report content

Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
    No notes for slide

Transcript

  • 1. 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
  • 2. 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