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<math>Y^tY = (N\times p)(p\times N) = N \times N</math> | <math>Y^tY = (N\times p)(p\times N) = N \times N</math> | ||
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| + | since N<<P, the size of <math>Y^tY</math> is smaller | ||
b) <math>YY^t = U \Sigma V^t V \Sigma U^t = U\Sigma^2 U^t</math> and <math>(YY^t)^t = (U\Sigma^2 U^t)^t = U\Sigma^2 U^t = YY^t </math>. Therefore, <math>YY^t</math> is symmetric | b) <math>YY^t = U \Sigma V^t V \Sigma U^t = U\Sigma^2 U^t</math> and <math>(YY^t)^t = (U\Sigma^2 U^t)^t = U\Sigma^2 U^t = YY^t </math>. Therefore, <math>YY^t</math> is symmetric | ||
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d)<math>U\Sigma V^t=Y \Longrightarrow U = Y(\Sigma V^t)^{-1} = Y(D^{1/2}T^t)^{-1}</math> | d)<math>U\Sigma V^t=Y \Longrightarrow U = Y(\Sigma V^t)^{-1} = Y(D^{1/2}T^t)^{-1}</math> | ||
| − | e)<math>YY^t = U\Sigma U^t = E\ | + | e)<math>YY^t = U\Sigma^2 U^t = E\Lambda E^t</math> |
<math>E = U = Y(D^{1/2}T^t)^{-1}</math> | <math>E = U = Y(D^{1/2}T^t)^{-1}</math> | ||
f)The name we give to the column of U is eigenimages | f)The name we give to the column of U is eigenimages | ||
Latest revision as of 19:45, 9 July 2019
Communication, Networking, Signal and Image Processing (CS)
Question 5: Image Processing
August 2016 (Published in Jul 2019)
Problem 1
a) $ \lambda_n^c=\lambda_n^b-\lambda_n^d $
b) $ G_n = \frac{d\lambda_n^c}{dx}=-\mu (x,y_0+n\Delta d)\lambda_n^c $
c) $ \lambda_n = \lambda_n^c e^{-\int_{0}^{x}\mu(t)dt} \Longrightarrow \hat{P}_n = \int_{0}^{x}\mu(t)dt= -ln(\frac{\lambda_n}{\lambda_n^c}) = -ln(\frac{\lambda_n}{\lambda_n^b-\lambda_n^d}) $
d) $ \hat{P}_n = \int_{0}^{T_n}\mu_0dt = \mu_0 T_n $
A straight line with slope $ \mu_0 $
Problem 2
a)Since U is $ p \times N $, $ \Sigma $ and V are $ N \times N $]
$ Y = U \Sigma V^t = p \times N $
$ YY^t = (p\times N)(N\times p) = p \times p $
$ Y^tY = (N\times p)(p\times N) = N \times N $
since N<<P, the size of $ Y^tY $ is smaller
b) $ YY^t = U \Sigma V^t V \Sigma U^t = U\Sigma^2 U^t $ and $ (YY^t)^t = (U\Sigma^2 U^t)^t = U\Sigma^2 U^t = YY^t $. Therefore, $ YY^t $ is symmetric
For an arbitrary x, $ x^tYY^tx = x^t U\Sigma \Sigma U^t x=(\Sigma U^t x)^t\Sigma U^t x=\|\Sigma U^t x\|^2 \geq 0 $. Therefore, $ YY^t $ is positive semi-definite.
Similarly, $ Y^tY = V \Sigma^2 V^t $ and $ (Y^tY)^t = (V \Sigma^2 V^t)^t = V \Sigma^2 V^t = Y^tY $, $ Y^tY $ is symmetric
For an arbitrary x, $ x^tY^tYx = x^t V\Sigma \Sigma V^t x=(\Sigma V^t x)^t\Sigma V^t x=\|\Sigma V^t x\|^2 \geq 0 $. Therefore, $ Y^tY $ is positive semi-definite.
c) From B, obtain that $ Y^tY = V\Sigma^2 V^t $ while $ Y^tY = TDT^t $. $ V = T $ and $ \Sigma = D^{1/2} $
d)$ U\Sigma V^t=Y \Longrightarrow U = Y(\Sigma V^t)^{-1} = Y(D^{1/2}T^t)^{-1} $
e)$ YY^t = U\Sigma^2 U^t = E\Lambda E^t $
$ E = U = Y(D^{1/2}T^t)^{-1} $
f)The name we give to the column of U is eigenimages
