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 $
```

I obtained the same answer as Sige did. However, I am not sure if we did part (d) correctly. -YC

## 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

I obtained the same answer as Sige did. -YC