## Attachment 'sheet05_02.m'

Download```
1 function sheet05_02
2 % Principal Component Analysis
3
4 % simple data set
5 X = randn(500, 2) * diag([1, 2]) * rotmat(0.3*pi);
6 X = X + repmat([3, -4], 500, 1);
7
8 plot(X(:,1), X(:,2), '.');
9 axis equal
10 grid
11
12 % first PCA without centering
13 figure(1)
14 p = pca(X, 0)
15 plot_pca(X, p);
16 title('PCA without centering')
17
18 % now, with proper centering
19 figure(2)
20 p = pca(X)
21 plot_pca(X, p);
22 title('PCA with centering')
23
24 % and now, we add further, far away point
25 figure(3)
26 X = [X; -50, -50];
27 p = pca(X);
28 plot_pca(X, p);
29 title('PCA is not very stable with respect to outliers')
30
31 % how about other data?
32 figure(4)
33 phi = rand(500, 1) * 4 * pi;
34 X = [phi, cos(phi) + (phi.^3)/1e2] + 0.5*randn(500, 2);
35
36 p = pca(X);
37 plot_pca(X, p);
38
39 function R = rotmat(phi)
40 R = [cos(phi), sin(phi); -sin(phi), cos(phi)];
41
42 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
43 % Fill in your solutions below
44
45 % 1. Compute the principal components. If the second argument is given and
46 % non-zero, properly center the data. Otherwise, don't center. Return a
47 % struct with fields "U", "D", and "M" with
48 %
49 % U is the matrix whose columns are the principal components
50 % D is a vector with the variances
51 % M is the mean vector
52 function result = pca(X, center)
53 if nargin == 1
54 center = 1;
55 end
56 % ...
57 result.U = U;
58 result.D = D;
59 result.M = M;
60
61 % 2. Plot the principal components for the PCA. First, plot the data as
62 % points. Then, draw an ellipsis whose radii are given as 2*sqrt(D) such
63 % that the ellipsis contains the data more or less. Add a grid and make sure
64 % axes are sized equally.
65 function plot_pca(X, p)
66 % ...
```

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