#### Transcript SVD and PCA

```SVD and PCA
COS 323, Spring 05
SVD and PCA
• Principal Components Analysis (PCA):
approximating a high-dimensional data set
with a lower-dimensional subspace
* *
Second principal component
* * First principal component
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* * *
* * *
** * *
Original axes
*
*
* * *
** *
*
Data points
SVD and PCA
• Data matrix with points as rows, take SVD
– Subtract out mean (“whitening”)
• Columns of Vk are principal components
• Value of wi gives importance of each
component
PCA on Faces: “Eigenfaces”
Average
face
First principal component
Other
components
For all except average,
“gray” = 0,
“white” > 0,
“black” < 0
Uses of PCA
• Compression: each new image can be
approximated by projection onto first few
principal components
• Recognition: for a new image, project onto first
few principal components, match feature vectors
PCA for Relighting
• Images under different illumination
[Matusik & McMillan]
PCA for Relighting
• Images under different illumination
• Most variation captured
by first 5 principal
components – can
re-illuminate by
combining only
a few images
[Matusik & McMillan]
PCA for DNA Microarrays
• Measure gene activation under different conditions
[Troyanskaya]
PCA for DNA Microarrays
• Measure gene activation under different conditions
[Troyanskaya]
PCA for DNA Microarrays
• PCA shows patterns of correlated activation
– Genes with same pattern might have similar function
[Wall et al.]
PCA for DNA Microarrays
• PCA shows patterns of correlated activation
– Genes with same pattern might have similar function
[Wall et al.]
Multidimensional Scaling
• In some experiments, can only measure
similarity or dissimilarity
– e.g., is response to stimuli similar or different?
• Want to recover absolute positions in kdimensional space
Multidimensional Scaling
• Example: given pairwise distances between cities
– Want to recover locations
[Pellacini et al.]
Euclidean MDS
• Formally, let’s say we have n  n matrix D
consisting of squared distances dij = (xi – xj)2
• Want to recover n  d matrix X of positions
in d-dimensional space

0

 ( x1  x2 ) 2
D
2
 ( x1  x3 )


( x1  x2 ) 2
( x1  x3 ) 2
0
( x2  x3 ) 2
( x2  x3 ) 2
0
 ( x1 ) 


X   ( x2 ) 












Euclidean MDS
• Observe that
dij2  ( xi  x j )2  xi  2 xi x j  x j
2
2
• Strategy: convert matrix D of dij2 into
matrix B of xixj
– “Centered” distance matrix
– B = XXT
Euclidean MDS
• Centering:
– Sum of row i of D = sum of column i of D =
si 
2
2
2
d

x

2
x
x

x
 ij  i i j j
j
j
 nxi2  2 xi  x j   x 2j
j
j
– Sum of all entries in D =


s   si  2n x  2  xi 
i
i
 i 
2
i
2
Euclidean MDS
• Choose xi = 0
– Solution will have average position at origin
– Then,
si  nxi2   x 2j , s  2n x 2j
j
j
dij2  1n si  1n s j  n12 s  2 xi x j
• So, to get B:
–
–
–
–
compute row (or column) sums
compute sum of sums
apply above formula to each entry of D
Divide by –2
Euclidean MDS
• Now have B, want to factor into XXT
• If X is n  d, B must have rank d
• Take SVD, set all but top d singular values to 0
–
–
–
–
Eliminate corresponding columns of U and V
Have B3=U3W3V3T
B is square and symmetric, so U = V
Take X = U3 times square root of W3
Multidimensional Scaling
• Result (d = 2):
[Pellacini et al.]
Multidimensional Scaling
• Caveat: actual axes, center not necessarily
what you want (can’t recover them!)
• This is “classical” or “Euclidean” MDS [Torgerson 52]
– Distance matrix assumed to be actual Euclidean distance
• More sophisticated versions available
– “Non-metric MDS”: not Euclidean distance,
sometimes just relative distances
– “Weighted MDS”: account for observer bias
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