Foundations of Data Science

etween vertices in different blocks is equivalent to finding a vector x with coordinates
±1 of which half of its coordinates are +1 and half of which are –1 that minimizes
E cut = 1 4
∑
(i,j)∈E
(x i − x j ) 2
Let A be the adjacency matrix of G. Then
x T Ax = ∑ a ij x i x j = 2 ∑ x i x j
(
ij
edges
) ( number of edges
number of edges
= 2 ×
− 2 ×
(
within components
) (
between components
)
total number
number of edges
= 2 ×
− 4 ×
of edges
between components
)
Maximizing x T Ax over all x whose coordinates are ±1 and half of whose coordinates are
+1 is equivalent to minimizing the number of edges between components.
Since finding such an x is computational difficult, replace the integer condition on the
components of x and the condition that half of the components are positive and half of the
∑
components are negative with the conditions n ∑
x 2 i = 1 and n x i = 0. Then finding the
optimal x gives us the second eigenvalue since it is easy to see that the first eigenvector
Is along 1
x T Ax
λ 2 = max ∑
x⊥v 1 x
2
i
∑
Actually we should use
n n∑
x 2 i = n not x 2 i = 1. Thus nλ 2 must be greater than
( ) i=1 ( i=1 )
total number
number of edges
2 ×
− 4 ×
since the maximum is taken over
of edges
between components
a larger set of x. The fact that λ 2 gives us a bound on the minimum number of cross
edges is what makes it so important.
12.7.8 Distance between subspaces
Suppose S 1 and S 2 are two subspaces. Choose a basis of S 1 and arrange the basis
vectors as the columns of a matrix X 1 ; similarly choose a basis of S 2 and arrange the
basis vectors as the columns of a matrix X 2 . Note that S 1 and S 2 can have different
dimensions. Define the square of the distance between two subspaces by
i=1
i=1
dist 2 (S 1 , S 2 ) = dist 2 (X 1 , X 2 ) = ||X 1 − X 2 X T 2 X 1 || 2 F
Since X 1 − X 2 X T 2 X 1 and X 2 X T 2 X 1 are orthogonal
‖X 1 ‖ 2 F = ∥ ∥ X1 − X 2 X T 2 X 1
∥ ∥
2
F + ∥ ∥ X2 X T 2 X 1
∥ ∥
2
F
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