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This means that the sum of the squared coefficients will be equal to one for each PC.
Substituting Equation 6.11 into Equation 6.10 yields Var( Y j )
5 S , demonstrating that the
constraint has been met.
The next constraint is the requirement that principal component axes have covariances
of zero. This means that the axes must be orthogonal . More formally stated, this constraint
is the requirement that the dot product of any two axes must be zero. For the first two
PCs, the constraint is expressed as:
X A 1 i A 2 i
A j A 2 5
0
(6.12)
5
The general requirement that the products of corresponding coefficients must be zero
for any pair of PCs is expressed as:
A j A j 5
0
(6.13)
The requirements imposed by Equations 6.11 and 6.13 indicate that we are solving for
an orthonormal basis . A basis is the smallest number of vectors necessary to describe a vec-
tor space (a matrix). An orthogonal basis is one in which each vector is orthogonal to every
other, so that a change in the value of one does not necessarily imply a change in the value
of another
in other words, all the variables are independent, or have zero covariance
( Equation 6.13 ) in an orthogonal basis. An orthonormal basis is an orthogonal basis in
which each axis has the same unit length. This very particular kind of normality was
imposed by the first requirement ( Equation 6.11 ). In an orthonormal basis, a distance or
difference of one unit on one axis is equivalent to a difference of one unit on every other
axis; consecutive steps of one unit on any two axes would describe two sides of a square.
So far, we have defined important relationships among the values of A i :
There is an
infinite number of possible orthonormal bases that we could construct to describe the orig-
inal data. The third constraint imposed above defines the relationship of the new basis
vectors to the original vector space of the data. Specifically, this constraint is the require-
ment that the variance of PC1 is maximized, and that the variance of each subsequent
component is maximized within the first two constraints.
We begin with the variance of PC1. From Equation 6.10 we know that:
ðA 1 5 A 1 SA 1
Var
ðY 1 Þ 5
Var
(6.14)
The matrix S can be reduced to:
2
3
λ
1 00
0
?
4
5
0
λ
0
0
?
2
Λ 5
00
0
^^^& ^
000
λ
(6.15)
?
3
? λ
p
λ i is an eigenvalue , a number that is a solution of the characteristic equation :
where each
S 2 λ
i I 5
0
(6.16)
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