Two-way tables: biplots associated with correspondence analysis - Understanding Biplots - page 297

2

In order to maximize ρ we first differentiate (7.26) with respect to z 1 and z 2 giving the

derivatives

2 ( z 1 Xz 2 )( Xz 2 )( z 1 Rz 1 ) − 1 ( z 2 Cz 2 ) − 1 − 2 ( z 1 Rz 1 ) − 2 ( Rz 1 )( z 1 Xz 2 )

2

( z 2 Cz 2 ) − 1

and

2 ( z 2 X z 1 )( Xz 1 )( z 1 Rz 1 ) − 1 ( z 2 Cz 2 ) − 1 − 2 ( z 2 Cz 2 ) − 2 ( Cz 2 )( z 2 X z 1 )

2

( z 1 Rz 1 ) − 1 .

Equating these derivatives to zero and rearranging, we obtain

( z 1 Rz 1 ) Xz 2 = ( z 1 Xz 2 ) Rz 1

(7.27)

and

z 2 Cz 2 ) X z 1 = ( z 1 Xz 2 )

( Cz 2 . (7.28)

On normalizing z 1 Rz 1 = 1and z 2 Cz 2 = 1, it follows from (7.27) and (7.28) that (7.26)
is maximized when

Rz 1 ,
X z 1 = ρ Cz 2 ,

Xz 2 = ρ

(7.29)

the familiar equations for canonical correlation. These may be rewritten as

) C 1 / 2 z 2 = ρ R 1 / 2 z 1 ,
( C − 1 / 2 X R − 1 / 2

( R − 1 / 2 XC − 1 / 2

(7.30)

) R 1 / 2 z 1 = ρ C 1 / 2 z 2 .

But

( R − 1 / 2 XC − 1 / 2 ) C 1 / 2 1 = R − 1 / 2 X1 = 1 R 1 / 2 1

and

( C − 1 / 2 X R − 1 / 2 ) R 1 / 2 1 = C − 1 / 2 X 1 = 1 C 1 / 2 1 ,

from which it is evident that R 1 / 2 1 and C 1 / 2 1 are a singular vector pair of R − 1 / 2 XC − 1 / 2
with unity as corresponding singular value. Since the elements of R 1 / 2 1 , C 1 / 2 1 and
R − 1 / 2 XC − 1 / 2 are all nonnegative, it follows from the Frobenius theorem (see Gower
and Hand, 1996, Appendix A.11) that all other singular values of R − 1 / 2 XC − 1 / 2 must be

smaller than unity. Therefore, it follows that

= R 1 / 2 11 C 1 / 2

R − 1 / 2 XC − 1 / 2

/ n + ρ R 1 / 2 z 1 z 2 C 1 / 2

+ ...

(7.31)

are the first terms in the SVD of R − 1 / 2 XC − 1 / 2 , provided R 1 / 2 z 1 and C 1 / 2 z 2 are normal-
ized so that z 1 Rz 1 = 1and z 2 Cz 2 = 1.
The above development is in terms of uncentred variables, but correlation
requires centring. The changes are straightforward, replacing G 1 G 2 , G 1 G 1 , G 2 G 2 by
G 1 ( I − N ) G 2 , G 1 ( I − N ) G 1 , G 2 ( I − N ) G 2 ,where N is the n × n centring matrix with
1 / n in every position. Because 1 G 1 = 1 R and 1 G 2 = 1 C ,wehavethat
G 1 ( I − N ) G 2 = X − R11 C / n = X − E ,
G 1 ( I − N ) G 1 = R − R11 R / n ,
G 2 ( I − N ) G 2 = C − C11 C / n ,

(7.32)