9.4. DIMENSIONALITY REDUCTION
343
This sum simpliﬁes to
(4 −x)2 + (1 −x)2 + (3 −x)2 + (3 −x)2 + (2 −x)2
We want the value of x that minimizes the sum, so we take the derivative and
set that equal to 0, as:
−2 ×
(4 −x) + (1 −x) + (3 −x) + (3 −x) + (2 −x)

= 0
or −2 × (13 −5x) = 0, from which it follows that x = 2.6.


2.6
1
1
1
1
1
1
1
1
1


×

1
1
1
1
1
1
1
1
1
1

=


3.6
3.6
3.6
3.6
3.6
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2


Figure 9.12: The best value for u11 is found to be 2.6
Figure 9.12 shows U and V after u11 has been set to 2.6. Note that the sum
of the squares of the errors in the ﬁrst row has been reduced from 18 to 5.2, so
the total RMSE (ignoring average and square root) has been reduced from 75
to 62.2.


2.6
1
1
1
1
1
1
1
1
1


×

y
1
1
1
1
1
1
1
1
1

=


2.6y + 1
3.6
3.6
3.6
3.6
y + 1
2
2
2
2
y + 1
2
2
2
2
y + 1
2
2
2
2
y + 1
2
2
2
2


Figure 9.13: v11 becomes a variable y
Suppose our next entry to vary is v11. Let the value of this entry be y, as
suggested in Fig. 9.13. Only the ﬁrst column of the product is aﬀected by y, so
we need only to compute the sum of the squares of the diﬀerences between the
entries in the ﬁrst columns of M and UV . This sum is
5−(2.6y +1)
2 +
3−(y +1)
2 +
2−(y +1)
2 +
2−(y +1)
2 +
4−(y +1)
2
This expression simpliﬁes to
(4 −2.6y)2 + (2 −y)2 + (1 −y)2 + (1 −y)2 + (3 −y)2
As before, we ﬁnd the minimum value of this expression by diﬀerentiating and
equating to 0, as:
−2 ×
2.6(4 −2.6y) + (2 −y) + (1 −y) + (1 −y) + (3 −y)

= 0
