9.4. DIMENSIONALITY REDUCTION
345


2.6
1
1
1
1.178
1
1
1
1
1


×
 1.617
1
1
1
1
1
1
1
1
1

=


5.204
3.6
3.6
3.6
3.6
2.617
2
2
2
2
2.905
2.178
2.178
2.178
2.178
2.617
2
2
2
2
2.617
2
2
2
2


Figure 9.16: Replace z by 1.178
that M is an n-by-m utility matrix with some entries blank, while U and V
are matrices of dimensions n-by-d and d-by-m, for some d. We shall use mij,
uij, and vij for the entries in row i and column j of M, U, and V , respectively.
Also, let P = UV , and use pij for the element in row i and column j of the
product matrix P.
Suppose we want to vary urs and ﬁnd the value of this element that mini-
mizes the RMSE between M and UV . Note that urs aﬀects only the elements
in row r of the product P = UV . Thus, we need only concern ourselves with
the elements
prj =
d
X
k=1
urkvkj =
X
k̸=s
urkvkj + xvsj
for all values of j such that mrj is nonblank. In the expression above, we have
replaced urs, the element we wish to vary, by a variable x, and we use the
convention
• P
k̸=s is shorthand for the sum for k = 1, 2, . . ., d, except for k = s.
If mrj is a nonblank entry of the matrix M, then the contribution of this
element to the sum of the squares of the errors is
(mrj −prj)2 =
mrj −
X
k̸=s
urkvkj −xvsj
2
We shall use another convention:
• P
j is shorthand for the sum over all j such that mrj is nonblank.
Then we can write the sum of the squares of the errors that are aﬀected by
the value of x = urs as
X
j
mrj −
X
k̸=s
urkvkj −xvsj
2
Take the derivative of the above with respect to x, and set it equal to 0, in
order to ﬁnd the value of x that minimizes the RMSE. That is,
X
j
−2vsj
mrj −
X
k̸=s
urkvkj −xvsj

= 0
