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7.8 Least Squares Vectors and Data Fitting 233
is a matrix of numbers, then
AX = x 1 C 1 + x 2 C 2 + ··· + x m C m = B
exactly when B is a linear combination of the columns of A, hence is in S.
The following lemma reveals a connection between the least squares vectors for AX=B and
orthogonal projections, as suggested by the inequality (7.2).
LEMMA 7.1
Let B be an n-vector. Then an m-vector X is a least squares vector for AX = B if and only if
∗
∗
AX = B S ,
where B S is the orthogonal projection of B onto S.
∗
Proof Suppose first that AX = B S . Then
∗
B − B S = B − AX
≤ B − C
for all vectors C in S, because B S is the vector in S closest to B. But the vectors C in S are exactly
m
the vectors AX for X in R ,so
∗
B − AX ≤ B − AX
for every m-vector X, and this proves that X is a least squares vector for AX = B.
∗
Conversely, suppose X is a lease squares vector for AX = B. Then
∗
∗
AX − B ≤ AX − B
∗
for all X in S. But then AX is the vector in S closest to B. Because B S is the unique vector with
this property, then AX = B S . This completes the proof.
∗
We are now able to completely characterize the least squares vectors of AX = B as the
solutions of a system of linear equations obtained using A.
THEOREM 7.17 Least Squares Vectors for AX = B
An m-vector X is a least squares vector of AX = B if and only if X is a solution of the system
A AX = A B.
t
t
Proof Suppose first that X is a least squares vector of AX = B. By the lemma,
∗
AX = B S .
∗
⊥
∗
⊥
We know that B − B S is in S ,so B − AX is in S . This means that the columns of A are
∗
∗
orthogonal to B − AX . Writing the dot product of column j of A with B − AX as a matrix
product, this orthogonality means that
t
∗
(C j ) (B − AX ) = 0.
t
t
Now (C j ) is row j of A ,so
t
∗
A (B − AX ) = O
in which O is the m × 1 zero matrix. But this equation can be written
t
t
A AX = A (B)
∗
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October 14, 2010 14:23 THM/NEIL Page-233 27410_07_ch07_p187-246