Page 254 - Advanced engineering mathematics

234 CHAPTER 7 Matrices and Linear Systems

t
and this means that X is a solution of the system A AX = A B.
t
∗
To prove the converse, suppose X is a solution of this system. Reversing part of the
∗
argument just given shows that B − AX is in S . But then
∗
⊥
∗
∗
B = AX + (B − AX )
is a decomposition of B into a sum of a vector in S and a vector in S . Since this decomposition
⊥
is unique, then AX must be the orthogonal projection of B onto S:
∗
∗
AX = B S .
By the lemma, X is a least squares vector for AX = B.
∗
Theorem 7.17 provides a way of obtaining all least squares vectors for AX = B. These are
t
t
the solutions of the linear system A AX=A B. Since we know how to solve linear systems,
this provides a computable method for finding least squares vectors. For this reason, we
will call the system
t
t
A AX = A B
the auxiliary lsv system of AX = B.
In addition to providing a method for finding all least squares vectors for a system, the
auxiliary lsv system tells us when a system has only one least squares vector. This occurs exactly
T
when the auxiliary system has a unique solution, which in turn occurs when A A is nonsingular.
In this event, the least squares vector for AX = B is
t
t
−1
∗
X = (A A) A B.
This proves the following.
COROLLARY 7.7
AX = B has a unique least squares vector if A A is nonsingular.
t

EXAMPLE 7.31
Let
⎛ ⎞
−1 −2
A = ⎝ 1 4 ⎠
2 2
and
⎛ ⎞
3
.
⎝
B = −2 ⎠
7
We will find all of the least squares vectors for AX = B. Compute

6 10
t
A A = .
10 24
This is nonsingular, and we find that

12/22 −5/22
t −1
(A A) = .
−5/22 3/22

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October 14, 2010 14:23 THM/NEIL Page-234 27410_07_ch07_p187-246