Page 200 - Advanced engineering mathematics

180 CHAPTER 6 Vectors and Vector Spaces

We will now show that u S has a remarkable property—it is the unique vector in S that is
closest to u. That is, the distance between u and u S is less than or equal to the distance between
u and v for every v in S:
u − u S < u − v for every v in S.

THEOREM 6.8
n
Let S be a nontrivial subspace of R and let u be in R . Then, for all vectors v in S different
n
from u S ,
u − u S < u − v .

Proof If u is in S, then u = u S and u − u S = 0. Clearly u is the unique vector in S closest to
itself.
Thus suppose that u is not in S.Let v be any vector in S different from u S . Write
u − v = (u − u S ) + (u S − v).
Now u S − v is in S, being a sum of vectors in S. And we know that u − u S is in S . Therefore
⊥
u S − v and u − u S are orthogonal. By the Pythagorean theorem,
2
2
2
u − v = u − u S + u S − v .
But u = u S ,so
u − u S > 0.
Therefore
2 2
u − v > u S − v
and this is equivalent to the conclusion of the theorem.

EXAMPLE 6.21
5
Let S be the subspace of R having orthogonal basis vectors
V 1 =< 1,0,0,0,0,0 >,V 2 =< 0,1,0,0,0,1 >,V 3 =< 0,1,0,0,0,−1 >.
Let u=<1,−1,4,1,2,−5>. We will find the vector in S closest to u. We may also think of this
as the distance between u and S. First, the orthogonal projection of u onto S is
1 1
u S = (u · v 1 )v 1 + (u · v 2 )v 1 + (u · v 3 )v 3
2 2
= v 1 − 3v 2 + 2v 3
=< 1,−1,0,0,0,−5 >.
Then
√
u − u S = 21.
This is the distance between u and the vector in S closest to u.

Because the distance between two vectors is the square root of a sum of squares, use of
Theorem 6.8 to find a vector at minimum distance from a given vector is called the method of
least squares. We will pursue the idea of least squares approximations in the next section and in
Section 7.8.

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October 14, 2010 14:21 THM/NEIL Page-180 27410_06_ch06_p145-186