Page 198 - Advanced engineering mathematics

178 CHAPTER 6 Vectors and Vector Spaces

Next we will show that linear combinations of vectors in S are in S . Suppose u and v are
⊥
⊥
in S . Then u and v are orthogonal to every vector in S.If c and d are real numbers and w is in
⊥
S, then
w · (cu + dv) = cw · u + dw · v = 0 + 0 = 0.
n
Therefore w is orthogonal to cu + dv,so cu + dv is in S and S is a subspace of R .
⊥
⊥
Certainly O is in both S and S .If u is in both S and S , then u is orthogonal to itself, so
⊥
⊥
2
u · u = u = 0
and then u = O.
n
We will now show that, given a subspace S of R , containing nonzero vectors, then each
n
vector in R has a unique decomposition into the sum of a vector in S and a vector in S .This
⊥
decomposition will prove useful in developing approximation techniques in Section 7.8.
THEOREM 6.7

n
n
Let S be a nontrivial subspace of R and let u be in R . Then there is exactly one vector u S in S
⊥
⊥
and exactly one vector u in S such that
u = u S + u .
⊥
Proof We know that we can produce an orthogonal basis V 1 ,··· ,V m for S. Define
u · V 1 u · V 2 u · V m
u S = V 1 + V 2 + ··· + V m
V 1 2 V 2 2 V m 2
m
u · V j

= V j .
V j · V j
j=1
u S is the sum of the projections of u onto each of the orthogonal basis vectors V 1 ,··· ,V m , and
is in S because this is a linear combination of the basis vectors of S.Nextset
⊥
u = u − u S .
⊥
Certainly u = u S + u . All that remains to show is that u is in S . To show this, we must
⊥
⊥
⊥
show that u is orthogonal to every vector in S. Since every vector in S is a linear combination
⊥
of V 1 ,··· ,V m , it is enough to show that u is orthogonal to each V j . Begin with V 1 . Since
V 1 · V j = 0if j = 1,
⊥
u · V 1 = (u − u S ) · V 1

m

u · V j
= u · V 1 − V j · V 1
V j · V j
j=1
u · V 1
= u · V 1 − (V 1 · V 1 ) = 0.
V 1 · V 1
Similarly, u · V j = 0for j = 2,··· ,m. Therefore u is in S .
⊥
⊥
⊥
Finally, we must show that u can be written in only one way as the sum of a vector in S and
a vector in S . Suppose
⊥
⊥
⊥
u = u S + u = U + U ,
where U is in S and U is in S . Then
⊥
⊥
u S − U = u − U .
⊥
⊥
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October 14, 2010 14:21 THM/NEIL Page-178 27410_06_ch06_p145-186