Download MATH 304 Linear Algebra Lecture 29

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

Document related concepts
no text concepts found
Transcript
MATH 304
Linear Algebra
Lecture 29:
Orthogonal sets.
The Gram-Schmidt process.
Orthogonal sets
Let V be an inner product space with an inner
product h·, ·i and the induced norm k · k.
Definition. A nonempty set S ⊂ V of nonzero
vectors is called an orthogonal set if all vectors in
S are mutually orthogonal. That is, 0 ∈
/ S and
hx, yi = 0 for any x, y ∈ S, x 6= y.
An orthogonal set S ⊂ V is called orthonormal if
kxk = 1 for any x ∈ S.
Remark. Vectors v1 , v2 , . . . , vk ∈ V form an
orthonormal set if and only if
1 if i = j
hvi , vj i =
0 if i 6= j
Examples. • V = Rn , hx, yi = x · y.
The standard basis e1 = (1, 0, 0, . . . , 0),
e2 = (0, 1, 0, . . . , 0), . . . , en = (0, 0, 0, . . . , 1).
It is an orthonormal set.
• V = R3 , hx, yi = x · y.
v1 = (3, 5, 4), v2 = (3, −5, 4), v3 = (4, 0, −3).
v1 · v2 = 0, v1 · v3 = 0, v2 · v3 = 0,
v1 · v1 = 50, v2 · v2 = 50, v3 · v3 = 25.
Thus the set {v1 , v2 , v3 } is orthogonal but not
orthonormal. An orthonormal set is formed by
normalized vectors w1 = kvv11 k , w2 = kvv22 k ,
w3 = kvv33 k .
• V = C [−π, π], hf , g i =
Z
π
f (x)g (x) dx.
−π
f1 (x) = sin x, f2 (x) = sin 2x, . . . , fn (x) = sin nx, . . .
hfm , fn i =
Z
π
sin(mx) sin(nx) dx
−π
=
Z
1
cos(mx − nx) − cos(mx + nx) dx.
2
π
−π
Z
π
sin(kx) π
= 0 if k ∈ Z, k 6= 0.
k
x=−π
Z π
cos(kx) dx =
dx = 2π.
cos(kx) dx =
−π
k = 0 =⇒
Z
π
−π
−π
Z
1 π
cos(m − n)x − cos(m + n)x dx
hfm , fn i =
2 −π
π if m = n
=
0 if m 6= n
Thus the set {f1 , f2 , f3 , . . . } is orthogonal but not
orthonormal.
It is orthonormal with respect to a scaled inner
product
Z
1 π
hhf , g ii =
f (x)g (x) dx.
π −π
Orthogonality =⇒ linear independence
Theorem Suppose v1 , v2 , . . . , vk are nonzero
vectors that form an orthogonal set. Then
v1 , v2 , . . . , vk are linearly independent.
Proof: Suppose t1 v1 + t2 v2 + · · · + tk vk = 0
for some t1 , t2 , . . . , tk ∈ R.
Then for any index 1 ≤ i ≤ k we have
ht1 v1 + t2 v2 + · · · + tk vk , vi i = h0, vi i = 0.
=⇒ t1 hv1 , vi i + t2 hv2 , vi i + · · · + tk hvk , vi i = 0
By orthogonality, ti hvi , vi i = 0 =⇒ ti = 0.
Orthonormal bases
Let v1 , v2 , . . . , vn be an orthonormal basis for an
inner product space V .
Theorem Let x = x1 v1 + x2 v2 + · · · + xn vn and
y = y1 v1 + y2 v2 + · · · + yn vn , where xi , yj ∈ R. Then
(i) hx, yi = x1 y1 + x2 y2 + · · · + xn yn ,
p
(ii) kxk = x12 + x22 + · · · + xn2 .
Proof: (ii) follows from (i) when y = x.
+
*
+
* n
n
n
n
X
X
X
X
yj vj
xi vi ,
hx, yi =
xi vi ,
yj vj =
i=1
=
j=1
n
n
XX
xi yj hvi , vj i =
i=1 j=1
j=1
i=1
n
X
i=1
xi yi .
Let v1 , v2 , . . . , vn be a basis for an inner product
space V .
Theorem If the basis v1 , v2 , . . . , vn is an
orthogonal set then for any x ∈ V
hx, v1 i
hx, v2 i
hx, vn i
x=
v1 +
v2 + · · · +
vn .
hv1 , v1 i
hv2 , v2 i
hvn , vn i
If v1 , v2 , . . . , vn is an orthonormal set then
x = hx, v1 iv1 + hx, v2 iv2 + · · · + hx, vn ivn .
Proof: We have that x = x1 v1 + · · · + xn vn .
=⇒ hx, vi i = hx1 v1 + · · · + xn vn , vi i, 1 ≤ i ≤ n.
=⇒ hx, vi i = x1 hv1 , vi i + · · · + xn hvn , vi i
=⇒ hx, vi i = xi hvi , vi i.
Let V be a vector space with an inner product.
Suppose that v1 , . . . , vk ∈ V are nonzero vectors
that form an orthogonal set. Given x ∈ V , let
hx, vk i
hx, v1 i
v1 + · · · +
vk , o = x − p.
p=
hv1 , v1 i
hvk , vk i
Let W denote the span of v1 , . . . , vk .
Theorem (a) o ⊥ w for all w ∈ W (denoted o ⊥ W ).
(b) kok = kx − pk = min kx − wk.
w∈W
Thus p is the orthogonal projection of the vector
x on the subspace W . Also, p is closer to x than
any other vector in W , and kok = dist(x, p) is the
distance from x to W .
Orthogonalization
Let V be a vector space with an inner product.
Suppose x1 , x2 , . . . , xn is a basis for V . Let
v1 = x1 ,
hx2 , v1 i
v1 ,
hv1 , v1 i
hx3 , v1 i
hx3 , v2 i
v3 = x3 −
v1 −
v2 ,
hv1 , v1 i
hv2 , v2 i
.................................................
hxn , vn−1 i
hxn , v1 i
v1 − · · · −
vn−1 .
vn = xn −
hv1 , v1 i
hvn−1 , vn−1 i
v2 = x2 −
Then v1 , v2 , . . . , vn is an orthogonal basis for V .
The orthogonalization of a basis as described above
is called the Gram-Schmidt process.
Normalization
Let V be a vector space with an inner product.
Suppose v1 , v2 , . . . , vn is an orthogonal basis for V .
v1
v2
vn
Let w1 =
, w2 =
,. . . , wn =
.
kv1 k
kv2 k
kvn k
Then w1 , w2 , . . . , wn is an orthonormal basis for V .
Theorem Any finite-dimensional vector space with
an inner product has an orthonormal basis.
Remark. An infinite-dimensional vector space with
an inner product may or may not have an
orthonormal basis.
Related documents