Notes on Homework 11
1.
(a) No; let v = (0 1 0) T ; then hv, vi = 0, but v 6= 0.
(b) No; let v = (1 1 0) T ; then hv, vi = 0, but v 6= 0.
(c) Yes. I’ll just check positive definiteness. You should check each property. Suppose x ∈ R3 .
Then
h x, xi = 2x1 x1 + x2 x2 + 4x3 x3
= 2x21 + x22 + 4x23
a sum of positive multiples of squares, thus nonnegative. Moreover, the only way this
sum can be zero is if each xi is zero, i.e., only if x = 0.
(d) No. Let x = (0 1 0) T . Then
h2x, xi = h(0 2 0) T , (0 1 0)i
= 22 · 12
=4
but
2h x, xi = 2h(0 1 0) T , (0 1 0) T i
= 2 · 12
= 2.
2.
(a) Not orthonormal, since
  √3 
√
√
√
1
√3 
3
3
3

h 0  ,  − 33 i = 1
+ 0(−
)+0
√
3
3
3
3
0

3
6= 0.
Professor Dan Bates
Colorado State University
M369 Linear Algebra
Fall 2008


1

(b) Orthonormal. Let v1 =  0 , v2 = 
0

0
√
2
√2
2
2


0
√




, v2 =  − √22 . Then;
2
2
h v1 , v1 i = 12 + 02 + 02 = 1
√
√
2 2
2 2
2
) +(
)
h v2 , v2 i = 0 + (
2
2
=1
√
√
2 2
2 2
2
hv2 , v2 i = 0 + (−
) +(
)
2
2
=1
√
√
2
2
+0·
h v1 , v2 i = 1 · 0 + 0 ·
2
2
=0
√
√
2
2
)+0·
hv1 , v3 i = 1 · 0 + 0 · (−
2
2
=0
√
√ √
√
2
2
2 2
(−
)+
h v2 , v3 i = 0 · 0 +
2
2
2 2
= 0.
(c) Orthonormal. Omitted.
 
3

3. Let w =
4 .
1
(b) Since we have an orthonormal basis {v1 , v2 , v3 }, we know that
3
w=
∑ hw, vi ivi
i= 1
√
√
√
2
2
2
2
+1
)v2 + (3 · 0 + 4(−
)+1
) v3
= 3v1 + (3 · 0 + 4
2 √ 2
2
2
√
−3 2
5 2
v2 +
v3
= 3v1 +
2
2
√
(c) Let { z1 , z2 , z3 } be the orthonormal basis from part (c). Again,
3
w=
∑ hw, zi i zi
i= 1
√
√
√
7 10
26 35
5 14
=
z1 +
z2 −
z3 .
10
35
14
Professor Dan Bates
Colorado State University
M369 Linear Algebra
Fall 2008
4. Let v1 = 1, v2 = x, v3 = x2 . Apply the process iteratively:
(a) e1 = v1 / kv1 k. Since
k v1 k =
=
q
s
h1, 1i
Z 1
0
1 · 1dx
= 1,
we let e1 = 1.
(b) Calculate w2 by subtracting off the portion of v2 in the direction of w2 :
w 2 = v2 − h v2 , e1 i e1
1
= x− ·1
√ 2
3
kw2 k =
6
√
Then e2 = w2 / kw2 k = 2 3( x − 21 ).
(c) Calculate w3 similalry:
w 3 = v3 − h v3 , e1 i e1 − h v3 , e2 i e2
= v3 − (
=
kw3 k =
e3 =
=
Z 1
0
2
( x )(1)dx)1 − (
1
x + −x
6
√
5
30
w3
kw3 k
√
6 5( x2 + 1/6 − x)
Z 1
0
√
√
1
1
x2 2 3( x − )dx) · (2 3( x − ))
2
2
2
So the orthonormal basis, with respect to this inner product, is
√
√
1
{1, 2 3( x − ), 6 5( x2 + 1/6 − x)}.
2
5. Check the usual properties:
• For any u ∈ U, hu, 0i = 0. Therefore, 0 ∈ U ⊥ .
Professor Dan Bates
Colorado State University
M369 Linear Algebra
Fall 2008
• Suppose v, w ∈ U ⊥ ; we will show that v + w ∈ U ⊥ . Suppose u ∈ U. Then
hu, v + wi = hu, vi + hu, wi additivity
= 0 + 0 v,w ∈ U ⊥
=0
and v + w ∈ U ⊥ .
• Suppose v ∈ U ⊥ , λ ∈ R. If u ∈ U, then
hu, λvi = λhu, vi
= λ·0 = 0
and λv ∈ U ⊥ .
Professor Dan Bates
Colorado State University
M369 Linear Algebra
Fall 2008