1 Page of 8
Math 430
Siddharth Dahiya
[e-mail]
October 30, 2011
Homework # 8
3.1.3. Show that v, w = v1 w1 + v1 w2 + v2 w1 + v2 w2 does not dene an inner product on R2 .
By Inner Product Axioms, the inner product has to be positive denite i.e.
v , v > 0;
Assume, v =
v = 0
1
. Then using the denition of inner product provided,
1
v , v = v1 v1 + v1 v2 + v2 v1 + v2 v2
v , v = (1)(1) + (1)(1) + (1)(1) + (1)(1)
v, v = 1 1 1 + 1 = 0
The inner product is zero when v is not zero. That is, it is not positive denite.
Hence, v, w = v1 w1 + v1 w2 + v2 w1 + v2 w2 does not dene an inner product on R2 .
Math 430
Siddharth Dahiya
[e-mail]
October 30, 2011
Homework # 8
Page 2 of 8
3.1.5(b). The unit circle for an inner product on R2 is dened as the set of all vectors of unit length: v = 1.
w1
v1
.
,w =
Graph the unit of circles for the weighted inner product v, w = 2v1 w1 + 5v2 w2 , v =
w2
v2
2
2
v , v = 2v1 + 5v2 = 1
This gives us an eclipse of the dual radii, 1/ 2 and 1/ 5
Page 3 of 8
Math 430
Siddharth Dahiya
[e-mail]
October 30, 2011
Homework # 8
3.1.10. Let V be an inner product space.
(a) Prove that x, v = 0 for all v V if and only if x = 0.
Let v = x. Then
n
n
x2
i
xi xi =
x, x =
i=1
i=1
By denition of norm,
n
2
x
x2
i
=
i=1
x, x = x
2
=0
x =0
x=0
(b) Prove that x, v = y, v for all v V if and only if x = y.
Rewriting the equation, we get,
x, v y , v = 0
By denition of inner products, we get,
x y, v = 0
Assume, v = x y , then
n
n
x y, x y =
i=1
(x y )i (x y )i =
By denition of norm,
i=1
(x y )2
i
n
(x y )
2
=
i=1
(x y )2
i
(x y ), (x y ) = (x y )
(x y ) = 0
(x y ) = 0
x=y
2
=0
Page 4 of 8
Math 430
Siddharth Dahiya
[e-mail]
October 30, 2011
Homework # 8
(c) Let v1 , . . . , vn be a basis for V . Prove that x, vi = y, v i , i = 1, . . . , n, if and only if x = y.
Since, v1 , . . . , vn be a basis for V , then, v = v1 c1 + + cn vn . By bilinearity,
x, v = c1 x, v1 + + cn x, vn = 0
Now, this holds true for v V . Then,
x y, v = 0
And,
(x y ), v = c1 (x y ), v1 + + cn (x y ), vn = 0
n
n
i=1
ci (x y )vi = (x y )
ci vi = 0
i=1
However, ci = 0 and vi = 0, therefore,
xy =0x=y
Page 5 of 8
Math 430
Siddharth Dahiya
[e-mail]
October 30, 2011
Homework # 8
3.1.20. Let f (x) = x, g(x) = 1 + x2 . Compute f , g , f , and g for
(a) the L2 inner product
1
f, g =
f (x)g(x)dx,
0
1
x(x2 + 1)dx =
f, g =
3
4
0
1
f=
f, f =
0
1
xxdx =
3
1
g=
28
15
(1 + x2 )2 dx =
g, g =
0
(b) the L2 inner product
1
f, g =
f (x)g(x)dx,
1
1
x(x2 + 1)dx = 0
f, g =
1
f=
f, f =
xxdx =
2
3
1
1
g=
(1 + x2 )2 dx =
g, g =
1
56
15
Page 6 of 8
Math 430
Siddharth Dahiya
[e-mail]
October 30, 2011
Homework # 8
(c) the weighted inner product
1
f, g =
f (x)g(x)xdx,
0
1
x(x2 + 1)xdx =
f, g =
8
15
0
1
f=
(x)2 xdx =
x f, f =
1
2
0
1
g=
(1 + x2 )2 xdx =
x g, g =
0
7
6
Page 7 of 8
Math 430
Siddharth Dahiya
[e-mail]
October 30, 2011
Homework # 8
3.2.2(a). Find the Euclidean angle between the vectors (1, 1, 1, 1)T and (1, 1, 1, 1)T in R4 .
Let v1 = (1, 1, 1, 1)T and v2 = (1, 1, 1, 1) Euclidean Angle is:
cos =
v1 v2
v1 v2
1
(1)(1) + (1)(1) + (1)(1) + (1)(1)
2
cos =
==
4
2
( 12 + 12 + 12 + 12 )( 12 + 12 + 12 + (1)2 )
=
3
Page 8 of 8
Math 430
Siddharth Dahiya
[e-mail]
October 30, 2011
Homework # 8
3.2.7. Prove that
(a1 + a2 + + an )2 n(a2 + a2 + + a2 )
1
2
n
for any real numbers a1 , . . . , an . When does equality hold?
Proof. By Cauchy-Schwarz Inequality,
|v w| v
w
Assume, v = (a1 , a2 , . . . , an )T and w = (1, 1, . . . , 1)T . Then,
|v w| = |a1 + a2 + + an |
w
v=
n
a2 + a2 + a2
n
1
2
By Cauchy-Schwarz Inequality,
|a1 + a2 + + an |
n a2 + a2 + a2
n
1
2
Squaring both sides,
|a1 + a2 + + an |2 n(a2 + a2 + a2 )
1
2
n