Course 111: Algebra, 19th January 2007
To be handed in at tutorials on Jan 22nd and 23rd.
1. Given u = (u1 , u2 , . . . , un ) and v = (v1 , v2 , . . . , vn ) are two vectors in Rn
and that w1 , w2 , . . . , wn are positive real numbers (called the weights).
The weighted Euclidean inner product is defined
(u, v) = w1 u1 v1 + w2 u2 v2 + . . . wn un vn .
Show that this is indeed an inner product.
Solution: to show this is an inner product we need to show it satisfies
the following:
(i)(u, v) = (v, u)
We know that u and v in Rn then
(u, v) = w1 u1 v1 +w2 u2 v2 +. . .+wn un vn = w1 v1 u1 +w2 v2 u2 +. . .+wn vn un = (v, u)
since the ui and vi are real numbers which commute under multiplication.
(ii) (u, u) ≥ 0 and (u, u) = 0 iff u = 0
(u, u) = w1 u21 + w2 u22 + . . . + wn u2n ≥ 0
(1)
since the weights wi are all positive and the u2i are also positive.
Now consider (u, u) = 0. In this case since each of the terms in Eqn is
greater than or equal to zero, and the weights are all positive numbers
the only way (u, u) = 0 can be zero is for each u2i = 0 and therefore
ui = 0 so u = 0.
(iii) (αu + βv, y) = α(u, y) + β(v, y) where y is also a vector and α, β
are scalars.
Again, just using the definition of the inner product given
(αu + βv, y) = w1 (αu1 + βv1 )y1 + w2 (αu2 + βv2 )y2 + . . . + wn (αun + βvn )yn
= w1 αu1 y1 + w2 αu2 y2 + . . . + wn αun yn
+w1 βv1 y1 + w2 βv2 y2 + . . . + wn βvn yn
= α(w1 u1 y1 + w2 u2 y2 + . . . + wn un yn )
+β(w1 v1 y1 + w2 v2 y2 + . . . + wn vn yn )
= α(u, y) + β(v, y)
as required.
2. Compute (u, v), ||u|| and d(u, v) for the following vectors and inner
products
• u = (2, −1, 4), v = (3, 2, 0) in R3 with the standard Euclidean
inner product.
(u, v) = (2)(3) + (−1)(2) + (4)(0) = 4
1
||u|| = (u, u) 2 =
q
√
(2)2 + (−1)2 + (4)2 =
d(u, v) = ||u−v|| = ||(−1, −3, 4)|| =
q
21
(−1)2 + (−3)2 + (4)2 =
√
26
• u = (2, −1, 4), v = (3, 2, 0) in R3 with the weighted Euclidean
inner product and w1 = 2, w2 = 6, w3 = 1/5.
(u, v) = (2)(3)(2) + (−1)(2)(6) + (4)(0)
1
2
||u|| = (u, u) =
s
(2)2 (2) + (−1)2 (6) + (4)2
d(u, v) = ||u−v|| = ||(−1, −3, 4)|| =
1
=0
5
1
=
5
s
86 √
= 17.2
5
s
(−1)2 (2) + (−3)2 (6) + (4)2
1
=
5
s
296
5
• u = x, v = x2 in C[0, 1] the set of continous real-valued
functions
R1
on [0, 1] and an inner product, (f (t), g(t)) = 0 f (t)g(t)dt
(u, v) =
1
2
Z
||u|| = (u, u) =
1
2
x(x )dx =
0
s
Z
Z
1
0
1
x(x)dx =
0
2
d(u, v) = ||u−v|| = ||x−x || =
s
Z
1 4 1 1
x dx = x =
4 0 4
3
s
Z
1
x2 dx =
0
s
1
0
(x −
x2 )dx
=
1
1 3 1
x =√
3 0
3
s
1 5 1 4 1 3 1
1
x − x + x =√
5
2
3
30
0