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Mathematics 206
Solutions for HWK 17b
Section 5.2 p274
§5.2 p274. Problem 3. Given u = (−4,3), v = (0,5), and inner product hu,vi = 3u v +u v ,
1 1 2 2
find (a) hu,vi, (b) ||u||, and (c) d(u,v).
Solution.
(a)
hu,vi = 3(−4)(0)+3(5) = 15
(b) p p √
||u|| = hu,ui = 3(−4)2 +32 = 57
(c) p √ √
2 2
d(u,v) = ||v −u|| = ||(4,2)|| = 3(4 ) +2 = 52 = 2 13
§5.2 p274. Problem 9. Given the inner product on C[−1,1] that is defined by hf,gi =
Z 1 2 2
−1f(x)g(x)dx, and given f(x) = x , g(x) = x +1, find (a) hf,gi, (b) ||f||, and (c) d(f,g).
Solution.
(a) Z Z
1 1 2 2 16
2 2 4 2
hf,gi = −1x (x +1)dx= −1(x +x )dx= 5 + 3 = 15
(b)Since Z
1 2
4
hf,fi = −1x dx = 5
we have r √
||f|| = phf,fi = 2 = 10.
5 5
(c) Since Z
1
hg−f,g−fi= −11dx=2
we have p √
d(f,g) = ||g−f|| = hg−f,g−fi= 2.
Page 1 of 6 A. Sontag April 6, 2001
Math 206 HWK 17b Solns contd
5.2 p274
§5.2 p274. Problem 11. With the same inner product space as in Problem 9, and with
x
f(x) = x, g(x) = e , find (a) hf,gi, (b) ||f||, and (c) d(f,g).
Solution. R
x x x
(a) Integration by parts gives xe dx=xe −e +C,so
Z ¸
1 1
x x x −1
hf,gi = −1 xe dx = xe −e −1 =2e .
Z 1 2 2 r2 √6
(b) Since −1x dx= 3, we have ||f|| = 3 = 3 .
(c) Since Z Z
1 1 2 1
x 2 2 x 2x −1 2 −2
−1(x−e ) dx= −1(x −2xe +e )dx= 3 −4e +2(e −e )
we have sZ r
1 2 1
x 2 −1 2 −2
d(f,g) = ||f − g|| = −1(x−e ) dx= 3 −4e +2(e −e ).
§5.2 p274. Problem 13. With inner product on M2,2 defined by
hA,Bi=2a b +a b +a b +2a b
11 11 12 12 21 21 22 22
and with · ¸ · ¸
A= −1 3 ,B= 0 −2
4 −2 1 1
find hA,Bi, (b) ||A||, and (c) d(A,B).
Solution.
(a)
hA,Bi=2(−1)(0)+(3)(−2)+(4)(1)+2(−2)(1) =−6
2 2 2 2 p √
(b) From hA,Ai = 2(−1) +3 +4 +2(−2) = 35 find ||A|| = hA,Ai = 35.
(c) From · ¸
A−B= −1 5
3 −3
find
hA−B,A−Bi=2(−1)2+52+32+2(−3)2=2+25+9+18=54
so √ √
d(A,B) = 54 = 3 6.
Page 2 of 6 A. Sontag April 6, 2001
Math 206 HWK 17b Solns contd
5.2 p274
2
§5.2 p274. Problem 21. For u = (u ,u ) and v = (v ,v ) vectors in R , explain why the
1 2 1 2
formula hu,vi = u v does not give an inner product.
1 1
Solution. Since the proposed inner product totally ignores second entries, anticipate that
something will go wrong with positivity. Nonzero vectors will be able to have length zero. In fact,
if we use u = (0,b), where b is any nonzero number, then we’ll have hu,ui = 0 but u 6= 0. So the
positivity axiom (IP4) fails.
2 2 2 2
§5.2 p274. Problem 23. SamequestionasProblem21, butfor theformulahu,vi = u v +u v .
1 1 2 2
Solution. This time suspect that the fact that entries are squared may mess up homogeneity
or additivity. Here are some particular examples to show this suspicion is correct. With u =
(2,2) and c = 2, we would have h2u,ui = h(4,4),(2,2)i = 16(4) + 16(4) = 128, but 2hu,ui =
2(16 + 16) = 64, so homogeneity (IP3) fails. With this same choice for u, we would also have
hu,u+ui = h(2,2),(4,4)i = 128 but hu,ui + hu,ui = 64, so additivity (IP2) also fails. Lots of
other specific counterexamples could be used.
§5.2 p274. Problem 27. Find the angle between the vectors u = (1,1,1) and v = (2,−2,2) if
3
the inner product on R is given by hu,vi = u v +2u v +u v .
1 1 2 2 3 3
Solution. With the specified angle denoted by θ, we have
cosθ = h(1,1,1),(2,−2,2)i = 0 =0
so ||(1, 1,1)|| ||(2,−2,2)|| ||(1,1,1)|| ||(2,−2,2)||
θ = π.
2
§5.2 p274. Problem 29. With the same inner product on C[−1,1] as in Problems 9 and 11,
2
find the angle, θ, say, between the vectors f(x) = x and g(x) = x .
R1 3 π
Solution. From hf,gi = −1x dx = 0, conclude that θ = 2.
§5.2 p274. Problem 35. With inner product on C[−π,π] given by hf,gi = Z π f(x)g(x)dx
−π
Page 3 of 6 A. Sontag April 6, 2001
Math 206 HWK 17b Solns contd
5.2 p274
verify the Cauchy-Schwarz Inequality and the Triangle Inequality for the vectors given by f(x) =
sinx, g(x) = cosx.
Solution. To verify Cauchy-Schwarz, we need hf,gi and ||f||, k|g||. Find that
hf,gi = Z π sinxcosxdx = 1 Z π sin2xdx = 0.
−π 2 −π
Weneed go no further, since we have
0 = |hf,gi| ≤ ||f|| ||g||.
To verify the Triangle Inequality we need to verify that ||f + g|| ≤ ||f|| + ||g||. We have
Z π 2 Z π
hf+g,f+gi= −π(sinx+cosx) dx= −π(1+sin2x)dx=2π
hf,fi = Z π sin2xdx = 1 Z π (1−cos2x)dx = 1(2π) = π
−π 2 −π 2
hg,gi = Z π cos2xdx = Z π (1−sin2x)dx = 2π −π = π
−π −π
Since √ √ √ √
2π ≤ 2 π = π+ π
see that ||f + g|| ≤ ||f|| + ||g||, as anticipated.
§5.2 p274. Problem 37. With the same inner product space as in Problem 35, as well as the
same vectors f and g, show that f and g are orthogonal.
Solution. In the solution for Problem 35, we already showed that hf,gi = 0. Therefore f and g
are orthogonal.
2
§5.2 p274. Problem 41. For the Euclidean inner product on R , and for u = (1,2), v = (2,1),
find proj u and proj v and sketch both results.
v u
Solution. First, note that u·v = 4, u·u = 5, and u·v =. Therefore
proj u = u·vv = 4v = 4(2,1)
v u·v 5 5
Page 4 of 6 A. Sontag April 6, 2001
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