Homework 3 Solutions
5.4 # 2. Given x = [1 1 1 1]T and y = [8 2 2 0]T :
(a) Determine the angle θ between x and y.
(b) Find the projection p of x onto y.
(c) Verify x − p is orthogonal to p.
(d) Compute kx − pk2 , kpk2 , and kxk2 and verify the Pythagorean Law
is satisfied.
Solution. (a) First,
cos θ =
8+2+2
x·y
1
= √ √
=√
kxk2 kyk2
4 72
2
and, hence,
θ = arccos
µ
1
√
2
¶
=
π
.
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(b) By definition
p=
x·y
2y
kyk2
=
12
1
1
T
T
y = [8 2 2 0] = [4 1 1 0] .
72
6
3
(c) x − p = x− 16 y and p = 16 y so
¶
µ
´
1
1
1 ³
1
6x · y − kyk22 =
(72 − 72) = 0.
(x − p) · p = x− y · y =
6
6
36
36
(d) Finally,
2
kx − pk2
=
=
=
So
kpk2
2
=
kxk22
=
µ
¶ µ
¶
1
1
x− y · x− y
6
6
2
1
kxk22 − x · y+ kyk22
6
36
2
1
4 − (12) +
(72) = 2
6
36
1
1
1
y · y = (72) = 2
6
6
36
4.
kxk22 = 4 = 2 + 2 = kx − pk22 + kpk22 .
1
5.4 # 9. In C [−π, π] with inner product defined by (6), show that cos mx and sin nx
are orthogonal and that both are unit vectors. Determine the distance
between the two vectors.
Solution: Here are two ways to prove the orthogonality, depending on what you
know.
Orth. 1: The addition formulas for the sine and cosine are one key: If α and β
are any two angles, then
sin (α + β) = sin α cos β + cos α sin β
cos (α + β) = cos α cos β − sin α sin β.
Replace β by −β in the first identity to obtain (why)
sin (α − β) = sin α cos β − cos α sin β.
Add the first and third identities to get
sin α cos β = sin (α + β) + sin (α − β) .
Now consider
hsin nx, cos mxi =
1
π
Z
π
sin nx cos mx dx.
−π
If n = 0, then the integral is zero. (Why?) So assume n 6= 0. Let α = nx and
β = mx and use the last identity to find
Z
1 π
hsin nx, cos mxi =
(sin (n + m) x + sin (n − m) x) dx
π −π
R
½
1 π
(sin (m + m) x) dx = 0 if n = m,
π
−π
R
=
1 π
π −π (sin (n + m) x + sin (n − m) x) dx = 0 if n 6= m.
(Evaluate the integrals and check that the stated results are correct.) Thus,
hsin nx, cos mxi = 0 and cos mx and sin nx are orthogonal with respect to this
inner product.
Orth. 2: If you remember from calculus that the integral of an odd function
over a symmetric range is zero and you realize that sin nx cos mx is odd because
sin nx is odd and cos mx is even, then you get free of charge
Z
1 π
hsin nx, cos mxi =
sin nx cos mx dx = 0.
π −π
To find the lengths of the vectors set α = β in the first cosine identity to get
½
2 cos2 α − 1
cos (2α) = cos2 α − sin2 α =
1 − 2 sin2 α
2
or
cos2 α =
sin2 α =
1
(1 + cos 2α) ,
2
1
(1 − cos 2α) .
2
Use these formulas with α = nx (or mx which is really the same (why?) to find
¶
½
Z
Z µ
1 π 1 1
1 π
1 if m 6= 0
cos2 mx dx =
hcos mx, cos mxi =
+ cos 2mx dx =
,
2 if m = 0
π −π
π −π 2 2
¶
½
Z
Z µ
1 π
1 π 1 1
1 if n 6= 0
hsin nx, sin nxi =
sin2 nx dx =
− cos 2nx dx =
.
0 if n = 0
π −π
π −π 2 2
So both cos mx with and sin nx are orthogonal unit vectors. Finally, using this
information, if m and n are both different from 0, then
2
ksin nx − cos mxk
= hsin nx − cos mx, sin nx − cos mxi
= hsin nx, sin nxi − 2 hsin nx, cos mxi + hcos mx, cos mxi
= 1 − 2 (0) + 1 = 2
and, hence,
ksin nx − cos mxk =
√
2 if m, n 6= 0 or m, n = 0.
What do you get in the other cases?
P4 Prob. 1
Let V be an inner product space, S and S0 be subsets of V. Do the following.
(a) Prove: If S0 ⊂ S, then S ⊥ ⊂ S0⊥ .
¡ ¢⊥
¡ ¢⊥
(b) Prove that S ⊂ S ⊥
and deduce that, span (S) ⊂ S ⊥ .
First a general principle. To prove A ⊂ B you must show that if a ∈ A, then
a ∈ B. (Why?)
Proof of (a): Let s⊥ ∈ S ⊥ . We must prove that s⊥ ∈ S0⊥ . Now, by definition,
s ∈ S ⊥ means
­ ⊥ ®
s , s = 0 for every vector s ∈ S.
⊥
Since S0 ⊂ S it follows immediately that
­ ⊥ ®
s , s0 = 0 for every vector s0 ∈ S0 .
Therefore by definition s⊥ ∈ S0⊥ . So s⊥ ∈ S ⊥ =⇒ s⊥ ∈ S0⊥ and, hence,
S ⊥ ⊂ S0⊥ .
An alternative proof of (a) with fewer symbols follows: If a vector is perpendicular to all vectors in a given set (here S) it is automatically perpendicular to
all vectors in any subset of that set (here S0 ). That is, every vector in S ⊥ is in
S0⊥ ; thus, S ⊥ ⊂ S0⊥ .
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¡ ¢⊥
(b) Let s ∈ S. We must prove that s ∈ S ⊥ . Now by definition
¡ ⊥ ¢⊥ ©
­
®
ª
S
= v ∈ V : v, s⊥ = 0 for every vector s⊥ ∈ S ⊥ .
­
¡ ¢⊥
®
But s, s⊥ = 0 for every vector s⊥ ∈ S ⊥ by the definition of S ⊥ . So s ∈ S ⊥
¡ ¢⊥
¡ ¢⊥
¡ ¢⊥
by definition of S ⊥ . So s ∈ S =⇒ s ∈ S ⊥
and, hence, S ⊂ S ⊥ .
¡ ¢⊥
Finally, to prove span (S) ⊂ S ⊥ reason as follows. Let t ∈ span (S) . Then
¡ ¢⊥
Pn
t = i=1 αi si for some scalars αi and some vectors si ∈ S. Since S ⊂ S ⊥ ,
¡ ¢⊥
­
®
each si ∈ S ⊥ ; that is, si , s⊥ = 0 for every vector s⊥ ∈ S ⊥ . Then by
linearity of the inner product for any s⊥ ∈ S ⊥
+
* n
n
X
X
­ ⊥®
­
®
⊥
αi si , s
αi si , s⊥ = 0.
=
t, s =
i=1
i=1
¡ ¢⊥
¡ ¢⊥
and, hence, span (S) ⊂
Therefore, t ∈ S ⊥ . So t ∈ span (S) =⇒ t ∈ S ⊥
¡ ⊥ ¢⊥
.
S
An alternative proof of (b) can be given using results of other problems:
¡ ¢⊥
First, for any nonempty set T, T ⊥ is a subspace. So S ⊥
is a subspace of
V. Second, if any set U is contained in a subset, then span (U ) is contained in
¡ ¢⊥
that subspace. By (a) S is contained in the subspace S ⊥ ; hence, span (S) ⊂
¡ ⊥ ¢⊥
.
S
P4 Prob. 9 (Very Useful) Let W be a subspace of an inner product space
V and let w1 , ..., wn be a spanning set for W. Prove: for v ∈ V
v ⊥ W ⇐⇒ v ⊥ w1 , ..., v ⊥ wn .
Proof: =⇒ : Since v ⊥ W means v ⊥ w for each w ∈ W and w1 , ..., wn are all
elements of W = span (w1 , ..., wn ) , we deduce at once that v ⊥ w1 , ..., v ⊥ wn .
⇐= : Now assume v ⊥ w1 , ..., v ⊥ wn . Let w ∈ W = span (w1 , ..., wn ) . Then
w = α1 w1 + α2 w2 + · · · + αn wn
for some scalar α and by linearity of the inner product
hv, wi = hv, α1 w1 + α2 w2 + · · · + αn wn i
= α1 hv, w1 i + α2 hv, w2 i + · · · + αn hv, wn i
= α1 (0) + α2 (0) + · · · + αn (0) = 0.
So v ⊥ w for every w ∈ W ; therefore, v ⊥ W.
Comments:
4
1. Don’t put subscripts on variables unless the subscripts are needed in your
argument. Otherwise their use is distracting to a reader and suggests that
you don’t fully understand what you are doing.
2. In the notation hu, vi , u and v are vectors. The notation hS, vi where S
is some set or subspace of vectors is not defined.
3. Don’t write span (S) = α1 s1 +α2 s2 +· · ·+αn sn or something similar. This
syntax says a set, span (S) is equal to an element, α1 s1 +α2 s2 +· · ·+αn sn ,
of that set even if you intend a general element of the span. You should
use proper set notation:
span (S) = {α1 s1 + α2 s2 + · · · + αn sn : αk are scalars and sk ∈ S}
or something equivalent that equates two sets.
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