Solutions
Homework 1
1.1 Verify that hx, yik = nx1 y1 + (n − 1)x2 y2 + ... + xn yn defines an inner product on Rn .
Then compute the norm of u = (a1 , a2 , ..., an ).
Proof. The given inner product is positive as
hx, xik = nx21 + (n − 1)x22 + ... + x2n ≥ 0,
as it is a sum of non-negative numbers. Furthermore this inner product is definite as
each term in the sum is non-negative, and if any term is positive the resulting sum
will also be positive. Thus hx, xik = 0 implies ix2i = 0 for all i = 1, . . . , n. This, in
turn, implies each xi = 0, which implies the vector x has all zero entries or x = 0.
Additivity and homogeneity in the first slot follow from the distributive law, that is,
hx + y, zik = n(x1 + y1 )z1 + (n − 1)(x2 + y2 )z2 + ... + (xn + yn )zn ,
= nx1 z1 + ny1 z1 + (n − 1)x2 z2 + (n − 1)y2 z2 + ... + xn zn + yn zn ,
= hx, zik + hy, zik ,
and
hax, yik = nax1 y1 + (n − 1)ax2 y2 + ... + axn yn ,
= a(nx1 y1 + (n − 1)x2 y2 + ... + xn yn ),
= ahx, yik .
Finally symmetry of the given inner product follows from the commutivity of multiplication in R.
hx, yik = nx1 y1 + (n − 1)x2 y2 + ... + xn yn ,
= ny1 x1 + (n − 1)y2 x2 + ... + yn xn ,
= hy, xik .
Therefore hx, yik = nx1 y1 + (n − 1)x2 y2 + ... + xn yn defines an inner product on
Rn .
1.2 Given real numbers aij , i, j = 1, 2, define hx, yi = a11 x1 y1 +a12 x1 y2 +a21 x2 y1 +a22 x2 y2 ,
determine the conditions that make hx, yi an inner product on R2 .
Solution:
hx, yi = a11 x1 y1 + a12 x1 y2 + a21 x2 y1 + a22 x2 y2
determines an inner product on R2 if and only if the following conditions hold,
(a) a11 > 0,
(b) a11 a22 − a12 a21 > 0,
(c) a12 = a21 ,
Proof. (Necessity) First assume that h·, ·i defines an inner product, then by positivity
and definiteness
a11 = h(1, 0), (1, 0)i > 0.
Similarly,
h(−a12 , a11 ), (−a12 , a11 )i = a11 a212 − a12 a12 a11 − a21 a11 a12 + a22 a211
= a11 (a11 a22 − a12 a21 ) > 0.
1
2
Since a11 > 0 this implies that a11 a22 − a12 a21 > 0. Finally, by symmetry
a12 = h(1, 0), (0, 1)i = h(0, 1), (1, 0)i = a21 .
(Sufficiency) Next assume, conversely, that the real numbers aij satisfy conditions
a-c. Then positivity holds, as
hx, xi = a11 x21 + 2a12 x1 x2 + a22 x22 ,
"
#
2 2
a12
a12
a22 2
= a11
x1 +
x2 −
x22 +
x ,
a11
a11
a11 2
"
2 #
a12
a11 a22 − a212
= a11
x1 +
x2 +
x22 ≥ 0.
2
a11
a11
Indeed, this also shows definiteness as the second term will only be zero when x2 = 0,
and if x2 = 0 the first term can only be zero if x1 = 0. Additivity holds as
hx + y, zi = a11 (x1 + y1 )z1 + a12 (x1 + y1 )z2 + a12 (x2 + y2 )z1 + a22 (x2 + y2 )z2 ,
= a11 x1 z1 + a11 y1 z1 + a12 x1 z2 + a12 y1 z2
+ a12 x2 z1 + a12 y2 z1 + a22 x2 z2 + a22 y2 z2 ,
= hx, zi + hy, zi.
Homogeneity holds as
hcx, yi = a11 cx1 y1 + a12 cx1 y2 + a12 cx2 y1 + a22 cx2 y2 ,
= c(a11 x1 y1 + a12 x1 y2 + a12 x2 y1 + a22 x2 y2 ),
= chx, yi.
Finally symmetry also holds as
hx, yi = a11 x1 y1 + a12 x1 y2 + a12 x2 y1 + a22 x2 y2 ,
= a11 y1 x1 + a12 y2 x1 + a12 y1 x2 + a22 y2 x2 ,
= hy, xi.
1.3 Given an inner product on R2 , compute kx − yk. Hence use the cosine law to deduce
that hx, yi = kxkkyk cos θ, with θ equal to the angle between x and y, when hx, yi is
the dot product.
Proof. First
p
hx − y, x − yi,
p
= hx, x − yi − hy, x − yi,
p
= hx, xi − 2hx, yi + hy, yi,
p
= kxk2 − 2hx, yi + kyk2 .
kx − yk =
3
Recall the law of cosines, which states that for a triangle with sides of length a, b,
and c, with angle θ opposite to the side of length c and between the sides of length
a and b the following relation holds,
a2 + b2 − 2ab cos θ = c2 .
Now, for any two vectors, x and y, the vector x − y, when starting at y ends at x.
This is encoded in the vector equation y + (x − y) = x. Thus x, y, and x − y form a
triangle. If θ is the angle between x and y, then the law of cosines implies
kxk2 + kyk2 − 2kxkkyk cos θ = kx − yk2 .
Where k · k is the euclidean norm corresponding to the dot product as the inner
product on Rn . From the first computation this is the same as
kxk2 + kyk2 − 2kxkkyk cos θ = kxk2 − 2hx, yi + kyk2 ,
which, after cancellation, implies
kxkkyk cos θ = hx, yi.
6.2 Suppose u, v ∈ V . Prove that hu, vi = 0 if and only if
kuk ≤ ku + avk
for all a ∈ F
Proof. (Necessity) Assume hu, vi = 0 then for all a ∈ F
ku + avk2 = kuk2 + 2ahu, vi + a2 kvk2 ,
= kuk2 + a2 kvk2 ,
≥ kuk2 .
(Sufficiency) Assume kuk ≤ ku + avk for all a ∈ F. In particular, when a = − hu,vi
,
kvk2
kuk2 ≤ ku + avk2 = kuk2 + 2ahu, vi + a2 kvk2 ,
2
hu, vi
hu, vi
2
= kuk − 2
hu, vi +
kvk2 ,
kvk2
kvk2
2
hu, vi
2
= kuk −
,
kvk
≤ kuk2 .
This, of course, implies that hu, vi = 0.
6.3 Prove that
n
X
j=1
!2
aj b j
≤
n
X
!
ja2j
j=1
for all real numbers a1 , . . . , an and b1 , . . . , bn .
n
X
b2j
j=1
j
!
4
Proof. The Euclidean inner product on Rn is given by,
hx, yi =
n
X
xj yj ,
j=1
for vectors x = (x1 , . . . , xn ), y = (y1 , . . . , yn ) ∈ Rn . For j = 1, . . . , n, let ãj =
b
b̃j = √jj , ã = (ã1 , . . . , ãn ), and b̃ = (b̃1 , . . . , b̃n ) then
!2
!2
n
n
X
X
aj b j
=
ãj b̃j
= hã, b̃i2 ,
j=1
√
jaj ,
j=1
and
n
X
!
ja2j
j=1
n
X
b2j
j=1
!
j
=
n
X
j=1
!
ã2j
n
X
!
b̃2j
= kãk2 kb̃k2 .
j=1
This reduces the inequality to the Cauchy-Schwartz inequality.
6.4 Suppose u, v ∈ V are such that kuk = 3, ku + vk = 4, ku − vk = 6. What number
must kvk equal?
√
kvk = 17
Proof. By the Parallelogram Law
16 + 36 = 2(9 + kvk2 ).
Thus kvk2 = 17.
6.5 Prove or disprove: there is an inner product on R2 such that the associated norm is
given by
k(x1 , x2 )k = |x1 | + |x2 |.
Solution: There is no inner product on R2 with the given associated norm.
Proof. Consider the vectors in R2 , u = (1, 0) and v = (0, 1). Then u + v = (1, 1) and
u − v = (1, −1). With the given norm,
ku + vk2 + ku − vk2 = (1 + 1)2 + (1 + 1)2 = 8,
and
2(kuk2 + kvk2 ) = 2((1 + 0)2 + (1 + 0)2 ) = 4.
If the norm were associated to an inner product the Parallelogram Law would imply
these two quantities must be equal. Since they are not, there can be no such inner
product.
6.6 Prove that if V is a real inner-product space, then
hu, vi =
for all u, v ∈ V .
ku + vk2 − ku − vk2
,
4
5
Proof. Expand using bilinearity to get,
ku + vk2 − ku − vk2
hu + v, u + vi − hu − v, u − vi
=
,
4
4
kuk2 + 2hu, vi + kvk2 − (kuk2 − 2hu, vi + kvk2 )
=
.
4