1Normed vector spaces
II Linear Analysis
1.6 Finite-dimensional normed vector spaces
We are now going to look at a special case of normed vector spaces, where the
vector space is finite dimensional.
It turns out that finite-dimensional vector spaces are rather boring. In
particular, we have
(i) All norms are equivalent.
(ii) The closed unit ball is compact.
(iii) They are Banach spaces.
(iv) All linear maps whose domain is finite dimensional are bounded.
These are what we are going to show in this section.
First of all, we need to say what we mean when we say all norms are
“equivalent”
Definition
(Equivalent norms)
.
Let
V
be a vector space, and
k · k
1
,
k · k
2
be
norms on
V
. We say that these are equivalent if there exists a constant
C >
0
such that for any v ∈ V , we have
C
−1
kvk
2
≤ kvk
1
≤ Ckvk
2
.
It is an exercise to show that equivalent norms induce the same topology,
and hence agree on continuity and convergence. Also, equivalence of norms is an
equivalence relation (as the name suggests).
Now let
V
be an
n
-dimensional vector space with basis
{e
1
, ··· , e
n
}
. We
can define the `
n
p
norm by
kvk
`
n
p
=
n
X
i=1
|v
i
|
p
!
1/p
,
where
v =
n
X
i=1
v
i
e
i
.
Proposition.
Let
V
be an
n
-dimensional vector space. Then all norms on
V
are equivalent to the norm k · k
`
n
1
.
Corollary. All norms on a finite-dimensional vector space are equivalent.
Proof. Let k · k be a norm on V .
Let v = (v
1
, ··· , v
n
) =
P
v
i
e
i
∈ V . Then we have
kvk =
X
v
i
e
i
≤
n
X
i=1
|v
i
|ke
i
k
≤
sup
i
ke
i
k
n
X
i=1
|v
i
|
≤ Ckvk
`
n
1
,
where C = sup ke
i
k < ∞ since we are taking a finite supremum.
For the other way round, let
S
1
=
{v ∈ V
:
kvk
`
n
1
= 1
}
. We will show the
two following results:
(i) k · k : (S
1
, k · k
`
n
1
) → R is continuous.
(ii) S
1
is a compact set.
We first see why this gives what we want. We know that for any continuous map
from a compact set to
R
, the image is bounded and the infimum is achieved. So
there is some v
∗
∈ S
1
such that
kv
∗
k = inf
v∈S
1
kvk.
Since v
∗
6= 0, there is some c
0
such that kvk ≥ c
0
for all v ∈ S
1
.
Now take an arbitrary non-zero v ∈ V , since
v
kvk
`
n
1
∈ S
1
, we know that
v
kvk
`
n
1
≥ c
0
,
which is to say that
kvk ≥ c
0
kvk
`
n
1
.
Since we have found c, c
0
> 0 such that
c
0
kvk
`
n
1
≤ kvk ≤ ckv k
`
n
1
,
now let C = max
c,
1
c
0
> 0. Then
C
−1
kvk
2
≤ kvk
1
≤ Ckvk
2
.
So the norms are equivalent. Now we can start to prove (i) and (ii).
First, let v, w ∈ V . We have
kvk − kwk
≤ kv − wk ≤ Ckv − w k
`
n
1
.
Hence when
v
is close to
w
under
`
n
1
, then
kvk
is close to
kwk
. So it is continuous.
To show (ii), it suffices to show that the unit ball
B
=
{v ∈ V
:
kvk
`
n
1
≤
1
}
is compact, since
S
1
is a closed subset of
B
. We will do so by showing it is
sequentially compact.
Let {v
(k)
}
∞
k=1
be a sequence in B. Write
v
(k)
=
n
X
i=1
λ
(k)
i
e
i
.
Since v
(k)
∈ B, we have
n
X
i=1
|λ
(k)
i
| ≤ 1.
Consider the sequence λ
(k)
1
, which is a sequence in F.
We know that
|λ
(k)
1
| ≤
1. So by Bolzano-Weierstrass, there is a convergent
subsequence λ
(k
j
1
)
1
.
Now look at
λ
(k
j
1
)
2
. Since this is bounded, there is a convergent subsequence
λ
(k
j
2
)
2
.
Iterate this for all
n
to obtain a sequence
k
j
n
such that
λ
(k
j
n
)
i
is convergent
for all i. So v
(k
j
n
)
is a convergent subsequence.
Proposition.
Let
V
be a finite-dimensional normed vector space. Then the
closed unit ball
¯
B(1) = {v ∈ V : kvk ≤ 1}
is compact.
Proof. This follows from the proof above.
Proposition.
Let
V
be a finite-dimensional normed vector space. Then
V
is a
Banach space.
Proof.
Let
{v
i
} ∈ V
be a Cauchy sequence. Since
{v
i
}
is Cauchy, it is bounded,
i.e.
{v
i
} ⊆
¯
B
(
R
) for some
R >
0. By above,
¯
B
(
R
) is compact. So
{v
i
}
has a
convergent subsequence
v
i
k
→ v
. Since
{v
i
}
is Cauchy, we must have
v
i
→ v
.
So v
i
converges.
Proposition.
Let
V, W
be normed vector spaces,
V
be finite-dimensional. Also,
let T : V → W be a linear map. Then T is bounded.
Proof.
Recall discussions last time about
V
∗
for finite-dimensional
V
. We will
do a similar proof.
Note that since
V
is finite-dimensional,
im T
finite dimensional. So wlog
W
is finite-dimensional. Since all norms are equivalent, it suffices to consider the
case where the vector spaces have
`
n
1
and
`
m
1
norm. This can be represented by
a matrix T
ij
such that
T (x
1
, ··· , x
n
) =
X
T
1i
x
i
, ··· ,
X
T
mi
x
i
.
We can bound this by
kT (x
1
, ··· , x
n
)k ≤
m
X
j=1
n
X
i=1
|T
ji
||x
i
| ≤ m
sup
i,j
|T
ij
|
n
X
i=1
|x
i
| ≤ Ckxk
`
n
1
for some
C >
0, since we are taking the supremum over a finite set. This implies
that kT k
B(`
n
1
,`
m
1
)
≤ C.
There is another way to prove this statement.
Proof.
(alternative) Let
T
:
V → W
be a linear map. We define a norm on
V
by kvk
0
= kvk
V
+ kT vk
W
. It is easy to show that this is a norm.
Since
V
is finite dimensional, all norms are equivalent. So there is a constant
C > 0 such that for all v, we have
kvk
0
≤ Ckvk
V
.
In particular, we have
kT vk ≤ Ckvk
V
.
So done.
Among all these properties, compactness of
¯
B
(1) characterizes finite dimen-
sionality.
Proposition.
Let
V
be a normed vector space. Suppose that the closed unit
ball
¯
B(1) is compact. Then V is finite dimensional.
Proof. Consider the following open cover of
¯
B(1):
¯
B(1) ⊆
[
y∈
¯
B(1)
B
y,
1
2
.
Since
¯
B
(1) is compact, this has a finite subcover. So there is some
y
1
, ··· , y
n
such that
¯
B(1) ⊆
n
[
i=1
B
y
i
,
1
2
.
Now let
Y
=
span{y
1
, ··· , y
n
}
, which is a finite-dimensional subspace of
V
. We
want to show that in fact we have Y = V .
Clearly, by definition of Y , the unit ball
B(1) ⊆ Y + B
1
2
,
i.e. for every
v ∈ B
(1), there is some
y ∈ Y, w ∈ B
(
1
2
) such that
v
=
y
+
w
.
Multiplying everything by
1
2
, we get
B
1
2
⊆ Y + B
1
4
.
Hence we also have
B(1) ⊆ Y + B
1
4
.
By induction, for every n, we have
B(1) ⊆ Y + B
1
2
n
.
As a consequence,
B(1) ⊆
¯
Y .
Since
Y
is finite-dimensional, we know that
Y
is complete. So
Y
is a closed
subspace of V . So
¯
Y = Y . So in fact
B(1) ⊆ Y.
Since every element in
V
can be rescaled to an element of
B
(1), we know that
V = Y . Hence V is finite dimensional.
This concludes our discussion on finite-dimensional vector spaces. We’ll end
with an example that shows these are not true for infinite dimensional vector
spaces.
Example.
Consider
`
1
, and
e
i
= (0
,
0
, ··· ,
0
,
1
,
0
, ···
), where
e
i
is 1 on the
i
th
entry, 0 elsewhere.
Note that if i 6= j, then
ke
i
− e
j
k = 2.
Since
e
i
∈
¯
B
(1), we see that
¯
B
(1) cannot be covered by finitely many open balls
of radius
1
2
, since each open ball can contain at most one of {e
i
}.