The norm is a function, defined on a vector space, that associates to each vector a measure of its length. In abstract vector spaces, it generalizes the notion of length of a vector in Euclidean spaces.
There is a tight connection between norms and inner products, as every inner product can be used to induce a norm on its space.
Before reading this lecture you should be familiar with inner products and with the basics of the algebra of complex vectors and matrices.
Table of contents
We are going to give an abstract, axiomatic definition of norm. Later we will show some examples of norm to clarify its meaning.
Definition
Let
be a vector space. A norm on
is a function
that associates to each
a positive real number, denoted by
,
which has the following properties.
Definiteness:
Absolute
homogeneity:where
is the field over which the vector space is defined (i.e., the set of scalars
used for scalar multiplication);
denotes the absolute value if
and the modulus if
.
Triangle
inequality:
These properties are pretty intuitive.
As the norm is a measure of the length of a vector, it is reasonable to require that it should always be a positive number.
The definiteness property imposes that all vectors except the zero vector should have a strictly positive length.
Absolute homogeneity means that if you scale up (or down) a vector by a factor
,
then its length is re-scaled accordingly.
Finally, in the triangle inequality,
is to be interpreted as the side of a triangle, while
and
are the other two sides. A well-known fact from geometry is that the length of
one side of a triangle is less than the sum of the lengths of the other two
sides. The triangle inequality axiom extends this property to the notion of
length in an abstract vector space.
Before providing some examples of normed vector spaces, we need to introduce an important connection between inner products and norms.
Definition
Let
be a vector space and
an inner product on
.
Then, the function
defined
by
is
called an induced norm on
.
We will prove that an induced norm indeed satisfies all the properties of a norm. But before doing that, we need to prove some preliminary results.
We highly recommend reading the proofs of the next three propositions as not only they are very short and simple to understand, but also they are a good way to hone our skills in dealing with norms and inner products.
In order to understand the following generalization of the well-known Pythagoras' theorem, we need to remember that two vectors are said to be orthogonal if and only if their inner product is equal to zero.
Proposition
Let
be a vector space equipped with an inner product
and its induced norm
.
If two vectors
are orthogonal (i.e.,
),
then
The proof is as
follows:where:
in step
we have used the additivity of the inner product; in step
we have used the orthogonality of
and
;
in step
we have used the definition of induced norm.
Pythagoras' theorem says that the squared length of the hypotenuse
()
of a right triangle is equal to the sum of the squared lengths of the other
two sides of the triangle
(
).
The triangle is right because we assume orthogonality (a 90 degrees angle)
between the two sides
and
.
Given two vectors, we can always write the first as a scalar multiple of the second plus a third vector orthogonal to the second.
Proposition
Let
be a vector space equipped with an inner product
and its induced norm
.
Given two vectors
,
if
,
then there exist a scalar
and a vector
such
that
It suffices to verify the first
equality:and
the second
one:
where:
in step
we have used the additivity and homogeneity in the first argument of the inner
product; in step
we have used the definition of induced norm.
The inequality presented in the next proposition is known as Cauchy-Schwarz inequality.
Proposition
Let
be a vector space equipped with an inner product
and its induced norm
.
For any two vectors
,
the following inequality
holds:
If
,
then
,
and, as a consequence, the inequality holds. We can thus focus on the case in
which
.
In this case, we can use the orthogonal decomposition shown in the previous
section:
where
is orthogonal to
.
Thanks to the orthogonality of
,
we can apply Pythagoras'
theorem:
Since
,
the equality just derived implies
that
which
becomes, by taking square roots on both
sides,
If
we multiply both sides by
,
we obtain the desired inequality.
We now have almost all the tools that we need to prove that a norm induced by an inner product indeed satisfies all the properties of a norm.
We just need to remember a couple of facts about complex numbers.
First, if we add to a complex
numberits
complex
conjugate
we
get
where
is the so-called real part of the complex number.
Second, the modulus of
satisfies the inequality
because
We are now good to go!
Proposition
Let
be a vector space and
an inner product on
.
Then, the norm induced by the inner
product
is
a norm on
.
Let us first prove positivity and
definiteness. For any
,
by the positivity of inner products, we have
that
and
by the definiteness of inner products, we have
that
Now,
let us prove absolute homogeneity. Suppose
is defined over the complex field
.
Then, for every
,
we have
where:
in step
we have used the homogeneity in the first argument of the inner product; in
step
we have used the conjugate homogeneity in the second argument of the inner
product
(
is the complex conjugate of
);
in step
denotes the modulus of
.
Absolute homogeneity holds also when the vector space is over the real field
because
in that case, so that
is the absolute value of
.
Finally, let us prove the triangle inequality. For any
,
we have
that
where:
in step
we have used the additivity of the inner product in its two arguments; in step
we have used the definition of induced norm; in step
we have used the conjugate symmetry of the inner product; in step
and
we have used the elementary properties of complex numbers discussed in the
previous section; in step
we have used the Cauchy-Schwarz inequality. By taking the square root of both
sides of the inequality derived above, we
obtain
Wow! This was a long sequence of theoretical facts without ever presenting a single example. The wait was worthwhile because we can now make examples involving norms induced by inner products, which are the most practically relevant in linear algebra.
In this example we are going to discuss the norm of a
column vector having real entries.
Let
be the space of all
real vectors (on the real field
).
In the lecture on inner products,
we have explained that the most common way of defining an inner product
between two real
vectors
is
where
is the transpose of
,
are the
entries of
and
are the
entries of
.
The norm induced by this inner product
is
Thus, the norm of a real vector is equal to the square root of the sum of the squares of its entries. This is the ordinary way to compute the length of a vector in Euclidean space.
Example
DefineThen
Another important example is the norm of a
column vector having complex entries.
Let
be the space of all
complex vectors
(on the complex field
).
In the lecture on inner products,
the inner product between two
vectors
was defined to
be
where
is the conjugate transpose
of
,
are the
entries of
and
are the complex conjugates of the
entries of
.
The norm induced by this inner product
iswhere
we have used the fact that the modulus
is equal to the square root of
times its complex conjugate
.
Thus, the formula for the norm of a complex vector is not very different from the formula for real vectors: we just need to replace the absolute values of the entries of the vector with their moduli.
Example
DefineThen
Below you can find some exercises with explained solutions.
Let
be the space of all
real vectors.
For any
,
define
where
are the
entries of
.
Show that
is a norm.
The function
is positive since it is the sum of positive terms
.
It is definite because
if and only if
for
,
which happens if and only if
.
Absolute homogeneity is satisfied
because
Note
that for any couple of real numbers
,
we
have
Therefore,
for any two vectors
so
that the triangle inequality holds. Thus, all the properties of a norm are
satisfied.
Let
be the space of all
real vectors.
For any
,
define
where
are the
entries of
.
Show that
is a norm.
The function
is positive since it is the maximum over a set of positive terms
.
It is definite because
if and only if
for
,
which happens if and only if
.
Absolute homogeneity is satisfied
because
Note
that given two real numbers
,
we
have
For
any couple of vectors
,
we
have
so
that the triangle inequality holds. Thus, all the properties of a norm are
satisfied.
Please cite as:
Taboga, Marco (2021). "Norm of a vector", Lectures on matrix algebra. https://www.statlect.com/matrix-algebra/vector-norm.
Most of the learning materials found on this website are now available in a traditional textbook format.