Let be a vector space. Recall from Lecture 1 (together with Assignment 1, Problem 2) that to say
is
-dimensional means that
contains a linearly independent set of size
but does not contain a linearly independent set of size
Also recall from Lecture 1 that a basis of
is a subset
of
which is linearly independent and spans
The main goal of this lecture is to prove the following theorem.
Theorem 1: A vector space is
-dimensional if and only if it contains a basis
of size
Before going on to the proof of this theorem, let us pause to consider its ramifications. Importantly, Theorem 1 gives a method to calculate the dimension of a given vector space : all we have to do is find a basis
of
and then count the number of elements in that basis. As an example, let us compute the dimension of
This vector space is typically used to model the world in which we live, which consists of four physical dimensions: three for space and one for time. Let us verify that our mathematical definition of vector space dimension matches our physical understanding of dimension.
Let be a vector in
Then, we have
where
This shows that the set spans
Let us check that
is linearly independent. This amounts to performing the above manipulation in reverse. If
are numbers such that
then we have
which means that Thus
is a linearly independent set which spans
i.e. it is a basis of
Since
has size
we conclude from Theorem 1 that
Now let us prove Theorem 1. First, observe that we have already proved that if is
-dimensional then it contains a basis
of size
— this is Theorem 1 from Lecture 1. It remains to prove the converse, which we now state as a standalone result for emphasis and ease of reference.
Theorem 2: If contains a basis
of size
then
is
-dimensional.
The proof of Theorem 2 is quite subtle. In order to make it easier to understand, it is helpful to first prove the following lemma, in which the main difficulty is concentrated.
Lemma 1: If is a linearly independent set in a vector space
and
is a basis of
then
Proof: Suppose this were false, i.e. that there exists in the vector space a linearly independent set
and a basis
such that
We will see that this leads to a contradiction. The strategy is the following: we wish to demonstrate that the assumption
implies that we can replace
of the vectors in
with the vectors in
in such a way that the resulting set
is linearly independent. If we can show this, we will have obtained the desired contradiction: cannot possibly be independent, because
is a basis. In particular, each of the remaining vectors
is a linear combination of the vectors in
We will pursue the following strategy to show that the existence of the set as above follows from the hypothesis
For each
consider the following proposition: there exists a linearly independent set
in
of the form
where is a subset of
of size
and
Let us call this proposition
Now, if we can prove that
is true, and that
then we can conclude that is true for all
Indeed, we would then have
This is one version of a proof technique known as mathematical induction. But the statement true is exactly what we want, since it gives us a linearly independent set
consisting of some number of vectors from
together with all vectors in
which results in the contradiction explained above.
Let us now implement the above strategy. The first step is to prove that is true. To see that it is, consider the set
where
and
This set is of the required form, and since
is linearly independent so is
It remains to prove that if and
is true, then
is true. Given that
is true, there exists a linearly independent set
in
of the form
with
an -element subset of
and
Consider the set
If is linearly independent, then so is any subset of it, so in particular the subset
is linearly independent and is true. Now suppose that
is linearly dependent. Then, because
is linearly independent,
must be a linear combination of the vectors in
i.e
for some Moreover, there exists a number
such that
else
would be a linear combination of
which is impossible because
is a linearly independent set. We now claim that the set
is linearly independent. Indeed, if were linearly dependent, then we would have
where not all the numbers are zero. This means that the number
since otherwise the above would say that a subset of
is linearly dependent, which is false because
is linearly independent. Now, if we substitute the representation of
as a linear combination of elements
given above, this becomes a vanishing linear combination of the vectors in
in which the coefficient of
is
which contradicts the linear independence of
So,
must be linearly independent. — Q.E.D.
Let us note the following corollary of Lemma 1.
Corollary 1: If and
are two bases of a vector space
then
Proof: Since is linearly independent and
is a basis, we have
by Lemma 1. On the other hand, since
is linearly independent and
is a basis, we also have
by Lemma 1. Thus
— Q.E.D.
We now have everything we need to prove Theorem 2.
Proof of Theorem 2: Let be a finite-dimensional vector space which contains a basis
of size
We will prove that
is
-dimensional using the definition of vector space dimension (Definition 4 in Lecture 1) and Lemma 1. First, since
is a linearly independent set of size
we can be sure that
contains a linearly independent set of size
It remains to show that
does not contain a linearly independent set of size
This follows from Lemma 1: since
is a basis of size
every linearly independent set
in
must have size less than or equal to
— Q.E.D.
Another very important consequence of Theorem 2 is the following: it reveals that any two vector spaces of the same dimension can more or less be considered the same. Note that this is fundamentally new territory for us; so far, we have only considered one vector space at a time, but now we are going to compare two vector spaces.
To make the above precise, we need to consider functions between vector spaces. In fact, it is sufficient to limit ourselves to the consideration of functions which are compatible with the operations of vector addition and scalar multiplication. This leads to the definition of a linear transformation from a vector space to another vector space
If vector spaces are the foundation of linear algebra, then linear transformations are the structures we want to build on these foundations.
Defintion 1: A function is said to be a linear transformation if it has the following properties:
where
is the zero vector in
and
is the zero vector in
- For any vectors
and any scalars
we have
So, a linear transformation is a special kind of function from one vector space to another. The name “linear transformation” comes from the fact that these special functions are generalizations of lines in which pass through the origin
More precisely, every such line is the graph of a linear transformation
Indeed, a line through the origin in
of slope
is the graph of the linear transformation
defined by
The one exception to this is the vertical line in passing through
which has slope
. This line is not the graph of any function from
to
which should be clear if you remember the vertical line test.
To reiterate, a linear transformation from a vector space to
is a special kind of function
For linear transformations, it is common to write
instead of
and this shorthand is often used to implicitly indicate that
is a linear transformation. In the second half of the course, we will discuss linear transformations in great detail. At present, however, we are only concerned with a special type of linear transformation called an “isomorphism.”
Defintion 2: A linear transformation is said to be an isomorphism if there exists a linear transformation
such that
and
The word “isomorphism” comes from the Greek “iso,” which means “same,” and “morph” which means “shape.” It is not always the case that there exists an isomorphism from
to
; when an isomorphism does exist, one says that
and
are isomorphic. Isomorphic vector spaces
and
have the “same shape” in the sense that there is both a linear transformation
which transforms every vector
into a vector
and an inverse transformation
which “undoes”
by transforming
back into
To understand this, it may be helpful to think of isomorphic vector spaces
and
as two different languages. Any two human languages, no matter how different they may seem, are “isomorphic” in the sense that they describe exactly the same thing, namely the totality of human experience. The isomorphism
translates every word
in the language
to the corresponding word
in
which means the same thing. The inverse isomorphism
translates back from language
to language
On the other hand, if
is a human language and
is the language of an alien civilization, then
and
are not isomorphic, since the experience of membership in human society is fundamentally different from the experience of membership in a non-human society, and this difference is not merely a matter of language.
Theorem 2: Any two -dimensional vector spaces
and
are isomorphic.
Proof: Since is
-dimensional, it contains a basis
by Theorem 1. Likewise, since
is
-dimensional, it contains a basis
Now, since
is a basis in
for every
there exist unique scalars
such that
Likewise, since since is a basis in
for every
there exist unique scalars
such that
We may thus define functions
by
and
We claim that is a linear transformation from
to
To verify this, we must demonstrate that
has the two properties stipulated by Defintion 1. First, we have
which verifies the first property. Next, let be any two vectors, and let
be any two scalars. Let
be the unique representations of and
as linear combinations of the vectors in the basis
We then have
which verifies the second property. In the same way, one checks that is a linear transformation from
to
To prove that the linear transformation is an isomorphism, it remains only to prove that the linear transformation
undoes
To see this, let
be an arbitrary vector in expressed as a linear combination of the vectors in the basis
We then have
This completes the proof that and
are isomorphic vector spaces. — Q.E.D.
To continue with our linguistic analogy, Theorem 2 says that any two -dimensional vector spaces are just different languages expressing the same set of concepts. From this perspective, it is desirable to choose a “standard language” into which everything should be translated. In linguistics, such a language is called a lingua franca, a term which reflects the fact that this standard language was once the historical predecessor of modern French (these days the lingua franca is English, but this too may eventually change).
The lingua franca for -dimensional vector spaces is
and it is unlikely that this will ever change. Let
be an
-dimensional vector space, and let
be a basis in
. Consider the basis
of
in which
To be perfectly clear, for each the vector
has the number
in position
and a zero in every other position. The basis
is called the standard basis of
and you should immediately stop reading and check for yourself that it really is a basis. Assuming you have done so, we proceed to define an isomorphism
as follows. Given a vector let
be the unique representation of as a linear combination of vectors in
Now set
where the symbol “” means “equal by definition.” The fact that this really is an isomorphism follows from the proof of Theorem 2 above. The isomorphism
is called the coordinate isomorphism relative to the basis
and the geometric vector
is called the coordinate vector of
relative to the basis
Because of the special form of the standard basis of
we may write the coordinate vector of
more concisely as
When working with a given -dimensional vector space, it is often convenient to choose a basis
in
and use the corresponding coordinate isomorphism
to work in the standard
-dimensional vector space
rather than working in the original vector space
. For example, someone could come to you with a strange
-dimensional vector space
and claim that this vector space is the best model for spacetime. If you find that you disagree, you can choose a convenient basis
and use the corresponding coordinate isomorphism
to transform their model into the standard model of spacetime.
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