Linear Transformations

Author:Ku, Yin Bon (Albert)

Linear Transformations between Vector Spaces

We extend our definition of linear transformation to general vector spaces in a straightforward way: Definition: Let

and

be two vector spaces and

be a linear transformation from

for any

for any real number

and

Here are some examples of linear transformations: Example 1: Let

such that

for any

. Example 2: Let

such that

for any

. Example 3: Let

be the vector space of all continuous functions from

. We define

as follows:

for any

and

Example 4: Let

such that

is called the left shift operator of sequences. Compositions of two linear transformations is again a linear transformation. The definitions of injective, surjective, and bijective linear transformations are essentially the same as before. Bijective linear transformations are also called isomorphisms. Two vector spaces are said to be isomorphic if there exists an isomorphism between them.

Kernel and Image

Given a linear transformation, we define the following two important subspaces: Definition: Let

be a linear transformation. Then

is called the kernel of

, denoted by

. And

is called the Image of

, denoted by

is a subspace of

and

is a subspace of

. (Why?) By definition, a linear transformation

is surjective if and only if

. And the kernel is useful for detecting whether a linear transformation is injective thanks to the following theorem: Theorem:

is injective if and only if

. The proof is essentially the same as the proof of the similar theorem that we had before. A useful relation between the dimensions of

and

is given by the following theorem: Theorem: Suppose

is a linear transformation and

is finite-dimensional. Then

. The following is a useful corollary of the above theorem: Corollary: Suppose

is a linear transformation.

Suppose . Then is injective if and only if is surjective.
If is bijective, then

is called the rank of

and

is called the nullity of

. Hence, the above theorem is sometimes called the rank-nullity theoerm.

Proof of the Rank-nullity theorem

Since

is finite-dimensional, so is

. Let

. We can find a basis

for

such that

i.e.

is a basis for

. If

is the zero linear transformation and

. Hence the theorem is true. Assume

, we consider

. We will show that it is a basis for

: For any

, there exists

such that

. Write

. Then

since

for

. Hence

. Then we need to show that

is a linearly independent set. Suppose

for some real numbers

. Then we have

, which means

is in

. Therefore,

for some real numbers

. Rewrite it as follows:

Then

for

because

is a linearly independent set. In particular,

. Hence

is a linearly independent set. Therefore,

Exercise

For any positive integer , define such that for any in . Find . Hence, or otherwise, show that is surjective.

Let . Define such that for any in . Show that is a linear transformation. Find the rank and nullity of .

Linear Transformations