Fundamental Matrix Subspaces

Null Space

AKA The kernel

Definition

Let . The null space of is the subset of defined by:

(see Set Builder Notation, zero vector, matrix-vector product)

Intuition

The null space is the set of all vectors that, after having a linear transformation applied, get squished to the zero vector. This can only happen if the matrix is not full rank.

Column Space

Definition

Let . The column space of is the subset of defined by:

Intuition

The column space is the set of all possible outputs of . Since linear transformations must keep the origin, the zero vector is in the column space. Another way to think of it, is the space the columns span (as stated in the definition).

Recall that the columns of the matrix tell us where the basis vectors land.

Row Space

Definition

Let . The row space of is a subset of defined by:

Remark

I don't think we ever used row space again.

Intuition

The space the rows span... I guess?

Null, Row, and Column Subspaces

Theorem

Let . Then and are subspaces of , and is a subspace of .

Proof

By definition, is a subset of . We use the subspace sest to show is a subspace of . Since , then . For , we have that . Then

so is a subspace of .

Since and where are the columns of , is a subspace of by theorem span is a subspace.

Since and where are the rows of , is a subspace by theorem span is a subspace.

EROs Preserve Row Space

Theorem

Let . If is obtained from by a series of elementary row operations, then

Intuition

Since the span only cares about linearly independent anyways, EROs don't change anything since they're either simplifying existing vectors, or removing redundant ones.

EROs Preserve Column Summation

Theorem

Let and suppose is obtained from by a series of elementary row operations. Then for any , if and only if

Proof

Since can be obtained from by a series of EROs, the augmented matrix reduces to . Thus, is a solution to the system if and only if is a solution to . Hence, if and only if .

Intuition

As an example, if the first column of is the sum of the second and third columns of , then any matrix obtained from by EROs will have its first column be the sum of the second and third columns as well.

We know this must be the case, because if EROs change the span of your matrix, then you did something wrong.

Finding a Basis for Each Subspace

Abstract

Given a matrix , find the basis for , and state the dimension of each of these subspaces.

Carry to RREF
For null space, find the solution to
For column space, the basis is the set of linearly independent vectors, where each vector is a column of the matrix that has a leading entry
For Row space, the basis is the set of linearly independent vectors, where each vector is the row of the matrix that is also non-zero

Example

Let

Find a basis for , and state the dimension of each of these subspaces.

Carry to RREF:

The solution to the homogeneous system is:

So,

is a spanning set for , since we set .

Since each vector in has a where the other has a , is linearly independent, and is thus a basis for , so .

To find a basis for , we observe that the first two columns of the RREF have leading entries, so

Since the set is linearly independent (by inspection), is a basis for , and .

To find a basis for , we take each non-zero row of the RREF, and obtain that

is a basis for , and .