Null Space and Column Space

Author:Ku, Yin Bon (Albert)

Null Space and Column Space

Let

be a linear transformation and

be the m x n matrix such that

for any vector

. Suppose

, where

are the column vectors in

. Then we have the following definitions: Definition: The column space of

, denoted by

, is

, or equivalently,

. Definition: The null space of

, denoted by

, is

, or equivalently,

. Finding the null space of a matrix

means solving the homogeneous equation

. When A is row reduced to the matrix in reduced echelon form,

is the number of non-pivot positions (or free variables). By the rank-nullity theorem,

. Hence,

is the number of pivot positions (or basic variables) in the reduced echelon form. Definition: The rank (or column rank) of

. It is obvious that the rank of a m x n matrix must be smaller or equal to

. A matrix is said to have full rank if its rank equals

Basis for Nul A and Col A

We consider the following 4 x 5 matrix

Find the basis for

and

. Solution: Using the row reduction algorithm, A is row reduced to R in reduced echelon form:

To find

, we need to solve the homogeneous equation

, which is equivalent to

. The following is the corresponding linear system:

Then

are free variables. To express the solutions in parametric vector form, we set

and get

Hence the basis for

. And

. As for

, we already know that

. Therefore, we just need to select three column vectors in the given matrix

that are linearly independent. Claim: The column vectors in

corresponding to the pivot positions are linearly independent. Let

be the elementary matrices corresponding to the row reduction steps. Let

be the column vectors in

corresponding to the pivot positions. Then we have

Let

is clearly invertible i.e. the corresponding linear transformation is an isomorphism from

. In other words, the inverse transformation corresponding to

maps the set

to the set

. Since any isomorphism maps a linearly independent set to a linearly independent set (Why?),

is linearly independent. Hence the basis for

Row Space

Given an m x n matrix

, we can consider its rows instead of columns. Each row can be regarded as a vector in

. Let

be the row vectors of

. In the previous example, the row vectors for the 4 x 5 matrix are

Definition: The row space of

, denoted by

, is

. How can we find a basis for

? Again, we consider the previous example. By the row reduction algorithm, we have

. Notice that any row operation does not change the row space. (Why?) Therefore, we have

. In

, we can discard the row of zeros and collect the first three row vectors that span the row space. Obviously, they are linearly independent. Hence a basis for

. Remark: The above method still works if

is in echelon form instead of reduced echelon form i.e.

Then a basis for

. (Note: In general, there are many bases for a vector space so we won't expect the answer is unique.) In general, for any matrix

because they both equal the number of pivot positions in

Exercise

Compute the rank of matrix and find bases for the null space, column space and row space of . (You may want to use the row operator calculator.)

Null Space and Column Space