# Column and Row Spaces and Rank of a Matrix

     

We present the definitions of column and row spaces of a matrix using examples with detailed solutions.

## Column Space and Rank of a Matrix

Let $$A$$ be an $$m \times n$$ matrix.
The column space of matrix $$A$$, denoted by $$\text{Col} \; A$$, is the set of all linear combinations of the columns of matrix $$A$$. If $$c_1, c_2, ..., c_n$$ are the columns of matrix $$A$$, then
$\text{Col} \; A = \text{span} \; \{c_1, c_2, ..., c_n\}$
Note that the columns $$\{c_1, c_2, ..., c_n\}$$ may not be independent and in what follows we look at examples on how to find the basis of $$\text{Col} \; A$$ by selecting the independent columns only.
The rank of matrix $$A$$ is the dimension of $$\text{Col} \; A$$ which is given by the number of vectors in the basis of $$\text{Col} \; A$$.

Example 1
Find the basis of the column space and the rank of each of the matrices:
a) $$A_1 = \begin{bmatrix} 1 & 2 \\ 0 & 0 \end{bmatrix}$$ , b) $$A_2 = \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}$$ , c) $$A_3 = \begin{bmatrix} 1 & 0 & -6 & 0\\ 0 & 0 & 0 & 0\\ 0 & 2 & 0 & -9 \end{bmatrix}$$

Solution to Example 1
a)
Let $$c_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$$ and $$c_2 = \begin{bmatrix} 2 \\ 0 \end{bmatrix}$$ be the columns of matrix $$A_1$$.
Note that $$c_2 = 2 c_1$$ and therefore $$c_1$$ and $$c_2$$ are NOT independent. As a result, we need one column only to span $$\text{Col}\; A_1$$.
hence the basis $$B$$ of $$\text{Col} \; A_1$$ is given by: $$B = \{c_1\}$$ or $$B = \left\{ \begin{bmatrix} 1 \\ 0 \end{bmatrix} \right\}$$
$$\text{Col} \; A_1 = \text{span} \{c_1\}$$
The basis of $$\text{Col} \; A_1$$ has one vector and therefore $$\text{Rank} (A_1) = 1$$.

b)
Let $$c_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$$ and $$c_2 = \begin{bmatrix} 0 \\ 2 \end{bmatrix}$$ be the columns of matrix $$A_2$$. Note that $$c_1$$ and $$c_2$$ are independent. As a result, we need both columns $$c_1$$ and $$c_2$$ to span $$\text{Col}\; A_2$$.
The basis $$B$$ of $$\text{Col} \; A_2$$ is given by: $$B = \{c_1,c_2\}$$ or $$B = \left\{\begin{bmatrix} 1 \\ 0 \end{bmatrix} , \begin{bmatrix} 0 \\ 2 \end{bmatrix} \right\}$$
$$\text{Col} \; A_2 = \text{span} \{c_1,c_2\}$$
The basis of $$\text{Col} \; A_2$$ has two vectors and therefore $$\text{Rank} (A_2) = 2$$.

c)
Let $$c_1 = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$$, $$c_2 = \begin{bmatrix} 0 \\ 0 \\ 2 \end{bmatrix}$$, $$c_3 = \begin{bmatrix} -6 \\ 0 \\ 0 \end{bmatrix}$$ and $$c_4 = \begin{bmatrix} 0 \\ 0 \\ -9 \end{bmatrix}$$ be the columns of matrix $$A_3$$.
Note that $$c_1$$ and $$c_2$$ are independent. However $$c_3$$ and $$c_4$$ depend on $$c_1$$ and $$c_2$$ as follows: $$c_3 = - 6 c_1$$ and $$c_4 = - \dfrac{9}{2} c_2$$.
Hence the basis $$B$$ of $$\text{Col} \; A_3$$ is given by: $$B = \{c_1,c_2\}$$ or $$B = \left\{\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} , \begin{bmatrix} 0 \\ 0 \\ 2 \end{bmatrix} \right\}$$
$$\text{Col} \; A_3 = \text{span} \{c_1,c_2\}$$
The basis of $$\text{Col} \; A_3$$ has two vectors and therefore $$\text{Rank} (A_3) = 2$$.

Example 2
Find the basis of the column space and the rank of matrix
$$A = \begin{bmatrix} 1 & -2 & 0 & 4\\ -1 & 3 & 1 & 0\\ 0 & -1 & -1 & -4 \end{bmatrix}$$.

Solution to Example 2
As seen in example 1, the basis of $$Col \; A$$ is given by the linearly independent columns of matrix $$A$$. One way to find out which columns are linearly independent is to rewrite the given matrix in row echelon form (REF).
Rewrite matrix $$A$$ in row echelon form
$$\color{red}{\begin{matrix} \\ R_2+R_1\\ \\ \end{matrix}}$$ $$\begin{bmatrix} 1 & -2 & 0 & 4\\ 0 & 1 & 1 & 4\\ 0 & -1 & -1 & -4 \end{bmatrix}$$

$$\color{red}{\begin{matrix} \\ \\ R_3+R_2\\ \end{matrix}}$$ $$\begin{bmatrix} \color{blue}1 & -2 & 0 & 4\\ 0 & \color{blue}1 & 1 & 4\\ 0 & 0 & 0 & 0 \end{bmatrix}$$

The basis of $$\text{Col} \; A$$ is given by the column in the original matrix corresponding to the columns with pivot (the leading 1 in a row) in the row echelon form obtained.
The first and second column in the reduced matrix has a pivot each and therefore the first and second column in the original matrix form the basis $$B$$ of $$\text{Col}\ A$$ which is given by Hence
$$B = \left\{\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} , \begin{bmatrix} -2 \\ 3\\ -1 \end{bmatrix} \right\}$$

$$\text{Col}\ A = \text{span} \left\{ \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} , \begin{bmatrix} -2 \\ 3\\ -1 \end{bmatrix} \right\}$$
The basis of $$\text{Col} \; A$$ has two vectors and therefore $$\text{Rank} (A) = 2$$.

## Row Space and Rank of a Matrix

Let $$A$$ be an $$m \times n$$ matrix.
The row space of matrix $$A$$, denoted by $$\text{Row} A$$, is the set of all linear combinations of the rows of matrix $$A$$. If $$b_1, b_2, ..., b_m$$ are the rows of matrix $$A$$, then
$\text{Row} \; A = \text{span} \; \{b_1, b_2, ..., b_n\}$

Note that the rows $$\{b_1, b_2, ..., b_m\}$$may not be independent and in what follows we look at examples on how to find the basis of $$\text{Row} \; A$$ by selecting the independent rows only.
The rank of matrix $$A$$ is the dimension of $$\text{Row} \; A$$ which is given by the number of vectors in the basis of $$\text{Row} \; A$$.

Example 3
Find the basis of the row space and the rank of each of the matrices:
a) $$A_1 = \begin{bmatrix} 4 & 8 \\ - 2 & - 4 \end{bmatrix}$$ , b) $$A_2 = \begin{bmatrix} 1 & 0 & 5\\ 0 & 2 & 7 \end{bmatrix}$$ , c) $$A_3 = \begin{bmatrix} 1 & 0 & -6 & 0\\ 0 & 0 & 0 & 0\\ 0 & 2 & 0 & -9 \end{bmatrix}$$

Solution to Example 3
a)
Let $$b_1 = (4,8)$$ and $$b_2 = (-2,-4)$$ be the rows of matrix $$A_1$$. Note that $$b_2 = - \dfrac{1}{2} b_1$$ and therefore $$b_1$$ and $$b_2$$ are NOT independent. As a result, we need one row only to span $$\text{Row}\; A_1$$.
Hence the basis $$B$$ of $$\text{Row} \; A_1$$ is given by: $$B = \{ b_1\}$$ or $$B = \{ (4,8) \}$$
$$\text{Row} \; A_1 = \text{span} \{ (4,8) \}$$
The basis of $$\text{Row} \; A_1$$ has one vector and therefore $$\text{Rank} (A_1) = 1$$.

b)
Let $$b_1 = (1,0,5)$$ and $$b_2 = (0,2,7)$$ be the rows of matrix $$A_2$$. Note that $$b_1$$ and $$b_2$$ are independent. We therefore need both columns to span $$\text{Row} \; A_2$$.
The basis $$B$$ of $$\text{Row} \; A_2$$ is given by: $$B = \{ b_1 , b_2 \}$$ or $$B = \{ (1,0,5) , (0,2,7) \}$$
$$\text{Row} \; A_2 = \text{span} \{ (1,0,5) , (0,2,7) \}$$
The basis of $$\text{Row} \; A_2$$ has two vectors and therefore $$\text{Rank} (A_2) = 2$$.

c)
Let $$b_1 = (1,0,-6,0)$$, $$b_2 = (0,0,0,0)$$ and $$b_3 = (0,2,0,-9)$$ be the rows of matrix $$A_3$$.
Note that $$b_1$$ and $$b_3$$ are independent; $$b_2$$ is a zero vector which does not contribute in any linear combination and therefore is not included in the basis.
The basis $$B$$ of $$\text{Row} \; A_3$$ is given by: $$B = \{ b_1 , b_3 \}$$ or $$B = \{ (1,0,-6,0) , (0,2,0,-9) \}$$
$$\text{Row} \; A_3 = \text{span} \{ (1,0,-6,0) , (0,2,0,-9) \}$$
The basis of $$\text{Row} \; A_3$$ has two vectors and therefore $$\text{Rank} (A_3) = 2$$.

Example 4
Find the basis of the row space and the rank of the matrix
$$A = \begin{bmatrix} 1 & -2 & 0 & 4\\ 1 & 1 & -1 & - 8 \\ 2 & -1 & -1 & -4 \\ - 3 & 3 & 1 & 0 \end{bmatrix}$$.

Solution to Example 4
We need to find independent rows among the rows of the given matrix. This can be done by rewriting the given matix to row echelon form.
Rewrite matrix $$A$$ in row echelon form
$$\color{red}{\begin{matrix} \\ R_2 - R_1\\ R_3 - 2 R_1\\ R_4 + 3 R_1\\ \end{matrix}}$$ $$\begin{bmatrix} 1 & -2 & 0 & 4\\ 0 & 3 & -1 & -12\\ 0 & 3 & -1 & -12\\ 0 & -3 & 1 & 12 \end{bmatrix}$$

$$\color{red}{\begin{matrix} \\ \frac{1}{3} R_2\\ \\ \\ \end{matrix}}$$ $$\begin{bmatrix} 1 & -2 & 0 & 4\\ 0 & 1 & -1/3 & - 4\\ 0 & 3 & -1 & -12\\ 0 & -3 & 1 & 12 \end{bmatrix}$$

$$\color{red}{\begin{matrix} \\ \\ R_3 - 3 R_2\\ R_4 + 3 R_2\\ \end{matrix}}$$ $$\begin{bmatrix} 1 & -2 & 0 & 4\\ 0 & 1 & -1/3 & - 4\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{bmatrix}$$         (I)

The basis of $$\text{Row} \; A$$ is given by the nonzero rows in the row echelon form (matrix I) obtained. Hence
$$\text{Row}\ A = \text{span} \; \left\{ (1 , -2 , 0 , 4) , (0 , 1 , -1/3 , - 4) \right\}$$
The basis of $$\text{Row} \; A$$ has two vector and therefore $$\text{Rank} (A) = 2$$.

## Steps to Determine the Bases of the Column and Row Spaces of a Matrix and its Rank

If $$A$$ is an $$m \times n$$ matrix,
Step 1: Rewrite matrix $$A$$ in a row echelon form as matrix $$E$$
Step 2: The basis of $$\text{Row} \; A$$ is the set of all nonzero rows in matrix $$E$$ and $$\text{Row} \; A$$ is a subspace of $$\mathbb{R}^n$$
Step 3: The basis of $$\text{Col} \; A$$ is the set of all columns in $$A$$ corresponding to the columns with pivot in $$E$$ and $$\text{Col} A$$ is a subspace of $$\mathbb{R}^m$$
Step 4: Rank of A = dim$$\text{Row} \; A$$ = dim $$\text{Col} \; A$$

Example 5
Find the basis of the row space, the basis of the colum space and the rank of the matrix
$$A = \begin{bmatrix} -1 & 3 & 4 & -2\\ 1 & 2 & -2 & 0\\ 2 & -3 & 1 & 0\\ 0 & 5 & 2 & -2 \end{bmatrix}$$.

Solution to Example 5
Step 1: Write matrix $$A$$ in row echeclon form $$E$$
Note that since writing matrices in row echelon is not the main topic discussed here, we have used a row reduce calculator to obtain the row echelon fom $$E$$ of the given matrix.
$$E = \begin{bmatrix} 1 & -3 & -4 & 2\\ 0 & 1 & 2/5 & -2/5\\ 0 & 0 & 1 & -14/39\\ 0 & 0 & 0 & 0 \end{bmatrix}$$

Step 2: The basis $$B$$ of $$\text{Row} A$$ is the set of all nonzero rows in $$E$$.
Hence $$B = \{ (1 , -3 , -4 , 2) , (0 , 1 , 2/5 , -2/5) , (0 , 0 , 1 , -14/39) \}$$

Step 3: The basis $$C$$ of $$\text{Col} A$$ is the set of all columns in $$A$$ corresponding to the columns with pivots in $$E$$.
The first three columns in $$E$$ have a pivot hence the basis is the set of the first three columns in $$A$$.
$$C = \left\{ \begin{bmatrix} -1\\ 1\\ 2\\ 0 \end{bmatrix} , \begin{bmatrix} 3\\ 2\\ -3\\ 5 \end{bmatrix} , \begin{bmatrix} 4\\ -2\\ 1\\ 2 \end{bmatrix} \right\}$$

Step 4: The dimensions of $$\text{Row} \; A$$ and $$\text{Col} \; A$$ are given by
dim $$\text{Row} \; A$$ = number of rows in the basis $$B$$ of $$\text{Row} \; A$$ = 3
dim $$\text{Col} \; A$$ = number of vetors in the basis $$C$$ of $$\text{Col} \; A$$ = 3
Rank of $$A$$ = dim $$\text{Row} \; A$$ = dim $$\text{Col} \; A$$ = 3

## Questions with Solution

Given matrix $$A$$ and its row echelon form $$E$$, determine the column space, the row space and the rank of each matrix.

1. $$A = \begin{bmatrix} 1 & 0 & -3 \\ 1 & 0 & -2 \end{bmatrix}$$ , $$E = \begin{bmatrix} 1 & 0 & -3 \\ 0 & 0 & 1 \end{bmatrix}$$

2. $$A = \begin{bmatrix} 1 & -1 & 2 & -1\\ 1 & 0 & 4 & 0\\ 2 & -2 & 6 & -4 \\ \end{bmatrix}$$ , $$E = \begin{bmatrix} 1 & -1 & 2 & -1\\ 0 & 1 & 2 & 1\\ 0 & 0 & 1 & -1 \\ \end{bmatrix}$$

3. $$A = \begin{bmatrix} 1 & -1 & 0 & 0 & -1\\ -1 & 2 & 4 & -3 & 2\\ 2 & -2 & 0 & 1 & -3 \\ 3 & -3 & 0 & 0 & -3 \end{bmatrix}$$ , $$E = \begin{bmatrix} 1 & -1 & 0 & 0 & -1\\ 0 & 1 & 4 & -3 & 1\\ 0 & 0 & 0 & 1 & -1 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}$$

Solutions to the Above Questions

1. Basis $$B$$ of column space and basis $$C$$ of row space are given by:
$$B = \left\{ \begin{bmatrix} 1 \\ 1 \end{bmatrix} , \begin{bmatrix} -3 \\ -2\\ \end{bmatrix} \right\}$$
$$C = \{ (1 , 0 , -3) , (0,0,1) \}$$
Hence
$$\text{Col} \; A = \text{span} \left\{ \begin{bmatrix} 1 \\ 1 \end{bmatrix} , \begin{bmatrix} -3 \\ -2\\ \end{bmatrix} \right\}$$
$$\text{Row} \; A = \text{span} \left\{ ( 1 , 0 , -3) , ( 0 , 0 , 1) \right\}$$
Rank of $$A$$ = 2

2. Basis $$B$$ of column space and basis $$C$$ of row space are given by:
$$B = \left\{ \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix} , \begin{bmatrix} -1 \\ 0\\ -2 \end{bmatrix} , \begin{bmatrix} 2 \\ 4\\ 6 \end{bmatrix} \right\}$$
$$C = \{ (1 , -1 , 2 , -1) , (0 , 1 , 2 , 1) , ( 0 , 0 , 1 , -1) \}$$
Hence
$$\text{Col} \; A = \text{span} \left\{ \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix} , \begin{bmatrix} -1 \\ 0\\ -2 \end{bmatrix} , \begin{bmatrix} 2 \\ 4\\ 6 \end{bmatrix} \right\}$$
$$\text{Row} \; A = \text{span} \left\{ (1 , -1 , 2 , -1) , (0 , 1 , 2 , 1) , ( 0 , 0 , 1 , -1) \right\}$$
Rank of $$A$$ = 3

3. Basis $$B$$ of column space and basis $$C$$ of row space are given by:
$$B = \left\{ \begin{bmatrix} -1 \\ -1 \\ 2 \\ 3 \end{bmatrix} , \begin{bmatrix} -1 \\ 2 \\ -2 \\ -3 \end{bmatrix} , \begin{bmatrix} 0\\ -3\\ 1 \\ 0 \end{bmatrix} \right\}$$
$$C = \{ (1 , -1 , 0 , 0 , -1) , (0 , 1 , 4 , -3 , 1) , ( 0 , 0 , 0 , 1 , -1 ) \}$$
Hence
$$\text{Col} \; A = \text{span} \left\{ \begin{bmatrix} -1 \\ -1 \\ 2 \\ 3 \end{bmatrix} , \begin{bmatrix} -1 \\ 2 \\ -2 \\ -3 \end{bmatrix} , \begin{bmatrix} 0\\ -3\\ 1 \\ 0 \end{bmatrix} \right\}$$
$$\text{Row} \; A = \text{span} \left\{ (1 , -1 , 0 , 0 , -1) , (0 , 1 , 4 , -3 , 1) , ( 0 , 0 , 0 , 1 , -1 ) \right\}$$
Rank of $$A$$ = 3