# How to Find a Basis for the Nullspace, Row Space, and Range of a Matrix

## Problem 708

Let $A=\begin{bmatrix}

2 & 4 & 6 & 8 \\

1 &3 & 0 & 5 \\

1 & 1 & 6 & 3

\end{bmatrix}$.

**(a)** Find a basis for the nullspace of $A$.

**(b)** Find a basis for the row space of $A$.

**(c)** Find a basis for the range of $A$ that consists of column vectors of $A$.

**(d)** For each column vector which is not a basis vector that you obtained in part (c), express it as a linear combination of the basis vectors for the range of $A$.

Contents

- Problem 708
- Solution.
- (a) Find a basis for the nullspace of $A$.
- (b) Find a basis for the row space of $A$.
- (c) Find a basis for the range of $A$ that consists of column vectors of $A$.
- (d) For each column vector which is not a basis vector that you obtained in part (c), express it as a linear combination of the basis vectors for the range of $A$.

## Solution.

We first obtain the reduced row echelon form matrix corresponding to the matrix $A$.

We reduce the matrix $A$ as follows:

\begin{align*}

A=\begin{bmatrix}

2 & 4 & 6 & 8 \\

1 &3 & 0 & 5 \\

1 & 1 & 6 & 3

\end{bmatrix}

\xrightarrow{\frac{1}{2}R_1}

\begin{bmatrix}

1 & 2 & 3 & 4 \\

1 &3 & 0 & 5 \\

1 & 1 & 6 & 3

\end{bmatrix}\\[6pt]
\xrightarrow[R_3-R_1]{R_2-R_1}

\begin{bmatrix}

1 & 2 & 3 & 4 \\

0 &1 & -3 & 1 \\

0 & -1 & 3 & -1

\end{bmatrix}

\xrightarrow[R_3+R_2]{R_1-2R_2}

\begin{bmatrix}

1 & 0 & 9 & 2 \\

0 &1 & -3 & 1 \\

0 & 0 & 0 & 0

\end{bmatrix}.

\end{align*}

The last matrix is in reduced row echelon form. That is,

\[\rref(A)=\begin{bmatrix}

1 & 0 & 9 & 2 \\

0 &1 & -3 & 1 \\

0 & 0 & 0 & 0

\end{bmatrix}. \tag{*}\]

### (a) Find a basis for the nullspace of $A$.

By the computation above, we see that the general solution of $A\mathbf{x}=\mathbf{0}$ is

\begin{align*}

x_1&=-9x_3-2x_4\\

x_2&=3x_3-x_4,

\end{align*}

where $x_3$ and $x_4$ are free variables.

Thus, the vector form solution to $A\mathbf{x}=\mathbf{0}$ is

\begin{align*}

\mathbf{x}=\begin{bmatrix}

x_1 \\

x_2 \\

x_3 \\

x_4

\end{bmatrix}

=\begin{bmatrix}

-9x_3-2x_4 \\

3x_3-x_4 \\

x_3 \\

x_4

\end{bmatrix}

=x_3\begin{bmatrix}

-9 \\

3 \\

1 \\

0

\end{bmatrix}+x_4\begin{bmatrix}

-2 \\

-1 \\

0 \\

1

\end{bmatrix}.

\end{align*}

It follows that the nullspace of the matrix $A$ is given by

\begin{align*}

\calN(A)&=\left\{ \mathbf{x}\in \R^4 \quad \middle | \quad \mathbf{x}= x_3\begin{bmatrix}

-9 \\

3 \\

1 \\

0

\end{bmatrix}+x_4\begin{bmatrix}

-2 \\

-1 \\

0 \\

1

\end{bmatrix}, \text{ for all } x_3, x_4 \in \R^4 \right \}\\[6pt]
&= \Span \left\{ \begin{bmatrix}

-9 \\

3 \\

1 \\

0

\end{bmatrix}, \begin{bmatrix}

-2 \\

-1 \\

0 \\

1

\end{bmatrix} \right \}.

\end{align*}

Thus, the set

\[\left\{ \begin{bmatrix}

-9 \\

3 \\

1 \\

0

\end{bmatrix}, \begin{bmatrix}

-2 \\

-1 \\

0 \\

1

\end{bmatrix} \right \}\]
is a spanning set for the nullspace $\calN(A)$.

It is straightforward to see that this set is linearly independent, and hence it is a basis for $\calN(A)$.

### (b) Find a basis for the row space of $A$.

Recall that the nonzero rows of $\rref(A)$ form a basis for the row space of $A$.

Thus,

\[\left\{\begin{bmatrix}

1 \\

0 \\

9 \\

2

\end{bmatrix}, \quad \begin{bmatrix}

0 \\

1 \\

-3 \\

1

\end{bmatrix} \right \}\]
is a basis for the row space of $A$.

### (c) Find a basis for the range of $A$ that consists of column vectors of $A$.

Recall that by the leading 1 method, the columns of $A$ corresponding to columns of $\rref(A)$ that contain leading 1 entries form a basis for the range $\calR(A)$ of $A$.

From (*), we see that the first and the second columns contain the leading 1 entries. Thus,

\[\left\{\begin{bmatrix}

2 \\

1 \\

1

\end{bmatrix}, \begin{bmatrix}

4 \\

3 \\

1

\end{bmatrix}\right \}\]
is a basis for the range $\calR(A)$ of $A$.

### (d) For each column vector which is not a basis vector that you obtained in part (c), express it as a linear combination of the basis vectors for the range of $A$.

Let us write $A_1, A_2, A_3$, and $A_4$ for the column vectors of the matrix $A$.

In part (c), we showed that $\{A_1, A_2\}$ is a basis for the range $\calR(A)$.

Thus, we need to express the vectors $A_3$ and $A_4$ as a linear combination of $A_1$ and $A_2$, respectively.

A shortcut is to note that the entries of third column vector of $\rref(A)$ give the coefficients of the linear combination for $A_3$. That is, we have

\[A_3=9A_1-3A_2.\]
Similarly, the entries of the fourth column of $\rref(A)$ yield

\[A_4=2A_1+A_2.\]

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