# Row Equivalent Matrix, Bases for the Null Space, Range, and Row Space of a Matrix

## Problem 260

Let \[A=\begin{bmatrix}

1 & 1 & 2 \\

2 &2 &4 \\

2 & 3 & 5

\end{bmatrix}.\]

**(a)** Find a matrix $B$ in reduced row echelon form such that $B$ is row equivalent to the matrix $A$.

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

**(c)** Find a basis for the range of $A$ that consists of columns of $A$. For each columns, $A_j$ of $A$ that does not appear in the basis, express $A_j$ as a linear combination of the basis vectors.

**(d)** Exhibit a basis for the row space of $A$.

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Contents

- Problem 260
- Hint.
- Solution.
- Comment.

## Hint.

In part (c), you may use the following theorem.

**Theorem (leading-1 method)**: If $B$ is a matrix in reduced row echelon form that is row equivalent to $A$, then all the column vectors of $A$ whose corresponding columns in $B$ have leading 1’s form a basis of the range of $A$.

In part (d), you may use the following theorem.

**Theorem (Row-space method)**: If $B$ is a matrix in (reduced) row echelon form that is row equivalent to $A$, then the nonzero rows of $B$ form a basis for the row space of $A$.

## Solution.

### (a) Find a matrix $B$ in reduced roe echelon form such that $B$ is row equivalent to the matrix $A$.

We apply the elementary row operations to the matrix $A$ and obtain

\begin{align*}

A=\begin{bmatrix}

1 & 1 & 2 \\

2 &2 &4 \\

2 & 3 & 5

\end{bmatrix}

\xrightarrow{\substack{R_2-2R_1\\ R_3-2R_1}}

\begin{bmatrix}

1 & 1 & 2 \\

0 &0 &0 \\

0 & 1 & 1

\end{bmatrix}

\xrightarrow{R_2 \leftrightarrow R_3}\\

\begin{bmatrix}

1 & 1 & 2 \\

0 & 1 & 1\\

0 &0 &0

\end{bmatrix}

\xrightarrow{R_1-R_2}

\begin{bmatrix}

1 & 0 & 1 \\

0 & 1 & 1\\

0 &0 &0

\end{bmatrix}.

\end{align*}

The last matrix is in reduced row echelon form that is row equivalent to $A$.

Thus we set

\[B=\begin{bmatrix}

1 & 0 & 1 \\

0 & 1 & 1\\

0 &0 &0

\end{bmatrix}.\]

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

The null space $\calN(A)$ of the matrix is the set of solutions of the homogeneous system $A\mathbf{x}=\mathbf{0}$.

By part (a), the augmented matrix $[A\mid \mathbf{0}]$ is row equivalent to $[B \mid \mathbf{0}]$.

Thus, the solution $\mathbf{x}=\begin{bmatrix}

x_1 \\

x_2 \\

x_3

\end{bmatrix}$ must satisfy

\[x_1=-x_3 \text{ and } x_2=-x_3,\]
where $x_3$ is a free variable.

Thus the solutions are given by

\[\mathbf{x}=\begin{bmatrix}

-x_3 \\

-x_3 \\

x_3

\end{bmatrix}=x_3\begin{bmatrix}

-1 \\

-1 \\

1

\end{bmatrix}\]
for any number $x_3$.

Hence we have

\begin{align*}

\calN(A)&=\{ \mathbf{x}\in \R^3 \mid \mathbf{x}=x_3\begin{bmatrix}

-1 \\

-1 \\

1

\end{bmatrix} \text{ for any } x_3\in \R \} \\

&=\Span\left\{ \quad\begin{bmatrix}

-1 \\

-1 \\

1

\end{bmatrix}\quad \right \}.

\end{align*}

From this, we deduce that the set

\[\left\{ \quad\begin{bmatrix}

-1 \\

-1 \\

1

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

### (c) Find a basis for the range of $A$ that consists of columns of $A$. (Longer version)

Let us write

\[A=\begin{bmatrix}

1 & 1 & 2 \\

2 &2 &4 \\

2 & 3 & 5

\end{bmatrix}=[A_1, A_2, A_3],\]
where $A_1, A_2, A_3$ are the column vectors of the matrix $A$.

The range $\calR(A)$ of $A$ is the same as the column space of $A$.

Thus, it suffices to find the maximal number of linearly independent vectors among the column vectors $A_1, A_2, A_3$.

Consider the linear combination

\[x_1A_1+x_2A_2+x_3A_3=\mathbf{0}. \tag{*} \]
Then this is equivalent to the homogeneous system

\[A\mathbf{x}=\mathbf{0}\]
and we already found the solutions in part (b):

\[x_1=-x_3 \text{ and } x_2=-x_3.\]
This tells us the vectors $A_1, A_2, A_3$ are linearly dependent.

For example, $x_1=-1, x_2=-1, x_3=1$ is a nonzero solution of the system (*).

Thus, we have

\[\calR(A)=\Span(A_1, A_2, A_3)=\Span(A_1, A_2).\]

On the other hand, if we consider only $A_1$ and $A_2$, they are linearly independent.

(To see this, you just need to repeat the above argument without $A_3$. This amounts to just ignoring the third columns in the computations.)

Therefore, the set $\{A_1, A_2\}$ is a linearly independent spanning set for the range.

Hence a basis of the range consisting column vectors of $A$ is

\[\{A_1, A_2\}.\]

The only column vector which is not a basis vector is $A_3$.

We already found that $x_1=-1, x_2=-1, x_3=1$ is a nonzero solution of (*).

Thus we have

\[-A_1-A_2+A_3=\mathbf{0}.\]
Solving this for $A_3$, we obtain the linear combination for $A_3$ of the basis vectors:

\[A_3=A_1+A_2.\]

#### (c) A shorter solution using the leading 1 method

Here is a shorter solution which uses the following theorem.

Theorem (leading-1 method): If $B$ is a matrix in reduced row echelon form that is row equivalent to $A$, then all the column vectors of $A$ whose corresponding columns in $B$ have leading 1’s form a basis of the range of $A$.

Looking at the matrix $B$, we see that the first and the second columns of $B$ have leading 1’s. Thus the first and the second column vectors of $A$ form a basis for the range of $A$.

### (d) Exhibit a basis for the row space of $A$.

We use the following theorem.

Theorem (Row-space method): If $B$ is a matrix in (reduced) row echelon form that is row equivalent to $A$, then the nonzero rows of $B$ form a basis for the row space of $A$.

We have already found such $B$ in part (a), and the first and the second row vectors are nonzero. Thus they form a basis for the row space of $A$.

Hence a basis for the row space of $A$ is

\[\left\{\,\begin{bmatrix}

1 \\

0 \\

1

\end{bmatrix}, \begin{bmatrix}

0 \\

1 \\

1

\end{bmatrix} \, \right \}.\]

## Comment.

The longer solution of part (c) is essentially the proof of Theorem (leading 1 method).

It is good to know where the theorem came from, but when you solve a problem you may forget and just use the theorem like the shorter solution of (c).

Add to solve later

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## 2 Responses

[…] that these vectors are linearly independent, thus a basis for the range. (Basically, this is the leading 1 method.) Hence we have [calR(T)=calR(A)=Span left{begin{bmatrix} 1 \ 1 \ 0 end{bmatrix}, […]

[…] Since the both columns contain the leading $1$’s, we conclude that [left{, begin{bmatrix} 1 \ 0 \ 1 end{bmatrix}, begin{bmatrix} -1 \ 1 \ 1 end{bmatrix} ,right}] is a basis of the range of $A$ by the leading $1$ method. […]