# 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 $A$.

## 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 \} \\
-1 \\
-1 \\
1
\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).

### 2 Responses

1. 03/16/2017

[…] 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}, […]

2. 06/25/2017

[…] 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. […]

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