# Describe the Range of the Matrix Using the Definition of the Range

## Problem 703

Using the definition of the range of a matrix, describe the range of the matrix
$A=\begin{bmatrix} 2 & 4 & 1 & -5 \\ 1 &2 & 1 & -2 \\ 1 & 2 & 0 & -3 \end{bmatrix}.$

## Solution.

By definition, the range $\calR(A)$ of the matrix $A$ is given by
$\calR(A)=\left \{ \mathbf{b} \in \R^3 \quad \middle | \quad A\mathbf{x}=\mathbf{b} \text{ for some } \mathbf{x} \in \R^4 \right \}.$

Thus, a vector $\mathbf{b}=\begin{bmatrix} b_1 \\ b_2 \\ b_3 \end{bmatrix}$ in $\R^3$ is in the range $\calR(A)$ if and only if the system $A\mathbf{x}=\mathbf{b}$ is consistent.
So, let us find the conditions on $\mathbf{b}$ so that the system is consistent.

To do this, we consider the augmented matrix of the system and reduce it as follows.
\begin{align*}
\left[\begin{array}{rrrr|r}
2 & 4 & 1 & -5 & b_1\\
1 &2 & 1 & -2 & b_2 \\
1 & 2 & 0 & -3 &b_3
\end{array}\right] \xrightarrow{R_1 \leftrightarrow R_2}
\left[\begin{array}{rrrr|r}
1 &2 & 1 & -2 & b_2 \\
2 & 4 & 1 & -5 & b_1\\
1 & 2 & 0 & -3 &b_3
\end{array}\right] \xrightarrow[R_3-R_1]{R_2-2R_1}\6pt] \left[\begin{array}{rrrr|r} 1 &2 & 1 & -2 & b_2 \\ 0 & 0 & -1 & -1 & b_1-2b_2\\ 0 & 0 & 0 & -1 & b_3-b_2 \end{array}\right] \xrightarrow{-R_2} \left[\begin{array}{rrrr|r} 1 &2 & 1 & -2 & b_2 \\ 0 & 0 & 1 & 1 & -b_1+2b_2\\ 0 & 0 & -1 & -1 & b_3-b_2 \end{array}\right]\\[6pt] \xrightarrow[R_3+R_2]{R_1-R_2} \left[\begin{array}{rrrr|r} 1 &2 & 0 & -3 & b_1-b_2 \\ 0 & 0 & 1 & 1 & -b_1+2b_2\\ 0 & 0 & 0 & 0 & -b_1 +b_2 +b_3 \end{array}\right]. \end{align*} Note that if the (3, 5)-entry -b_1+b_2+b_3 is not zero, then the system A\mathbf{x}=\mathbf{0} is inconsistent because this implies 0=1. On the other hand, if -b_1+b_2+b_3=0, then we see that the system is consistent. Hence, the vector \mathbf{b} is in the range \calR(A) if and only if -b_1+b_2+b_3=0. In summary, we have \[\calR(A)=\left\{ \begin{bmatrix} b_1 \\ b_2 \\ b_3 \end{bmatrix}\in \R^3 \quad \middle | \quad -b_1+b_2+b_3=0 \right \}.

### Spanning set for the range

With a little bit additional computation, we can find the spanning set for the range as follows.

Thus, $\mathbf{b} \in \calR(A)$ if and only if
\begin{align*}
\mathbf{b}=\begin{bmatrix}
b_1 \\
b_2 \\
b_3
\end{bmatrix}=\begin{bmatrix}
b_2+b_3 \\
b_2 \\
b_3
\end{bmatrix}=b_2\begin{bmatrix}
1 \\
1 \\
0
\end{bmatrix}+b_3\begin{bmatrix}
1 \\
0 \\
1
\end{bmatrix}.
\end{align*}

In summary, we have
\begin{align*}
\calR(A)&=\left\{ \mathbf{b} \in \R^3 \quad \middle | \quad \mathbf{b}=b_2\begin{bmatrix}
1 \\
1 \\
0
\end{bmatrix}+b_3\begin{bmatrix}
1 \\
0 \\
1
\end{bmatrix} \right \}\6pt] &=\Span\left\{ \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} \right \}. \end{align*} Hence, the spanning set is \[ \left\{ \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} \right \}.

This spanning set is linearly independent, hence it’s a basis for the range.

Thus, the dimension of the range is $2$.

The following problems are True or False. Let $A$ and $B$ be $n\times n$ matrices. (a) If $AB=B$, then $B$...