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

Vector Space Problems and Solutions

Problem 703

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

 
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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 $2$.


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