# Find a Linear Transformation Whose Image (Range) is a Given Subspace

## Problem 392

Let $V$ be the subspace of $\R^4$ defined by the equation

\[x_1-x_2+2x_3+6x_4=0.\]
Find a linear transformation $T$ from $\R^3$ to $\R^4$ such that the null space $\calN(T)=\{\mathbf{0}\}$ and the range $\calR(T)=V$. Describe $T$ by its matrix $A$.

## Solution.

Any vector

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

x_1 \\

x_2 \\

x_3 \\

x_4

\end{bmatrix}\in V\]
can be written as

\begin{align*}

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

x_2-2x_3-6x_4 \\

x_2 \\

x_3 \\

x_4

\end{bmatrix}\\[6pt]
&=x_2\begin{bmatrix}

1 \\

1 \\

0 \\

0

\end{bmatrix}+x_3\begin{bmatrix}

-2 \\

0 \\

1 \\

0

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

-6 \\

0 \\

0 \\

1

\end{bmatrix}.

\end{align*}

Let $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$ be the vectors appearing in the above linear combination of $\mathbf{x}$.

Then it is straightforward to see that the set $B=\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}$ is a basis of $V$.

We define the linear transformation $T:\R^3\to \R^4$ by

\begin{align*}

T(\mathbf{x})=A\mathbf{x},

\end{align*}

where

\[A=[\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3]=\begin{bmatrix}

1 & -2 & -6 \\

1 &0 &0 \\

0 & 1 & 0 \\

0 & 0 & 1

\end{bmatrix}.\]

Since $B$ is a basis of $V$, in particular it is a linearly independent set. Thus, the columns of $A$ is linearly independent.

It follows that the null space $\calN(T)=\calN(A)=\{\mathbf{0}\}$.

Also, the range of $T$ is the same as the range of $A$, which is spanned by the columns of $A$.

Thus, the range $\calR(T)=\Span(B)=V$.

By our definition of $T$, the matrix representation of $T$ is $A$.

A more explicit formula for $T$ is given by

\begin{align*}

T\left(\, \begin{bmatrix}

x_1 \\

x_2 \\

x_3

\end{bmatrix} \,\right)=

\begin{bmatrix}

1 & -2 & -6 \\

1 &0 &0 \\

0 & 1 & 0 \\

0 & 0 & 1

\end{bmatrix}

\begin{bmatrix}

x_1 \\

x_2 \\

x_3

\end{bmatrix}

=\begin{bmatrix}

x_1-2x_2-6x_3 \\

x_1\\

x_2 \\

x_3

\end{bmatrix}.

\end{align*}

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