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 $A$.

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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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