Give the Formula for a Linear Transformation from $\R^3$ to $\R^2$

Problem 156

Let $T: \R^3 \to \R^2$ be a linear transformation such that
$T(\mathbf{e}_1)=\begin{bmatrix} 1 \\ 4 \end{bmatrix}, T(\mathbf{e}_2)=\begin{bmatrix} 2 \\ 5 \end{bmatrix}, T(\mathbf{e}_3)=\begin{bmatrix} 3 \\ 6 \end{bmatrix},$ where
$\mathbf{e}_1=\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \mathbf{e}_2=\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}, \mathbf{e}_3=\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$ are the standard unit basis vectors of $\R^3$.
For any vector $\mathbf{x}=\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}\in \R^3$, find a formula for $T(\mathbf{x})$.

Contents

Hint.

Recall the definition of a linear transformation $T: \R^3 \to \R^2$. A map $T$ is a linear transformation if the map $T$ satisfies:

1. $T(\mathbf{u}+\mathbf{v})=T(\mathbf{u})+T(\mathbf{v})$ for any $\mathbf{u}, \mathbf{v}\in \R^3$, and
2. $T(c\mathbf{v})=cT(\mathbf{v})$ for any $\mathbf{v} \in \R^3$ and $c\in \R$.

Solution.

Using the standard unit basis vectors, any vector $\mathbf{x}=\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}\in \R^3$ can be expressed as a linear combination
$\mathbf{x}=x_1\mathbf{e}_1+x_2\mathbf{e}_2+x_3\mathbf{e}_3.$ Since $T$ is a linear transformation, we have
\begin{align*}
T(\mathbf{x})&=T(x_1\mathbf{e}_1+x_2\mathbf{e}_2+x_3\mathbf{e}_3)\\
&=x_1T(\mathbf{e}_1)+x_2T(\mathbf{e}_2)+x_3T(\mathbf{e}_3)\6pt] &=x_1\begin{bmatrix} 1 \\ 4 \end{bmatrix}+x_2\begin{bmatrix} 2 \\ 5 \end{bmatrix}+x_3\begin{bmatrix} 3 \\ 6 \end{bmatrix}\\[6pt] &=\begin{bmatrix} x_1+2x_2+3x_3 \\ 4x_1+5x_2+6x_3 \end{bmatrix}. \end{align*} Therefore the formula is given by \[T(\mathbf{x})=\begin{bmatrix} x_1+2x_2+3x_3 \\ 4x_1+5x_2+6x_3 \end{bmatrix}.

Let $A$ be an $m \times n$ matrix. Let $\calN(A)$ be the null space of $A$. Suppose that \$\mathbf{u} \in...