Solution: Recall that if $V$ and $W$ are vector spaces over $\mathbb{R} $, then $T:V\to W$ is a linear transformation if
\[
T\left( \alpha v_1+v_2 \right) = \alpha T\left( v_1 \right) + T\left( v_2 \right)
\]
for $v_1,v_2\in V$ and $\alpha \in \mathbb{R} $.
-
Let $\mathbf{x}, \mathbf{y} \in \mathbb{R}^3$ and $\alpha \in \mathbb{R} $ be given. Then
\begin{align*}
T(\alpha \mathbf{x} + \mathbf{y} ) & = T \left(
\alpha
\begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
\end{bmatrix} +
\begin{bmatrix}
y_1 \\
y_2 \\
y_3 \\
\end{bmatrix}
\right) \\[1ex]
& = T \left(
\begin{bmatrix}
\alpha x_1 + y_1 \\
\alpha x_2 + y_2\\
\alpha x_3 + y_3\\
\end{bmatrix}
\right) \\
& = \begin{bmatrix}
\alpha x_1 + y_1 + 2\left( \alpha x_2 + y_2 \right) \\
3\left( \alpha x_1 + y_1 \right) - 2\left( \alpha x_2 + y_2 \right) \\
\end{bmatrix} \\[1ex]
& = \begin{bmatrix}
\alpha x_1 + 2\alpha x_2 \\
3\alpha x_1 - 2\alpha x_2 \\
\end{bmatrix} +
\begin{bmatrix}
y_1 + 2y_2 \\
3y_1 - 2y_2 \\
\end{bmatrix} \\
& = \alpha T \left(
\begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
\end{bmatrix}
\right) + T \left(
\begin{bmatrix}
y_1 \\
y_2 \\
y_3 \\
\end{bmatrix}
\right).
\end{align*}
Thus, $T$ is a linear transformation.
-
We need to find the matrix of $T$ with respect to the standard basis. We denote the standard basis of $R^n$ by $\mathbf{e}_1, \mathbf{e} _2,\dots, \mathbf{e} _n$, where $\mathbf{e} _i$ has $1$ only at the $i$th place and other coordinates are zero. We have
\begin{align*}
T \left( \mathbf{e} _1 \right) & =
\begin{bmatrix}
1 \\
3 \\
\end{bmatrix} = 1\cdot \mathbf{e} _1 + 3 \mathbf{e} _2 \\[1ex]
T\left( \mathbf{e} _2 \right) & =
\begin{bmatrix}
2 \\
-2 \\
\end{bmatrix} = 2 \mathbf{e} _1 - 2 \mathbf{e} _2 \\[1ex]
T \left( \mathbf{e} _3 \right) & =
\begin{bmatrix}
0 \\
0 \\
\end{bmatrix} = 0 \cdot \mathbf{e} _1 + 0 \cdot \mathbf{e} _2.
\end{align*}
Thus, the matrix of $T$ with respect to the standard basis will be
\[
[T] =
\begin{bmatrix}
1 & 2 & 0 \\
3 & -2 & 0 \\
\end{bmatrix}.
\]