Answer
See explanation
Work Step by Step
Given :
Vectors $\mathbf{u}$ and $\mathbf{v}$ in $\mathbb{R}^{3}(\text { linearly independent })$
Plane $P(\text { through } \mathbf{u}, \mathbf{v} \text { and } \mathbf{0})$
Linear Transformation $T: \mathbb{R}^{3} \mapsto \mathbb{R}^{3}$
Parametric Equation $(\text { of } P):$ $\mathbf{x}=s \mathbf{u}+t \mathbf{v},(s, t)$ in $\mathbb{R}$
2
Goal :
Show that $T$ maps $P$ onto a plane through 0 or onto a line through 0 or onto the origin in $\mathbb{R}^{3}$.
Determine what must be true about $T(\mathbf{u})$ and $T(\mathbf{v})$ in order for the image of the plane $P$ to be a plane.
3
Affine Transformation Properties of Linear Transformation
\[
\operatorname{Span}\{\mathbf{u}, \mathbf{v}\}
\]
Plan:
Verify that $T(\mathbf{x})$ satisfies property $T(s \mathbf{x}+t \mathbf{y})=$ $s T(\mathbf{x})+t T(\mathbf{y})$ of the definition of linear transformation Verify that $T(\mathbf{x})$ goes through the origin.
Solve:
\[
\begin{array}{c}
T(\mathbf{x})=T(s \mathbf{u}+t \mathbf{v}) \\
=T(s \mathbf{u})+T(t \mathbf{v})=s T(\mathbf{u})+t T(\mathbf{v}) \\
T(\mathbf{0})=s T(\mathbf{0})+t T(\mathbf{0}) \\
=\mathbf{0}
\end{array}
\]$T(\mathbf{x})$ goes through the origin.
Result:
$T(\mathbf{x})$ is a plane in $\mathbb{R}^{3}$ through 0 in the case when $T((\mathbf{u})) \neq t T((\mathbf{v}))$ and $T((\mathbf{v}) \neq s T(\mathbf{u})$
$T(\mathbf{x})$ is a line in $\mathbb{R}^{3}$ through $\mathbf{0}$ in the case when $T((\mathbf{u}))=t T((\mathbf{v}))$ or $T((\mathbf{v})=s T(\mathbf{u})$
$T(\mathbf{x})$ is a zero vector $\mathbb{R}^{3}$ in the case when $T(\mathbf{u})=$ $T \mathbf{v}=\mathbf{0}$
