MHB Linear Transformation of a Plane

[bwpbruce]

$\textbf{Problem}$
Let $\textbf{u}$ and $\textbf{v}$ be linearly independent vectors in $\mathbb{R}^3$, and let $P$ be the plane through $\textbf{u}, \textbf{v}$ and $\textbf{0}.$ The parametric equation of $P$ is $\textbf{x} = s\textbf{u} + \textbf{v}$ (with $s$, $t$ in $\mathbb{R}$). Show that a linear transformation $T: \mathbb{R}^3 \mapsto \mathbb{R}^3$ maps $P$ onto a plane through $\textbf{0}$ or onto a line through $\textbf{0}$ or onto just the origin $\mathbb{R}^3$. What must be true about $T\textbf{u}$

$\textbf{Solution}$

\begin{align*}T(\textbf{x}) &= T(s\textbf{u} + t\textbf{v}) \\&=T(s\textbf{u}) + T(t\textbf{u}) = sT\textbf{(u)} + tT\textbf{(v)}\end{align*}

\begin{align*}T(\textbf{0}) &= sT(\textbf{0}) + tT(\textbf{0})\\&=\textbf{0}\end{align*}

$T(\textbf{x})$ goes through the origin.

$T(\textbf{x})$ is a plane in $\mathbb{R}^3$ through $\textbf{0}$ in the case when $T(\textbf{(u)}) \ne tT(\textbf{(v)})$ and $T(\textbf{(v)} \ne sT\textbf{(u)}$.

$T(\textbf{x})$ is a line in $\mathbb{R}^3$ through $\textbf{0}$ in the case when $T(\textbf{(u)}) = tT(\textbf{(v)})$ or $T(\textbf{(v)} = sT\textbf{(u)}$

$T(\textbf{x})$ is a zero vector $\mathbb{R}^3$ in the case when $T(\textbf{u}) = T\textbf{v} = \textbf{0}$.

Can someone please provide feedback on this solution? Thanks.

 

[Evgeny.Makarov]

Show that a linear transformation $T: \mathbb{R}^3 \mapsto \mathbb{R}^3$ maps $P$ onto a plane through $\textbf{0}$ or onto a line through $\textbf{0}$ or onto just the origin $\mathbb{R}^3$. What must be true about $T\textbf{u}$

It may be a question of terminology, but I think that the phrase "A function $f$ maps a set $A$ onto a set $B$" means that $f$ is onto, i.e., a surjection. Thus, it is important to understand that you need to prove not only that that the image of every point from the original plane lies on the plain (or line, or point) generated by $0$, $T(\mathbf{u})$ and $T(\mathbf{v})$, but also that every point on the second plain is the image. I don't think you need to change your solution, but this should be kept in mind.

$T(\textbf{x})$ is a plane in $\mathbb{R}^3$ through $\textbf{0}$ in the case when $T(\textbf{(u)}) \ne tT(\textbf{(v)})$ and $T(\textbf{(v)} \ne sT\textbf{(u)}$.

This can be stated by saying that $T(\mathbf{u})$ and $T(\mathbf{v})$ are linearly independent.

$T(\textbf{x})$ is a line in $\mathbb{R}^3$ through $\textbf{0}$ in the case when $T(\textbf{(u)}) = tT(\textbf{(v)})$ or $T(\textbf{(v)} = sT\textbf{(u)}$

This does not prevent the case when both $T(\mathbf{u})$ and $T(\mathbf{v})$ are zero vectors.

Is there any significance in double parentheses?

 

[bwpbruce]

Is there any significance in double parentheses?

No, not really. I appreciate all your contributions.