Transformation of a Line Segment

• MHB
• bwpbruce
Now I get it -- there was a typo in the left side of the equation. That expression should be T((1 - t)p + tq).f

bwpbruce

$\textbf{Problem}$
The line segment from $\textbf{p}$ to $\textbf{q}$ is the set of points of the form $(1 - t)\textbf{p} + t\textbf{q}$ for $0 \le t \le 1$ (as shown in the figure below). Show that a linear transformation, $T$, maps this line segment onto a line segment or onto a single point.

$\textbf{Solution}$
$\textbf{x} = (1 - t)\textbf{p} + t\textbf{q}$ and $0\le t \le1$
\begin{align*}T\textbf{(x)} &= T((1 - t\textbf{p}) + t\textbf{q}) \\&=T(\textbf{p} + t\textbf{p}) + Tt\textbf{q} \\&=T\textbf{p} + Tt\textbf{p} + Tt\textbf{q}, \text{ when } 0 < t \le 1 \\&=T\textbf{p} + T(0)\textbf{p} + T(0)\textbf{q}, \text{when } t = 0\\&=T\textbf{p}\end{align*}

Conclusion:
$\textbf{x}$ maps onto line $T\textbf{p} + tT\textbf{p} + tT\textbf{q}$ when $0 < t \le1$ or point $T\textbf{p}$ when $t = 0$

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@bwpbruce, I don't follow why T((1 - tp) + tq) = T(p + tp) + T(tq).

@bwpbruce, I don't follow why T((1 - tp) + tq) = T(p + tp) + T(tq).
Thanks for pointing this out seven years later.

Thanks for pointing this out seven years later.
I didn't see it when you posted it. We're in the process of responding to very old posts that haven't received any replied, and I happened upon this one.

I didn't see it when you posted it. We're in the process of responding to very old posts that haven't received any replied, and I happened upon this one.
I'm just glad you guys are still here.

Greg Bernhardt
$\textbf{Problem}$
The line segment from $\textbf{p}$ to $\textbf{q}$ is the set of points of the form $(1 - t)\textbf{p} + t\textbf{q}$ for $0 \le t \le 1$ (as shown in the figure below). Show that a linear transformation, $T$, maps this line segment onto a line segment or onto a single point.

View attachment 308170
$\textbf{Solution}$
$\textbf{x} = (1 - t)\textbf{p} + t\textbf{q}$ and $0\le t \le1$
\begin{align*}T\textbf{(x)} &= T((1 - t\textbf{p}) + t\textbf{q}) \\&=T(\textbf{p} + t\textbf{p}) + Tt\textbf{q} \\&=T\textbf{p} + Tt\textbf{p} + Tt\textbf{q}, \text{ when } 0 < t \le 1 \\&=T\textbf{p} + T(0)\textbf{p} + T(0)\textbf{q}, \text{when } t = 0\\&=T\textbf{p}\end{align*}

Conclusion:
$\textbf{x}$ maps onto line $T\textbf{p} + tT\textbf{p} + tT\textbf{q}$ when $0 < t \le1$ or point $T\textbf{p}$ when $t = 0$

Of course $T(\mathbf{x}(0))$ must be a single point; that's how functions work.
We have $$\begin{split} T(\mathbf{x}(t)) &= T((1-t)\mathbf{p} + t\mathbf{q}) \\ &= T(\mathbf{p}) - tT(\mathbf{p}) + tT(\mathbf{q}). \end{split}$$ What happens if $T(\mathbf{p}) = T(\mathbf{q})$, ie. when $T(\mathbf{q} - \mathbf{p}) = \mathbf{0}$?