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$\textbf{Problem}$

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 $t = 0$

Please check for discrepancies?

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 $t = 0$

Please check for discrepancies?

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