- #1
bwpbruce
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$\textbf{Problem}$
Let $\textbf{u}$ and $\textbf{v}$ be vectors in $\mathbb{R}^n$. It can be shown that the set $P$ of all points in the parallelogram determined by $\textbf{u}$ and $\textbf{v}$ has the form $a\textbf{u} + b\textbf{v}$, for $0 \le a \le 1, 0 \le b \le 1$. Let $T: \mathbb{R}^n \rightarrow \mathbb{R}^m$ be a linear transformation. Explain why the image of a point in $P$ under transformation $T$ lies in the parallelogram determined by $T\textbf{(u)}$ and $T(\textbf{(v)}$.
$\textbf{Solution}$
Let
$P_n = a\textbf{u} + b\textbf{v}$ represent a point in set $P$ where $0 \le a \le 1$ and $0 \le b \le 1$.
$P'$ represent the image of $P$.
$P'_n$ represent a point in the image of $P'$
Then:
By property $(i)$ of the definition of linear transformation:
\begin{align*}P' &= (a\textbf{u} + b\textbf{v})' \\&= a\textbf{u}' + b\textbf{v}', 0 \le a \le 1, 0 \le b \le 1\end{align*}
The set $P'$ is a parallelogram if $\textbf{u}' \ne b\textbf{v}'$ and $\textbf{v}' \ne a\textbf{v}'$.
The set $P'$ is a line segment if $\textbf{u}' = \textbf{0}$ or $\textbf{v}' = \textbf{0}$
The set $P'$ is the zero vector if $\textbf{u}' = \textbf{v}' = \textbf{0}$.
In each case $P'_n$ is a point in $P'$.
Can someone please provide some feedback on my solution? Thanks?
Let $\textbf{u}$ and $\textbf{v}$ be vectors in $\mathbb{R}^n$. It can be shown that the set $P$ of all points in the parallelogram determined by $\textbf{u}$ and $\textbf{v}$ has the form $a\textbf{u} + b\textbf{v}$, for $0 \le a \le 1, 0 \le b \le 1$. Let $T: \mathbb{R}^n \rightarrow \mathbb{R}^m$ be a linear transformation. Explain why the image of a point in $P$ under transformation $T$ lies in the parallelogram determined by $T\textbf{(u)}$ and $T(\textbf{(v)}$.
$\textbf{Solution}$
Let
$P_n = a\textbf{u} + b\textbf{v}$ represent a point in set $P$ where $0 \le a \le 1$ and $0 \le b \le 1$.
$P'$ represent the image of $P$.
$P'_n$ represent a point in the image of $P'$
Then:
By property $(i)$ of the definition of linear transformation:
\begin{align*}P' &= (a\textbf{u} + b\textbf{v})' \\&= a\textbf{u}' + b\textbf{v}', 0 \le a \le 1, 0 \le b \le 1\end{align*}
The set $P'$ is a parallelogram if $\textbf{u}' \ne b\textbf{v}'$ and $\textbf{v}' \ne a\textbf{v}'$.
The set $P'$ is a line segment if $\textbf{u}' = \textbf{0}$ or $\textbf{v}' = \textbf{0}$
The set $P'$ is the zero vector if $\textbf{u}' = \textbf{v}' = \textbf{0}$.
In each case $P'_n$ is a point in $P'$.
Can someone please provide some feedback on my solution? Thanks?
Last edited:
