Transformation of a Parallelogram

[bwpbruce]

$\textbf{Problem}$
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?

 

[Evgeny.Makarov]

Let
$P_n = a\textbf{u} + b\textbf{v}$ represent a point in set $P$ where $0 \le a \le 1$ and $0 \le b \le 1$.

Usually when you add a subscript $n$, this signifies a sequence where $n$ ranges over natural numbers.

$P'_n$ represent a point in the image of $P'$

Any point?

Then:
By property $(i)$ of the definition of linear transformation:

For those who don't know the particular version of the definition you use, it is a good idea to state this property.

\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*}

Usually a prime is a part of a name of a point, set, or other object. It's like in a programming language an identifier may consist of letters, digits and underscore. The identifier [m]point2[/m] does not mean the second element of an array [m]point[/m] or something like that; the whole word [m]point2[/m] is a single name. Similarly, unless prime has an established meaning like the derivative in calculus, adding it to a name does not signify applying an operation; a prime is simply a part of a name just like a letter. You wrote $(a\textbf{u} + b\textbf{v})'$, where prime is used an an operation, which you have not defined. Also, $P'$ is a set, but the right-hand side of the equation, i.e., $(a\textbf{u} + b\textbf{v})'$, is a single point.

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 $\textbf{u}' = \textbf{v}' = \textbf{0}$.

These cases are not mutually exclusive and do not cover all possibilities.

 

[bwpbruce]

Usually when you add a subscript $n$, this signifies a sequence where $n$ ranges over natural numbers.
Any point?

You wrote $(a\textbf{u} + b\textbf{v})'$, where prime is used an an operation, which you have not defined. Also, $P'$ is a set, but the right-hand side of the equation, i.e., $(a\textbf{u} + b\textbf{v})'$, is a single point.

These cases are not mutually exclusive and do not cover all possibilities.

Using prime is more convenient. I don't see the harm in it. I've seen others use it. What's the best variable to use in place of $n$? How do I fix the mutually exclusive part?

 

[Evgeny.Makarov]

Using prime is more convenient. I don't see the harm

There is nothing convenient about using undefined operations. Besides, it must be a synonym for $T$. Why introduce another notation? I want to stress that even though you are free to introduce prime as a synonym for $T$, you have to explicitly define it this way.

What's the best variable to use in place of $n$?

What is your intention behind it? Why do you want to use a variable?

How do I fix the mutually exclusive part?

Think why they are not exclusive.

 

[blue_raver22]

Your solution is well-written and explains the reasoning behind why the image of a point in $P$ under transformation $T$ lies in the parallelogram determined by $T(\textbf{u})$ and $T(\textbf{v})$. However, you could provide more explanation on why the set $P'$ is a parallelogram if $\textbf{u}' \ne b\textbf{v}'$ and $\textbf{v}' \ne a\textbf{v}'$. This is because if $\textbf{u}'$ and $\textbf{v}'$ are not parallel, then they will form two sides of a parallelogram. Similarly, if $\textbf{u}'$ and $\textbf{v}'$ are not perpendicular, they will form two non-parallel sides of a parallelogram. You could also mention that the set $P'$ is a line segment if $\textbf{u}'$ or $\textbf{v}'$ is a scalar multiple of the other, and the set $P'$ is the zero vector if both $\textbf{u}'$ and $\textbf{v}'$ are zero vectors. Overall, your solution is clear and concise, but adding a bit more explanation on these points could enhance your explanation.