Show that a Parametric Equation Maps To Another Line By Linear Transformation.

Click For Summary
SUMMARY

The discussion focuses on demonstrating that a parametric equation, represented as $\textbf{x} = \textbf{p} + t\textbf{v}$, maps to another line or a single point through a linear transformation $T: \mathbb{R}^n \rightarrow \mathbb{R}^n$. The solution correctly applies the linear transformation property, resulting in $\textbf{y} = T(\textbf{p}) + tT(\textbf{v})$. It concludes that $\textbf{y}$ can either represent another line or a single point when $T(\textbf{v}) = \textbf{0}$. The final conclusion confirms that $\textbf{x}$ maps to $\textbf{y}$ via the transformation $T$.

PREREQUISITES
  • Understanding of linear transformations in vector spaces.
  • Familiarity with parametric equations in $\mathbb{R}^n$.
  • Knowledge of the properties of linear mappings.
  • Basic proficiency in mathematical notation and vector algebra.
NEXT STEPS
  • Study the properties of linear transformations in detail.
  • Explore examples of parametric equations in higher dimensions.
  • Investigate degenerate cases in linear transformations.
  • Learn about the implications of linear mappings on geometric interpretations.
USEFUL FOR

Mathematicians, students of linear algebra, and anyone interested in understanding the behavior of linear transformations and parametric equations in vector spaces.

bwpbruce
Messages
60
Reaction score
1
$\textbf{Problem}$
Given $\textbf{v} \ne \textbf{0}$ and $\textbf{p}$ in $\mathbb{R}^n$, the line through $\textbf{p}$ in the direction of $\textbf{v}$ is given by $\textbf{x} = \textbf{p} + t\textbf{v}$. Show that linear transformation $T: \mathbb{R}^n \rightarrow \mathbb{R}^n$ maps this line onto another line or onto a single point.

$\textbf{My Solution}$:
$\textbf{x} = \textbf{p} + t\textbf{v}$

By Linear Transformation Property:
$T(\textbf{x}) =T(\textbf{p} + t\textbf{v})$
$T(\textbf{x}) =T\textbf{p} + t(T\textbf{v})$

Let $T(\textbf{x}) = \textbf{y}, T\textbf{p} = \textbf{q}, T\textbf{v} = \textbf{w}$
Then $\textbf{y} = \textbf{q} + t\textbf{w}$ is another parametric equation and $\textbf{y}$ is the other line that $\textbf{x}$ maps to except in the case where $\textbf{w} = \textbf{0}$. Then $\textbf{y} = \textbf{q}$

Result:
$\textbf{y} = \textbf{q} + t\textbf{w}$

or

$\textbf{y} = \textbf{q}$

Conclusion:
$\textbf{x} \mapsto \textbf{y}$ by $T$.

Check my solution please?
 
Last edited:
Physics news on Phys.org
Hi,

It's almost correct, but $y$ doesn't need to be a line, read again the statement and fill the degenrate case.
 
Fallen Angel said:
Hi,

It's almost correct, but $y$ doesn't need to be a line, read again the statement and fill the degenrate case.

Is that better?
 
Yes, now is completely correct :D
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 0 ·
Replies
0
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K