Let $f:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ such that $f(x,y)=(\frac{x}{3}+11y^2,-2y)$.
Let $A=\{(3y^2,y)|y\in \mathbb{R}\}$.
Show that $A$ is invariant under $f$.
What is it asking?
Asked
Active
Viewed 162 times
2
-
It means the set does not change under f! – smanoos Nov 15 '13 at 01:52
1 Answers
1
See that
$f(3y^2,y)=(\frac{3y^2}{3}+11y^2,-2y)=(12y^2,-2y)=(3(-2y)^2,(-2y))$
which is obiviously in the set A.
smanoos
- 3,061
-
in $A$ because if we put $Y=-2y$ then $f(3y^2,y)=(3Y^2,Y)$ has the general element's of $A$ form. – Mohamed Nov 15 '13 at 02:17
-
-
Thanks smanoos. My comment was for help more the questionner. Your answer is complete. – Mohamed Nov 15 '13 at 02:23