Linear transformation easy question

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
4 replies · 3K views
synkk
Messages
216
Reaction score
0
EJZ2J.png


Sorry I feel like an idiot for asking this but why is part c and b not a linear transformation? The origin would still be (0,0) and it's an expression in x and y terms so I'm confused?

thanks
 
Physics news on Phys.org
They are not "linear transformations" because they are not linear. Specifically "[itex]x^2[/itex]" and "[itex]xy[/itex]" are not linear.

More precisely, a linear transformation, L, must satisfy L(x+ y)= L(x)+ L(y) and L(ax)= aLx for a any number. For b,
[tex]L\left(\begin{pmatrix}x \\ y\end{pmatrix}+ \begin{pmatrix}u \\ v\end{pmatrix}\right)= \begin{pmatrix}(x+u)^2 \\ y+ v\end{pmatrix}= \begin{pmatrix}x^2+ 2u+ u^2 \\ y+ v\end{pmatrix}[/tex]
not
[tex]L\left(\begin{pmatrix}x \\ y\end{pmatrix}\right)+ L\left(\begin{pmatrix}u \\ v\end{pmatrix}\right)= \begin{pmatrix} x^2 \\ y\end{pmatrix}+ \begin{pmatrix}u^2 \\ v\end{pmatrix}= \begin{pmatrix}x^2+ u^2\\ y+ v\end{pmatrix}[/tex]

Similarly,
[tex]L\left(a\begin{pmatrix}x \\ y \end{pmatrix}\right)= L\left(\begin{pmatrix}ax \\ ay\end{pmatrix}\right)= \begin{pmatrix}a^2x^2\\ ay\end{pmatrix}[/tex]
which is not the same as
[tex]aL\left(\begin{pmatrix}x \\ y\end{pmatrix}\right)= \begin{pmatrix}ax^2 \\ ay\end{pmatrix}[/tex]
and the same argument for (c).
 
HallsofIvy said:
They are not "linear transformations" because they are not linear. Specifically "[itex]x^2[/itex]" and "[itex]xy[/itex]" are not linear.

More precisely, a linear transformation, L, must satisfy L(x+ y)= L(x)+ L(y) and L(ax)= aLx for a any number. For b,
[tex]L\left(\begin{pmatrix}x \\ y\end{pmatrix}+ \begin{pmatrix}u \\ v\end{pmatrix}\right)= \begin{pmatrix}(x+u)^2 \\ y+ v\end{pmatrix}= \begin{pmatrix}x^2+ 2u+ u^2 \\ y+ v\end{pmatrix}[/tex]
not
[tex]L\left(\begin{pmatrix}x \\ y\end{pmatrix}\right)+ L\left(\begin{pmatrix}u \\ v\end{pmatrix}\right)= \begin{pmatrix} x^2 \\ y\end{pmatrix}+ \begin{pmatrix}u^2 \\ v\end{pmatrix}= \begin{pmatrix}x^2+ u^2\\ y+ v\end{pmatrix}[/tex]

Similarly,
[tex]L\left(a\begin{pmatrix}x \\ y \end{pmatrix}\right)= L\left(\begin{pmatrix}ax \\ ay\end{pmatrix}\right)= \begin{pmatrix}a^2x^2\\ ay\end{pmatrix}[/tex]
which is not the same as
[tex]aL\left(\begin{pmatrix}x \\ y\end{pmatrix}\right)= \begin{pmatrix}ax^2 \\ ay\end{pmatrix}[/tex]
and the same argument for (c).
Hi thank you for the response, I should of mentioned I did use to properties of linear transformations as you did in order to see that they are not linear but in the solutions they simple state that: B) is not ∵x→x^2 is not linear and c) is not ∵y→x+xy is not linear. I just don't understand how it doesn't make it linear (of course using the properties it shows it's not linear) but I'm trying to understand it in plain words of why the transformation of y --> x +xy is not linear.

thank you again
 
Then I have no clue what you mean by "in plain words". The "plainest" words I can use are what I said before: [itex]x^2[/itex] is not "linear" because [itex](a+ b)^2= a^2+ 2ab+ b^2\ne a^2+ b^2[/itex] and [itex]xy[/itex] is not linear because [itex](a+ b)(c+ d)= ac+ ad+ bc+ bd\ne ac+ bd[/itex].
 
HallsofIvy said:
Then I have no clue what you mean by "in plain words". The "plainest" words I can use are what I said before: [itex]x^2[/itex] is not "linear" because [itex](a+ b)^2= a^2+ 2ab+ b^2\ne a^2+ b^2[/itex] and [itex]xy[/itex] is not linear because [itex](a+ b)(c+ d)= ac+ ad+ bc+ bd\ne ac+ bd[/itex].

Brilliant thank you.