Linear transformation easy question

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Part B and C are not considered linear transformations because they do not satisfy the properties required for linearity. Specifically, the expressions x^2 and xy fail to meet the criteria that L(x+y) must equal L(x) + L(y) and L(ax) must equal aL(x). For instance, the transformation L applied to (x+u) results in (x+u)^2, which expands to x^2 + 2xu + u^2, differing from the sum of L(x) and L(u). Similarly, for the transformation involving xy, the product does not adhere to the linearity conditions. Thus, both transformations are classified as non-linear.
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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
 
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They are not "linear transformations" because they are not linear. Specifically "x^2" and "xy" 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,
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}
not
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}

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

Similarly,
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}
which is not the same as
aL\left(\begin{pmatrix}x \\ y\end{pmatrix}\right)= \begin{pmatrix}ax^2 \\ ay\end{pmatrix}
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: x^2 is not "linear" because (a+ b)^2= a^2+ 2ab+ b^2\ne a^2+ b^2 and xy is not linear because (a+ b)(c+ d)= ac+ ad+ bc+ bd\ne ac+ bd.
 
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: x^2 is not "linear" because (a+ b)^2= a^2+ 2ab+ b^2\ne a^2+ b^2 and xy is not linear because (a+ b)(c+ d)= ac+ ad+ bc+ bd\ne ac+ bd.

Brilliant thank you.
 
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