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I'm studying for a test, and I need help with this problem. I am not sure how to prove that this is not linear due to the notation. The comma is throwing me off.

Show that the transformation $T$ defined by $T(x_1,x_2)$ = $(x_1^2-2x_2, x_1+5x_2)$ is not linear.

I know that the definition of a linear transformation involves:

  1. $T(u+v)=T(u)+T(v)$ for all $u, v$ in the domain of $T$.

  2. $T(c*u) = c*T(u)$ for all scalars $c$ and all $u$ in the domain of $T$.

  3. $T(cu + dv) = cT(u) + dT(v)$ for all vectors $u, v$ in the domain of $T$ and all scalars $c, d$

  4. $T(0) = 0$ if $T$ is linear

However, I'm not sure how to use this definition with the specific function given.

Would $T(x_1 + x_2)$ = $T(x_1) + T(x_2)$ work?

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    The $x_1^2$ tells you it won't be linear, and suggests some numbers to try to show non-linearity.2012-09-12

3 Answers 3

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Try testing $T((x_1,x_2)+(y_1,y_2))$ and see if it equals $T((x_1,x_2))+T((y_1,y_2))$.

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    Okay, fixed it.2012-09-12
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$T(4,2)=(12,14)$, $T(2,1)=(2,7)$,

$2T(2,1)\neq T(4,2)$

so T is not linear.

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    @LearningPython yes it is right.2012-09-12
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The transformation $T$ in fact may be linear. We are not told what the domain of $T$ and the field of scalars is! If $F$ is a field of characteristic 2, then $(a+b)^2=a^2+b^2$ holds for all $a,b\in F$. However, $(c\cdot x)^2=c\cdot x^2$ with $x\ne0$ implies $c=0\lor c=1$ in any field. Hence we see that $T\colon F\times F\to F\times F$ is not $F$-linear, but atleast it is $\mathbb F_2$-linear.