0
$\begingroup$

I'm asked to show that transformation $ T(a_1, a_2) = (a_1, a_1^2)$ is not linear.

My attempt at was by showing that it is not closed under addition. That is; $T[(a_1, a_2)+(b_1,b_2)] \ne T(a_1, a_2)+T(b_1, b_2)$

and I got $ \begin{matrix}a_1+b_1 \\a_1^2+b_1^2 \end{matrix} \ \ \ne \ \ \begin{matrix}a_1+b_1 \\a_1^2 + 2a_1b_1+b_1^2 \end{matrix} $

Is this a good way to show it's not linear? Is it valid?

  • 1
    "closed under addition" is not the right terminology (ther is no set to be closed here). You should say either "additive" or "compatible with addition" or something like that.2012-10-05

2 Answers 2

2

You know that if the function $T: \mathbb R^n\to\mathbb R^m$ wants to be a linear transformation; it should satisfy:

$T(u + v) = T(u) + T(v)$ and $T(cu) = cT(u)$ for all $u,v\in\mathbb R^n$.

Let your function is a linear transformation so we have $T(cu) = cT(u)$ where in $u=(x,y)\in\mathbb R^2$ and $c$ is an arbitrary constant in our field $\mathbb R$. Therefore: $T(cu)=T(cx,cy)=(cx,c^2x^2)$ should be equal to $cT(x,y)=c(x,x^2)=(cx,cx^2)$ for any scalar $c\in \mathbb R$. Or $(cx,c^2x^2)=(cx,cx^2)$ for any scalar $c\in \mathbb R$. But it is obviously not true for all $c$. So your function is not a linear transformation in $\mathrm{Hom}(\mathbb R^2,\mathbb R^2)$.

  • 0
    @amWhy: Nice dreams I had. I am sure you made them for me. ;-)2013-03-23
1

I thought of another one: $ T(-x_1, -x_2) = (x_1, x_1^2) \ \ne -T(x_1, x_2) = -(x_1, x_1^2)=(-x_1, -x_1^2)$

  • 1
    I like your Icon and (+1). :-)2012-10-06