# Showing that a transformation is not linear

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?

• That looks fine. Just make sure you actually give values for what $a_1, b_1, a_2, b_2$ should be. – Zach L. Oct 5 '12 at 0:40
• What do you mean "give values"? Even in an abstract question like this? – CodyBugstein Oct 5 '12 at 0:41
• Yes. To prove it was linear, you'd have to do it abstractly as you have, but to prove it's not, you need to demonstrate one explicit counterexample. As you've written it now, we have no way to guarantee $a_1^2+b_1^2\neq a_1^2+2a_1b_1+b_1^2$ (for instance, what if $a_1=0?$) – Kevin Carlson Oct 5 '12 at 1:36
• @Imray: See my attempt. :) – mrs Oct 5 '12 at 8:14
• "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. – Marc van Leeuwen Oct 5 '12 at 9:40

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 $\mathrm{Hom}(\mathbb R^2,\mathbb R^2)$.

• Awesome - that's much more efficient! Thank you – CodyBugstein Oct 5 '12 at 14:17
• Efficient, indeed! I trust you're sound asleep as I write this. (I hope your dreams are blissful Perhaps my note will greet you upon awakening or near the start of a new day!) $\;\;\ddot\smile\;\;$ – Namaste Mar 23 '13 at 0:47
• @amWhy: Nice dreams I had. I am sure you made them for me. ;-) – mrs Mar 23 '13 at 6:06

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

• I like your Icon and (+1). :-) – mrs Oct 6 '12 at 5:53