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I have the following transformation:

$T : \Bbb R^3 → \Bbb R^2 , T (x_1 , x_2 , x_3 ) = (x_1 − x_2 , 2x_2 )$

I need to determine whether it's a linear transformation or not.

I understand that for a transformation to be linear, I should have: \begin{equation*} T(av + bu) = aT(v) + bT(u) \end{equation*}

Now for the question above let's say I do the following:

$u = ax_1 + ax_2 + ax_3$

$v = bx_1 + bx_2 + bx_3$

So now, I have to show that $T(au+bv) = aT(u) + bT(v)$. I'm quite stuck, and I'm not sure how to proceed. Could anyone guide me on where I'm wrong and how to proceed?

Thanks.

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  • $\begingroup$ You want $u$ and $v$ to be vectors (3 components) not scalars... $\endgroup$ – TravisJ May 28 '15 at 14:13
  • $\begingroup$ T is linear if and only if there is a matrix $A$ with $Av=T(v)$ $\endgroup$ – Peter May 28 '15 at 14:15
  • $\begingroup$ Just do the footwork, write $T(au+bv)$ expand it, then go "BUT WAIT! This $=aT(u)+bT(v)$ - literally it. Or find a matrix then go "all matrices are linear maps!" BAM done. $\endgroup$ – Alec Teal May 28 '15 at 14:15
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$v=(x_1,x_2,x_3)$ & $w=(y_1,y_2,y_3)$

We have then $$T(av+bw)=(ax_1+by_1-(ax_2+by_2),2(ax_2+by_2))=$$ $$=(ax_1+by_1-ax_2-by_2,2ax_2+2by_2)$$ $$=(ax_1-ax_2+by_1-by_2,2ax_2+2by_2)$$ $$=(ax_1-ax_2,2ax_2)+(by_1-by_2,2by_2)$$ $$=(a(x_1-x_2),2ax_2)+(b(y_1-y_2),2by_2)$$ $$=a(x_1-x_2,2x_2)+b(y_1-y_2),2y_2)$$ $$=aT(v)+bT(w)$$

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  • $\begingroup$ Where does the $y_a$ come from? I apologize for not being able to see it, but how to you reach the first line as well? $\endgroup$ – nTuply May 28 '15 at 14:20
  • $\begingroup$ Sorry it was a typo, it was meant to be $y_1$ $\endgroup$ – Zelos Malum May 28 '15 at 14:21
  • $\begingroup$ As for how I reach it, it is simple vector addition and scalar multiplicatoin where $a(x_1,x_2,x_3)=(ax_1,ax_2,ax_3)$ and vector addition being component wise. $\endgroup$ – Zelos Malum May 28 '15 at 14:22
  • $\begingroup$ I now understand it perfectly. Thanks a lot. $\endgroup$ – nTuply May 28 '15 at 14:24
  • $\begingroup$ Just glad to help :) That's what we are here for! $\endgroup$ – Zelos Malum May 28 '15 at 14:56
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You would do better to say

$\mathbf{v} = (v_1,v_2,v_3)$

$\mathbf{u} = (u_1,u_2,u_3)$

so you can say $a\mathbf{v}+b\mathbf{u} = (av_1+bu_1, av_2+bu_2, av_3+bu_3)$.

Now state what $T(a\mathbf{v}+b\mathbf{u})$ is and what $aT(\mathbf{v})+bT(\mathbf{u})$ is and whether they are equal.

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My first inclination for your approach would be to do exactly as Zelos Malum has done in their answer.

Alternatively, instead of trying to tackle the "linear-combination-preservation" property all at once for a transformation to determine whether it's linear, it might be helpful just to break that property down into

  1. Closure under vector addition Here you need to show that $T(\vec{u} + \vec{v}) = T(\vec{u}) + T(\vec{v})$.
  2. Closure under scalar multiplication (where your scalar is defined over the appropriate field ($\mathbb{Q}, \mathbb{R}, \mathbb{C}$, etc.). This means you need to show that if $k$ is a scalar, then $T(k\vec{v}) = kT(\vec{v})$.

So I'll demonstrate by letting $\vec{u}, \vec{v} \in \mathbb{R}^3$ such that $\vec{u} = (u_1, u_2, u_3)$ and $\vec{v} = (v_1, v_2, v_3)$. Then we have $$ T(\vec{u} + \vec{v}) \\ = T((u_1 + v_1), (u_2 + v_2), (u_3 + v_3)) \\ = ((u_1 + v_1) - (u_2+v_2), 2(u_2+v_2)) \\ = ((u_1-u_2)+(v_1-v_2), 2u_2+2v_2) \\ = (u_1-u_2, 2u_2) + (v_1-v_2, 2v_2) \\ = T(\vec{u}) + T(\vec{v}). $$

Thus, we can conclude that $T$ is closed under vector addition.

Next, let $k$ be a scalar from the same field as the components of $\vec{v}$; that is, let $k \in \mathbb{R}$. Then we have $$ T(k\vec{v}) \\ = T((kv_1, kv_2, kv_3)) \\ = (kv_1-kv_2, 2kv_2) \\ = (k(v_1-v_2), k(2v_2))\\ = k(v_1-v_2, 2v_2) \\ = kT(\vec{v}). $$

This means that $T$ is closed under scalar multiplication. This logically equivalent to saying that $T$ preserves linear combinations of vectors and so $T$ is linear.

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  • $\begingroup$ Their? :P I be a he! $\endgroup$ – Zelos Malum May 28 '15 at 15:15
  • $\begingroup$ @ZelosMalum Just being politically correct...Besides, I'd rather make the number disagree (plural vs. singular) instead of referring to you as an "it"! Because this just doesn't sound right: "...exactly as Zelos Malum has done in its answer." :P $\endgroup$ – Xoque55 May 28 '15 at 17:48
  • $\begingroup$ Political correctness is incorrect :P $\endgroup$ – Zelos Malum May 28 '15 at 17:49
  • $\begingroup$ @ZelosMalum I agree, though I'd rather walk on eggshells when it comes to stuff like that and not risk getting lectured on the sexist connotations when using the "generic he/him/his." $\endgroup$ – Xoque55 May 28 '15 at 17:53
  • $\begingroup$ I think Zelos is sufficiently masculine :P And there is nothing sexist using he or she. If they say it, they are idiots whose opinion is worthless $\endgroup$ – Zelos Malum May 28 '15 at 17:54

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