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?


  • $\begingroup$ 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

$v=(x_1,x_2,x_3)$ & $w=(y_1,y_2,y_3)$

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

  • $\begingroup$ 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

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.


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.

  • $\begingroup$ 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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