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Now I have:

$$\varphi\mathbf{x}=(x_2+x_3,2x_1+x_3,3x_1-x_2+x_3)$$

How do I check that this transformation is linear, and Also find its matrix?

As about matrix, the basis is not given, so it should be standard one: $$e_1=\langle1,0,0\rangle, e_2=\langle0,1,0\rangle, e_3=\langle0,0,1\rangle$$

As far as I know I have to put all the components into the matrix, like:

$$A =\begin{pmatrix}x_2+x_3 \\ 2x_1+x_3 \\ 3x_1-x_2+x_3 \end{pmatrix}$$

But I do not know how to proceed with given $e_{1,2,3}$ basis.

I have the final answer:

$$A_{\varphi} = \begin{pmatrix} 0 & 1 & 1 \\ 2 & 0 & 1 \\ 3 & -1 & 1 \end{pmatrix}$$

But asd I've said above, I do not know how to get the right result

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  • $\begingroup$ First tell what is the definition space and what the codomain...! And "to find its matrix"...with respect to what basis ? $\endgroup$ – DonAntonio Jun 9 '17 at 12:35
  • $\begingroup$ @Bye_World Good...and I guess the space is $\;\Bbb R^3\;$ both as domain and codomain. Let us let him confirm. Poor worded questions cause problems... $\endgroup$ – DonAntonio Jun 9 '17 at 12:39
  • $\begingroup$ @Bye_World And I don't see why you keep on intervening if the OP hasn't even said half a word, and I don't care who and why gave him the question: it is still a poorly worded one. Shall we wait until the OP address the questions or you intend to continue writing? $\endgroup$ – DonAntonio Jun 9 '17 at 12:41
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On linearity

In general, to show that a function $T:V\to W$ between real vector spaces $V$ and $W$ is linear, you need to show that it is

  • Homogeneous. I.e. for all $\mathbf v\in V$ and for all $k\in \Bbb R$, $T(k\mathbf v) = kT(\mathbf v)$.
  • Additive. I.e. for all $\mathbf v_1, \mathbf v_2 \in V$, $T(\mathbf v_1+\mathbf v_2) = T(\mathbf v_1) + T(\mathbf v_2)$.

In this particular case, showing homogeneity means proving (or disproving) that for all real $k$: $$\varphi(k\mathbf x) = \varphi\big(k(x_1,x_2,x_3)\big) = k\varphi(\mathbf x)$$ and showing additivity means (dis)proving that $$\varphi(\mathbf x +\mathbf y) = \varphi\big((x_1,x_2,x_3)+(y_1,y_2,y_3)\big) = \varphi(\mathbf x) + \varphi(\mathbf y)$$

On the Matrix Representation

Hint:

$$A =\begin{pmatrix}x_2+x_3 \\ 2x_1+x_3 \\ 3x_1-x_2+x_3 \end{pmatrix} = \pmatrix{ 0x_1 + 1x_2 + 1x_3 \\ 2x_1+0x_2+1x_3 \\ 3x_1-1x_2+1x_3}$$

Does that help?

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  • $\begingroup$ Aha. it explains the second part of question, first one is still unclear, how to check if transformation is linear or not $\endgroup$ – M.Mass Jun 9 '17 at 12:43
  • $\begingroup$ See if my edit helps. $\endgroup$ – user137731 Jun 9 '17 at 12:59
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I suppose that the transformations acts on the vector space $\mathbb{R}^3$ (over $\mathbb{R}$). If so you can prove linearity showing that: $$ \varphi(\mathbf x+a\mathbf y)=\varphi (\mathbf x)+a\varphi(\mathbf y) \quad \forall a\in\mathbb{R} \quad and \quad \forall \mathbf x,\mathbf y\in\mathbb{R}^3 $$

This is simple using the definition and the properties of addition and multiplication in $\mathbb {R}$.

To show that the matrix that you have found represents the transformation simply verify that:

$$A_{\varphi}\mathbf x^T =[\varphi (\mathbf x)]^T $$ That is obvious $$ \begin{pmatrix} 0 & 1 & 1 \\ 2 & 0 & 1 \\ 3 & -1 & 1 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}=\begin{pmatrix} x_1+x_2 \\ 2x_1+x_3 \\ 3x_1-x_2+x_3 \end{pmatrix}$$

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To find the matrix of a linear transformation $T$ with respect to the standard basis, first you have to know which space $T$ is going from (the "domain") and which space $T$ is sending vectors to (the "codomain"). Here you can see that the vectors in the domain have 3 components (they are $x_1,x_2,$ and $x_3$), and (if we're working with just real numbers, as is typical), and that the outputs have three components. So $T$ is a linear transformation from $ \mathbb{R}^3$ to $\mathbb{R}^3$. What you do is you find $T(e_1)$, $T(e_2)$, and $T(e_3)$. Those appear as the columns of the matrix.

So in this example, $T(e_1)=(0,2,3)$ (you just plug in $x_1=1,x_2=0,x_3=0$). This goes in the first column. Then find $T(e_2)$. That goes in the 2nd column. Then find $T(e_3)$. That goes in the third column.


How to check something is a linear transformation:

Let's review what the term linear transformation means. If $V$ and $W$ are vector spaces, a linear transformation from $V$ to $W$ is a map $T$ from $V$ to $W$ such that for every two scalars $c$ and $d$, and every two vector $v_1$ and $v_2$ in $V$, $T(cv_1+dv_2)=cT(v_1)+dT(v_2)$.

(If you're taking linear algebra, it would be a good idea to memorize this.)

As Bye_World says, you can also break this down into two criteria. First, check that for every single scalar $c$ and every single vector $v$ in $V$, $T(cv)=cT(v)$. Then check that for every two vectors $v_1$ and $v_2$ in $V$, $T(v_1+v_2)=T(v_1)+T(v_2)$. I'm going to walk you down how I would do the first thing to check. In parentheses I'll write down my mental picture.

If $\textbf{v}=(v_1,v_2,v_3)$ and $c$ is any scalar, does $T(c\textbf{v})=cT(\textbf{v})$?

(I have to convert the thing on the left of the equals sign to the thing on the right. First I'll write $c\textbf{v}$ in coordinates:)

$c(v_1,v_2,v_3)=(cv_1,cv_2,cv_3)$.

(Now I need to plug that into $T$. They tell you what $T$ does, so I use that.)

$T(cv_1,cv_2,cv_3)=(cv_2+cv_3, 2cv_1+cv_3, 3cv_1-cv_2+cv_3)$.

(OK. Now I'm trying to check that this is equal to $cT(\textbf{v})$. Let me write that in coordinates.)

$cT(\textbf{v})=cT(v_1,v_2,v_3)=c(v_2+v_3,2v_1+v_3,3v_1-v_2+v_3) ...

(So I rewrote the left hand side out and I rewrote the right hand side and I have to now check that I got the same thing. You can see that this is the same by distributing the $c$ across)

$=(cv_2+cv_3,2cv_1+cv_3,3cv_1-cv_2+cv_3)$.

(Now conclude:)

So $T(c\textbf{v})=cT(\textbf{v})$.

That's how you do the first part of checking that something is a linear transformation. I'll let you do the second part.

By the way, what, in words, is going on? Think of $T$ as a function, except instead of taking numbers to other numbers (like $f(x)=x^2$), it takes vectors to other vectors. The condition that $T(c\textbf{v})=cT(\textbf{v})$ can be visualized. If $\textbf{v}$ was a purple vector, then $T(\textbf{v})$ is where the purple vector goes. The statement says that, any scalar multiple of the purple vector goes to the corresponding scalar multiple of where the purple vector goes. So for instance, 2 times the purple vector has to go to 2 times where the purple vector goes, etc. I think some linear algebra students find this explanation helpful.

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  • $\begingroup$ Technically we don't know that $\operatorname{dom} \varphi = \Bbb R^3$, it's just implied (which is why Don was getting all bent out of shape in the comments). It could very well be $\varphi(x_1,x_2,x_3,x_4)=\dots$ where the map just doesn't use $x_4$. But good answer. +1 I like how you walked the reader through the proof that $T(cv) = cT(v)$. Just a note on formatting, tho. If you add double dollar signs instead of single ones you get $$\text{Math Mode}$$ which can help break up your answer a little more so it doesn't look like such a wall of text. Not a big deal tho. As I said, good answer. $\endgroup$ – user137731 Jun 9 '17 at 13:27

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