Permutation as Linear Transformation

Question

Could you help me please to move forward with the problem.

I'm trying to show that a function $\varphi_{\sigma }: \mathbb{R}^{n} \rightarrow \mathbb{R}^{n}$

$\varphi_{\sigma }(x_{1}, x_{2}, ... x_{n}) = (x_{i_{1}}, x_{i_{2}}, ... x_{i_{n}})$ is a linear transformation.

Where $\sigma =\begin{bmatrix} 1 &2 & ... &n \\ i_{1} & i_{2} & ... & i_{n} \end{bmatrix}$ from $S_{n}$ is a permutation.

To show it I need to check 2 conditions: 1. that sum of vectors under transformation and sum of transformations of two vectors are equal; 2. and product with scalar of a vector under transformation is equal to the product of the transformation of the vector with scalar.

So I have chosen some other vector y and try to test first condition.

I got $\varphi_{\sigma }(x_{1} + y_{1}, x_{2} + y_{2}, ... , x_{n} + y_{n}) = ..$ and It must be equal to $(x_{i_{1}} + y_{i_{1}}, x_{i_{2}} + y_{i_{2}}, ... x_{i_{n}} + y_{i_{n}}) $ But can I just put "=" between them? I don't know how to make this step. Is it that obvious? I mean, I can say, let's say $x_{1} + y_{1} = z_{1}$ which goes to $z_{i_{1}}$ but how should I then show that it equals to the sum of transformations of x and y?

Answer 1

Given $x=(x_1,\dots, x_n)$ and $y=(y_1,\dots, y_n)$, you can call $z=x+y$. That is, as you have said, $z=(z_1,\dots, z_n)$ where $z_i=x_i+y_i$ for $i=1,\dots,n$. Now by definition of $\varphi_\sigma$, we have $$\varphi_\sigma(z)=\varphi_\sigma(z_1,\dots, z_n)=(z_{i_1},\dots, z_{i_n})=(x_{i_1}+y_{i_1},\dots, x_{i_n}+y_{i_n})$$ since $z_i=x_i+y_i$ for all $i$.

On the other hand, by by definition of $\varphi_\sigma$, we also have $$\varphi_\sigma(x)=\varphi_\sigma(x_1,\dots, x_n)=(x_{i_1},\dots, x_{i_n}), \varphi_\sigma(y)=\varphi_\sigma(y_1,\dots, y_n)=(y_{i_1},\dots, y_{i_n}),$$ which implies that $$\varphi_\sigma(x)+\varphi_\sigma(y)=(x_{i_1},\dots, x_{i_n})+(y_{i_1},\dots, y_{i_n})= (x_{i_1}+y_{i_1},\dots, x_{i_n}+y_{i_n}).$$ Combining the above two equalities, we obtain $$\varphi_\sigma(z)=\varphi_\sigma(x+y)=\varphi_\sigma(x)+\varphi_\sigma(y)$$ as required.

Answer 2

If you're being explicitly asked to show it (I suppose it's a homework problem), you can't just claim that the result is obvious -- even if it is.

In order to structure the proof, you should give names to the entire vectors: Define $x=(x_1,\ldots,x_n)$, $y=(y_1,\ldots,y_n)$. Then set $z=\varphi_\sigma(x)+\varphi_\sigma(y)$ and $w=\varphi_\sigma(x+y)$. This gives you the vocabulary to speak about what it is you need to prove, namely for each $i$ the $i$th component of $z$ equals the $i$th component of $w$. Compute each of the components using the various definitions, and point out that they are equal as required.

Answer 3

Well, essentially, let $z = x + y$. Then, on one hand, $$ \begin{eqnarray} \varphi_{\sigma }(z_{1}, z_{2}, \ldots, z_{n}) &=& (z_{i_{1}}, z_{i_{2}}, \ldots, z_{i_{n}}) = (x_{i_{1}} + y_{i_{1}} , x_{i_{2}} + y_{i_{2}}, \ldots, x_{i_{n}} + y_{i_{n}}) \\ &=& (x_{i_{1}} , x_{i_{2}}, \ldots, x_{i_{n}}) + (y_{i_{1}} , y_{i_{2}}, \ldots, y_{i_{n}}) = \varphi_{\sigma }(x_{1}, x_{2}, \ldots, x_{n}) + \varphi_{\sigma }(y_{1}, y_{2}, \ldots, y_{n}) \end{eqnarray} $$ On another hand, $\varphi_{\sigma }(z_{1}, z_{2}, \ldots, z_{n}) = \varphi_{\sigma }(x_{1} + y_{1}, x_{2}+y_{2}, \ldots, x_{n}+y_{n})$. This establishes property 1. Property 2 is proved similarly.