Transforming a weighted sum to another one with different weights

Question

Let $x_1, x_2, \dots, x_n$ and $w_1, w_2, \dots, w_n$ be real numbers and $f_1 = x_1+\dots+x_n$ and $f_2 = w_1x_1+\dots+w_nx_n$. Is it possible to find a transformation $T$ which $T(f_1)=f_2$.

Answer 1

Only if all the $w_i$'s are equal.

In general you need to know all the $x_i$'s to multiply them by different numbers.

Answer 2

welcome to MSE. It is not clarified problem ,but I have a suggestion $$f_1=1_{1\times n}X_{n\times 1}=\begin{pmatrix} 1 & 1 & \cdots & 1 \end{pmatrix} \left( \begin{array}{ccc} x_1 \\x_2 & \\ x_3 \\ \vdots & \\x_n\end{array} \right) $$ so $$f_2=T(f_1)$$ where $$\begin{pmatrix} w_1 & 0 & \cdots & 0\\0&w_2 & \cdots & 0\\ \vdots & \ddots & \vdots\\ 0 & 0& \cdots & w_n \end{pmatrix} $$ so $$T(f_1)\\=1_{1 \times n}W_{n\times n}X_{n\times 1}\\= \begin{pmatrix} w_1 & w_2 & \cdots & w_n \end{pmatrix} X_{n\times 1}\\=w_1x_1+w_2x_2+...+w_nx_n$$ you can rewrite $$T(f_1)=(W_{n\times n}1_{n\times 1})^TX_{n\times 1}$$ I mean $^t$ transpose