{r setup, include=FALSE} knitr::opts_chunk$set(echo = TRUE)
Define the linear Transformation of
$T: , T:( \[\begin{bmatrix}x_1 & x_2 & x_3\end{bmatrix}\] ) = \[\begin{bmatrix}2x_1-x_2+5x_3 \\ -4x_1+2x_2-10x_3 \end{bmatrix}\]$
Veryfying that T is a linear Transformation if only these T repect this notation: \(T(\vec{u} +\vec{v}) = T(\vec{u}) + T(\vec{v})\) \(T(\beta\vec{u}) = \beta T(\vec{u})\)
Firt step Prove that \(T(\vec{u} +\vec{v}) = T(\vec{u}) + T(\vec{v})\)
$T( +) = \[\begin{bmatrix}2& -1 & 5 \\ -4& 2&-10 \end{bmatrix}\] ( \[\begin{bmatrix}x_1 + y_1 & x_2 + y_2& x_3 + y_3\end{bmatrix}\]) $
\(T(\vec{u} +\vec{v}) = \begin{bmatrix}2(x_1 +y_1)& -1(x_2 +y_2) & 5(x_3 +y_3) \\ -4(x_1 +y_1)& 2(x_2 +y_2)&-10(x_3 +y_3) \end{bmatrix}\)
\(T(\vec{u} +\vec{v}) = \begin{bmatrix}2x_1 + 2y_1-x_2-y_2+5x_3+5y_3 \\ -4x_1-4y_1+2x_2+2y_2-10x_3-10y_3 \end{bmatrix}\)
\(T(\vec{u} +\vec{v}) = T(\vec{u}) + T(\vec{v})\)
Second Steps
\(T(\vec{u}) + T(\vec{v}) = \begin{bmatrix}2& -1 & 5 \\ -4& 2&-10 \end{bmatrix} (\begin{bmatrix}x_1 & x_2 & x_3\end{bmatrix}) + \begin{bmatrix}2& -1 & 5 \\ -4& 2&-10 \end{bmatrix} (\begin{bmatrix}y_1 & y_2 & y_3\end{bmatrix})\) $T() + T()= \[\begin{bmatrix}2x_1-x_2+5x_3 \\ -4x_1+2x_2-10x_3 \end{bmatrix}\]3rd Steps
Prove that \(T(\beta\vec{u}) = \beta T(\vec{u})\)
$T() = \[\begin{bmatrix}2& -1 & 5 \\ -4& 2&-10 \end{bmatrix}\] ( \[\begin{bmatrix}\beta x_1 &\beta x_2 &\beta x_3\end{bmatrix}\]) $
$T() = \[\begin{bmatrix}2\beta x_1-\beta x_2+5 \beta x_3 \\ -4 \beta x_1+2\beta x_2-10\beta x_3 \end{bmatrix}\]$
$T() = \[\begin{bmatrix}2x_1-x_2+5x_3 \\ -4x_1+2x_2-10x_3 \end{bmatrix}\]$
\(T(\beta\vec{u}) = \beta T(\vec{u})\)
The transformation \(T\) takes a vector \(\mathbb{C^3}\) with components \((x_1 , x_2, x_3)\) in \(\mathbb{C^2}\) \(y_n\) is a linear combinaison of the input matrix x.