# Determine if F is a linear transformación f(x,y,z)=(x,x+y+z) ?

Jul 9, 2018

#### Explanation:

The tranformation is

$f \left(x , y , z\right) = \left(x , x + y + z\right)$

That is ${\mathbb{R}}^{3}$ to ${\mathbb{R}}^{2}$

Then ,

Calculate,

$f \left(x , y , z\right) + a f \left(x ' , y ' , z '\right) = \left(x , x + y + z\right) + a \left(x ' , x ' + y ' + z '\right)$

where $a \in \mathbb{R}$

$= \left(x , x + y + z\right) + \left(a x ' , a x ' + a y ' + a z '\right)$

$= \left(x + a x ' , x + y + z + a x ' + a y ' + a z '\right)$

$= \left(x + a x ' , x + a x ' + y + a y ' + z + a z '\right)$

$= f \left(x + a x ' , y + a y ' , z + a z '\right)$

As,

$f \left(x + a x ' , y + a y ' , z + a z '\right) = f \left(x , y , z\right) + a f \left(x ' , y ' , z '\right)$

The transformation is linear

Jul 9, 2018

This can be expressed as a matrix, here $M$, transformation so it is linear:

$F \left(\boldsymbol{x}\right) = \left(\begin{matrix}1 & 0 & 0 \\ 1 & 1 & 1\end{matrix}\right) \boldsymbol{x} = M \boldsymbol{x}$

Formal tests:

• Scaling

$\boldsymbol{F \left(\lambda \boldsymbol{x}\right)} = M \lambda \boldsymbol{x} = \lambda M \boldsymbol{x} \boldsymbol{= \lambda F \left(\boldsymbol{x}\right)}$

$\boldsymbol{F \left(\boldsymbol{x} + {\boldsymbol{x}}^{'}\right)} = M \left(\boldsymbol{x} + {\boldsymbol{x}}^{'}\right) = M \boldsymbol{x} + M {\boldsymbol{x}}^{'} \boldsymbol{= F \left(\boldsymbol{x}\right) + F \left({\boldsymbol{x}}^{'}\right)}$