Is $ T: \mathbb{R^3} -> \mathbb{R^2} : T(x,y,z) = (x+1, y+z)$ a linear transformation? Need help for my proof So I have
$$ T: \mathbb{R^3} \rightarrow\mathbb{R^2} $$
$$ T(x,y,z) = (x+1, y+z),$$
and I know that it is indeed a linear transformation if I show both cases:
$$  \forall v \in T :  F(a+b) =  F(a)+F(b),$$
and
$$  \forall k \in K: F(k \cdot a) = k \cdot F(a).$$
So I just start and I would like to know if what I'm doing is correct, I as am a little bit unsure if I'm not doing any mistakes:
$$ T(a+b)= \begin{pmatrix}  
x_1+1+x_2+1\\  
(y_1+z_1)+(y_2+z_2)  
\end{pmatrix}$$
$$ T(a)= \begin{pmatrix}  
x_1+1\\  
(y_1+z_1)  
\end{pmatrix}$$
$$ T(b)= \begin{pmatrix}  
x_2+1\\  
(y_2+z_2)  
\end{pmatrix}$$
For $T(a) + T(b)$ follows:
$$ \begin{pmatrix}  
x_1+1 + x_2+1\\  
(y_1+z_1)+(y_2+z_2)  
\end{pmatrix} = T(a+b)$$
Now for the second case, the skalar: Let $k = 2$
$$ F(k \cdot a) = \begin{pmatrix}  
2 \cdot (x_1 + 1)\\  
2 \cdot(y_1+z_1)  
\end{pmatrix} $$
now finally:
$$ k \cdot F( a) = 2 \cdot  \begin{pmatrix}  
 x_1+1\\  
(y_1+z_1)  
\end{pmatrix} =\begin{pmatrix}  
 x_1+1\\  
(y_1+z_1)  
\end{pmatrix} + \begin{pmatrix}  
 x_1+1\\  
(y_1+z_1)  
\end{pmatrix} = F(k \cdot a) $$
therefore
$$ T(x,y,z) = T(x+1, y+z),$$
is indeed a linear transformation.
I am correct or did I do something wrong? I am not sure about the skalar multiplication part but would appreciate it a lot if somebody could guide me if I'm on the right path.
Thanks in advance!
Greetings!
 A: *

*If you are to write a proof, you cannot choose the value of $k$. Your proof has to work for all $k$.

*If you want to show that it is not linear, then you can do so by picking a single value of $k$.

*Your notation is not good. You have $F$ and $T$ for the same object, and $\forall v\in T$ which makes no sense because $T$ is not a set and what you write after has no $v$, only $a$ and $b$.

*You computed $F(a+b)$ wrong. If $F(a) = \binom{F_1(a)}{F_2(a)}$, then $F(a+b)$ is to be computed as $\binom{F_1(a+b)}{F_2(a+b)}$. You are assuming what you are trying to prove.

*Same for the scecond part. You computed $F(2\cdot a)$ wrong.You should write $\binom{F_1(2\cdot a)}{F_2(2\cdot a)}$. These two things are equal only if $F$ is linear, and this is what you're trying to show.

Hint towards a solution: for $F_1(x):=x+1$, note that $F_1(x+y)=(x+y)+1=x+y+1$ which is not $F_1(x)+F_1(y)=(x+1)+(y+1)=x+y+2$.
A: Recall the definition of linear transformation:

Definition. Let $V$ and $W$ two vectors space over the same field $K$. A function $T: V\to W$ is said to be a linear transformation if and only if the following conditions it holds: 
i) (Additivity) For all $v_{1},v_{2}\in V$, we have $$T(v_{1}+v_{2})=T(v_{1})+T(v_{2})$$ii) (Homogeneity) For all $\alpha\in K$ and for all $v\in V$, we have $$T(\alpha v)=\alpha T(v)$$

So your function is not a linear transformation because taking in the definition $V=\mathbb{R}^{3}$ and $W=\mathbb{R}^{2}$ and field $K=\mathbb{R}$, the function $T:\mathbb{R}^{3}\to  \mathbb{R}^{2}$ defined by $T(x,y,z)=(x+1,y+z)$ doesn't satisfies at least one conditions of the definition. The problem is the number $1$ in the first coordinates of $(x+\color{red}{1},y+z)$ that kind of thing destroys linearity in a transformation with the conventionals operations $+$ and $\times$. So from the beginning we already know that $T$ cannot be a linear transformation if we work with the conventional operations.
Now, let's see this formally.

*

*Taking $v_{1}=\begin{bmatrix}x_1\\ y_1\\ z_1\end{bmatrix}$ and $v_{2}=\begin{bmatrix}x_2\\y_2\\z_2\end{bmatrix}$ with $v_{1},v_2\in \mathbb{R}^{3}$, then
\begin{align*}
T(v_{1}+v_{2})&=T\left(\begin{bmatrix}x_1\\ y_1\\ z_1\end{bmatrix}+\begin{bmatrix}x_2\\ y_2\\ z_2\end{bmatrix}\right),\\&=T\left(\begin{bmatrix} x_{1}+x_2,\\y_{1}+y_{2}\\z_{1}+z_{2}\end{bmatrix}\right),\\&=\begin{bmatrix} (x_1 + x_2)+\color{red}{1}\\(y_{1}+y_2)+(z_{1}+z_{2})\end{bmatrix},\\&\color{blue}{\not=}\begin{bmatrix} x_{1}+\color{red}{1}\\y_1 +z_1\end{bmatrix}+\begin{bmatrix}x_{2}+\color{red}{1}\\x_{2}+y_{2}\end{bmatrix},\\ &=T\left(\begin{bmatrix}x_1\\y_1 \\ z_1 \end{bmatrix}\right)+T\left(\begin{bmatrix}x_2\\ y_2 \\ z_2 \end{bmatrix}\right),\\ &=T(v_1)+T(v_2).
\end{align*}
So i) doesn't hold and therefore $T$ is not a linear transformation. Whether or not the second condition is holds is not important, note that the definition states that if at least one condition is not holds then automatically $T$ cannot be a linear transformation. However, if you wish, you can continue to see what happens in the second condition.


*Taking every $\alpha\in \mathbb{R}$ and $v=\begin{bmatrix}x\\y\\z\end{bmatrix}$ with $v\in \mathbb{R}^{3}$, then
\begin{align*}
T(\alpha v)&=T\left(\alpha\begin{bmatrix}x\\y\\z\end{bmatrix}\right),\\&=T\left(\begin{bmatrix}\alpha x\\ \alpha y\\ \alpha z\end{bmatrix}\right),\\&=\begin{bmatrix}\alpha x+\color{red}{1}\\\alpha y+\alpha z\end{bmatrix},\\&\color{blue}{\not=}\alpha\begin{bmatrix}x+\color{red}{1}\\ y+z\end{bmatrix},\\&=\alpha  T\left(\begin{bmatrix}x\\y\\z\end{bmatrix}\right),\\&=\alpha T(v).
\end{align*}
So in this case ii) also doesn't hold and again we can conclude that $T$ is not linear transformation.
Notice that number $\color{red}{1}$ and as this destroys the linearity of $T$. If you change the $1$ to a $\color{red}{0}$ then $T$ must be indeed a linear transformation, but if you change the $1$ to any number other than $0$ then $T$ cannot be a linear transformation. All this is so due to the behavior of the conventional operations that are being used.
A: Any linear transformation can be written in the form $$T(\overline{v})=M \overline{v}$$ where $M$ is a matrix of transformation. The transformation cannot be written in this form because
$$T(x,y,z)= \begin{pmatrix} x+1 \\ x+z
\end{pmatrix}  =\begin{pmatrix} 1 & 0 & 0 \\
1 & 0 & 1
\end{pmatrix}  \begin{pmatrix} x \\ y \\ z
\end{pmatrix} + \begin{pmatrix} 1 \\ 0
\end{pmatrix} =M\overline{v}+\overline{c}$$
Hence, it is not a linear transformation.
