Determining If a Function is a Linear Transformation I thought I understood this concept, as it's pretty similar to other ones I've done in linear algebra. Perhaps I'm tired, and just not seeing it right now.
Here's what I know:

For the vector spaces $V$ and $W$, the function $T: V\to W$  is a linear transformation of $V$ mapping into $W$ when two properties are true (for all vectors $u,v$ and any scalar $c$):
  
  
*
  
*$T(u+v) = T(u) + T(v)$   - Addition in $V$ to addition in $W$
  
*$T(cu) = cT(u)$ - Scalar multiplication in $V$ to SM in $W$
  

My book gives an example of proving $T(v_1, v_2) = (v_1 - v_2, v_1+2v_2)$ is a linear transformation. I won't copy the whole thing, but I actually follow it.
However, for practice, I have to determine if this function is a linear transformation from $\mathbb{R}^2$ to $\mathbb{R}^2$...

$T:\mathbb{R}^2 \to \mathbb{R}^2$ , $T(x,y) = (x,1)$

I tried doing it, but I honestly wasn't going anywhere or really understanding what I was doing. Can anyone please explain this problem to me? Appreciate any help.
 A: This is not a linear transformation.
Indeed,
$T(1,0)+T(1,0)=(1,1)+(1,1)=(2,2)$
On the other hand
$T(2,0)=(2,1)\neq (2,2)$
For T to be linear, these would have to be equal.
A: One consequence of the definition of a linear transformation is that every linear transformation must satisfy
$$ T(0_V)=0_W $$
where $0_V$ and $0_W$ are the zero vectors in $V$ and $W$, respectively. Therefore any function for which $T(0_V)\neq 0_W$ cannot be a linear transformation.
In your second example,
$$ T\Big(\begin{bmatrix}0\\0\end{bmatrix}\Big)=\begin{bmatrix}0\\1\end{bmatrix}\neq\begin{bmatrix}0\\0\end{bmatrix}$$
so this tells you right away that $T$ isn't linear.
A: Let $$v = (v_1,v_2) $$and $$w = (w_1, w_2)$$
From the definition of a linear transformation : $$T(v+w) = T(v) + T(w)$$
so, $$T(v) = [v_1,1]$$  and $$T(w) = [w_1,1]
T(v) + T(w) = [v_1+w_1, 2]$$
$$T(v+w)... v+w = (v_1+w_1,v_2+w_2)$$
$$T(v_1+w_1, v_2+w_2) = [v_1+w_1, 1] $$
$$T(v+w) = [v_1+w_1, 1]$$
so, $$T(v+w) \neq T(v) + T(w)$$
therefore, T is not a linear transformation.
