Proving a transformation is not a linear transformation I'm asked to prove if a transformation is linear or not. In the vector field $V=\{f(x)\colon \mathbb{R} \to\mathbb{R}\}$, so the transformation is $T\colon V \to V$ given by $T(f(x)) = (xf(x))+1$. I want to show it is not a linear transformation by a counter example. Could I use the function $f(x) = x+1$, then the transformation is $T(x+1) = x^2 + x + 1$. Then, having a transformation of $T(0)$ must equal zero, so in this I would show that $x=-1 \implies T(-1+1) = T(0) = (-1)^2 + (-1) + 1 = 1 \neq 0$, showing this is not linear. I'm thinking this is right, but am not positive if this is how you would go about proving a function transformation is not linear.
 A: show that it does not map the 0 function to the 0 function.
A: $T(0)$ is $T$ of the zero function which would give $T(0) = x \cdot 0 +1 = 0 + 1 = 1$.  You don't even need to specify what function $f(x)$ is because it has been specified by writing $T(0)$.
A: What you have done is show that the function $T(x+1)$ is not identically zero, because it evaluates to $1$ at $x=-1$.  This actually has nothing to do with $T(0)$.
The distinction between $T(0)$ and $T(x+1)$ evaluated at $x=-1$ is actually very large and you seem to be getting the two mixed up.  You're also getting confused at the dual use of $0$ to refer to the zero function, and $0$ as a real number.  So I'll use red when writing down a function from $\mathbb R$ to $\mathbb R$, and blue when writing down a number.
For instance, suppose that $T$ was instead the transformation
$$T(\color{red}{f(x)}) = \color{red}{f(x) - x - 1}.$$
Then $$T(\color{red}{x+1}) = \color{red}{x+1 - x - 1} = \color{red}{0},$$ and when you evaluate this at $\color{blue}{x=-1}$ you do get $\color{blue}{0}$.  However, the actual value of $T(\color{red}{0})$ is
$$T(\color{red}{0}) = \color{red}{0 - x - 1} = \color{red}{-x-1},$$
which is not equal to $\color{red}{0}$ (the function that is $\color{blue}{0}$ for all values of $\color{blue}{x}$).  It may be equal to $\color{blue}{0}$ for some values of $\color{blue}{x}$, but that is irrelevant because a linear transformation must satisfy $T(\color{red}{0}) = \color{red}{0}$.
So it is very incorrect to say that $T(\color{red}{0})$ can be computed just by looking that $T(\color{red}{x+1})$ and plugging in $\color{blue}{x=-1}$.
