Is this a correct way to prove T is not a linear transformation? I have the following transformation $T:\mathbb{R}^2 \longrightarrow \mathbb{R}^3$ defined by $T\left( x, y \right) = \left( y, x, x^2 + y^2 \right).$ I know the transformation is not linear but would like to prove it, so I deviced the following "proof."
We know every linear transformation $T$ has a unique matrix representation for the standard basis of $\mathbb{R}^2,$ which is given by $$A = \left[ \begin{array}{ccc}
T(\mathbf{e}_1) & T(\mathbf{e}_2) \\\end{array} \right],$$
and this matrix $A$ would move me back to the linear transformation by $T\left( \mathbf{x} \right) = A \mathbf{x}.$ 
So, I assume $T$ is a linear transformation and construct it standard matrix representation, which would be $$A = \left( \begin{array}{ccc}
0 & 1 \\
1 & 0 \\
1 & 1 \end{array} \right).$$
Now, to get my original transfomation back I would have to do $$T\left( \mathbf{x} \right) = A \mathbf{x} = \left( \begin{array}{ccc}
0 & 1 \\
1 & 0 \\
1 & 1 \end{array} \right) \cdot \left( \begin{array}{ccc}
x \\
y \end{array} \right) = \left( \begin{array}{ccc}
y \\
x \\
x+y \end{array} \right).$$
Since this transformation I got is not the original one, I conclude $T$ is not a linear transformation.
My question is, the above reasoning is correct?
And in general, can I apply this method to prove or disprove any transformation is a linear transformation?

EDIT:
Please do not sugegst alternative methods of proof; I know them well. All I need is to know if the method described works.

I find your introduction of the matrix $A$ difficult to follow since $T$ is non-linear.  I haven't yet satisfied myself that it makes sense; my questions would be:  i.) can you show all your steps in constructing $A$ from $T$?  ii.)  why introduce $A$ at all, when we can work directly from $T$?
A problem I have is that assuming $T$ is linear is not the same as assuming $T(x, y) = (x, y, x + y)$, which is what I think your matrix $A$ represents.  We would do better to take $T(x, y) = (x, y, c_x x + c_y y)$ for arbitrary $c_x, c_y \in \Bbb R$. Then what happens with $A$?  Setting $T(x, y) = (x, y, c_x x + c_y y)$ preserves the first two coordinates, and is the most general linear $T$ to do so.  
To show that $T$ is not linear, all we need do is show that it violates at least one axiom of linearity, which can be done without introducing $A$.  We work from $T$ directly:
$T(x, y) = (x, y, x^2 + y^2), \tag 1$
we consider $T(\alpha x, \alpha y)$ for $0, 1 \ne \alpha \in \Bbb R$.  We have:
$T(\alpha x, \alpha y) = (\alpha x, \alpha y, \alpha^2 x^2 + \alpha^2 y^2) = \alpha (x, y, \alpha x^2 + \alpha y^2)$
$\ne \alpha (x, y, x^2 + y^2) = \alpha T(x, y); \tag 2$
(2) holds as long as $\alpha \ne 0, 1$, i.e. for by for the majority of real $\alpha$; but (2) is a direct violation of the linearity axiom $T(\alpha x, \alpha y) = \alpha T(x, y)$.
Therefore, $T$ is not linear.
