Is this a correct way to prove T is not a linear transformation?

Question

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?

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EDIT:

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

Answer 1

Suppose instead of $\mathbb R$ we had used the field $\mathbb Z_2 = \mathbb Z/(2\mathbb Z),$ in other words, the set $\{0,1\}$ with the usual operations $+$ and $\cdot$ modulo $2.$

Now we would be asking about $T: \mathbb Z_2^2 \to \mathbb Z_2^3,$ with $T\left( x, y \right) = \left( y, x, x^2 + y^2 \right).$

Every part of your proof would then work just as well as it did for $T: \mathbb R^2 \to \mathbb R^3,$ with the exception of the conclusion. The conclusion would be false, because $x^2 + y^2 = x + y$ when $x, y \in \mathbb Z_2.$

I do not see any point in the proof where you invoke any property of $\mathbb R$ that $\mathbb Z_2$ does not have. Therefore I would say the proof is not valid.

In order to make a valid proof, you could invoke (for example) the fact that $\mathbb R$ contains an element named $2$ that is distinct from $0$ and $1,$ and you could have used the properties of that element to find a counterexample to the statement $x^2 + y^2 = x + y$ for $x,y \in \mathbb R.$

I think it is noteworthy that you used many more facts than you needed in this proof, which I think also is a bad thing to do in a proof, but of course that alone does not invalidate a proof.

Answer 2

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.

Answer 3

Hint

$T$ is linear iff $T(\alpha (x,y)+\beta (u,v))=\alpha T((x,y))+\beta T((u,v)).$

Is $T(2,2)=T(2(1,1)))=2T(1,1)?$

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Edit

I suggest my hint before the OP made his edit.

Answer 4

In proving that something isn't linear, it's good just to find a counterexample. The top comment says that your proof isn't valid because it doesn't work for all fields symbolically.

Try using numbers instead and see where your two axioms of linear transformations break down. You're already on track because you found a "matrix representation", now just show that it indeed predicts one thing by virtue of supposed linearity, but the actual transform is something different (violating linearity).

In general, it's "easy" to disprove linearity—all you need is one counterexample or a family of counterexamples.