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?


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

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    $\begingroup$ Your proof is correct if you find $(x,y)$ such that $x+y\ne x^2+y^2.$ $\endgroup$
    – mfl
    Oct 6, 2017 at 15:50

4 Answers 4


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.

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    $\begingroup$ The question now states that the transformation is over the reals, and the op did use a property of the reals by concluding that the matrix form and original form were different functions. I think this answer no longer matches the question as stated. $\endgroup$
    – Zach Boyd
    Oct 6, 2017 at 16:21
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    $\begingroup$ @ZachBoyd The question was always about a transformation over the reals. The fact that $x^2+y^2$ and $x+y$ are not identical over the reals is, of course, obvious to you, to me, and to the OP. But I think it's equally obvious that the original transformation $T$ is not linear--just looking at it, the last component of its output screams out this fact. My understanding of exercises like this is that we are meant to go back to basics when doing them. Otherwise, we could consider the problem statement itself to be a property of $\mathbb R,$ and write QED. $\endgroup$
    – David K
    Oct 6, 2017 at 16:50
  • $\begingroup$ That's a good rebuttal. I think you are right that this ultimately comes down to a question of how such questions should be understood. Perhaps it is more fair to say that the approach given by the OP can be simply extended to a complete argument by the reasoning proposed in your answer. I guess I just thought that saying the argument was incorrect hinted at a more fundamental flaw that would be hard to repair. $\endgroup$
    – Zach Boyd
    Oct 6, 2017 at 16:54
  • $\begingroup$ @ZachBoyd I agree, it's a small flaw, very easily repaired. Possibly the language I used obscured that point; I meant to suggest something like, "Have you looked at $T(2\mathbf e_1)$?" $\endgroup$
    – David K
    Oct 6, 2017 at 18:32

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.

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    $\begingroup$ Nonsense, it is perfectly OK to prove non-linearity by contradiction, assuming linearity. In that case a matrix must exist, and there is a standard way to obtain it, which is what OP does. The fact that that matrix does not describe the given map is the contradiction sought. $\endgroup$ Oct 7, 2017 at 8:00


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

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


I suggest my hint before the OP made his edit.

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    $\begingroup$ I appreciate your answer very much, but I just wanted to know if this was a valid method to apply to any transformation. $\endgroup$
    – Carl Rojas
    Oct 6, 2017 at 15:53

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.


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