# A certain kind of non-linear mapping from $\mathbb{R}^2$ to $\mathbb{R}^2$

Give an example of a map $T:\mathbb{R}^2\rightarrow\mathbb{R}^2$ with both of the following properties:

a. $T(kx)=kT(x)$ for all $x\in \mathbb{R}^2, k\in \mathbb{R}$

b. $T$ fails to be a linear transformation

I am really stuck on this. I can only find matrices that are not linear transformations and don't fulfill the first condition, or they are linear transformations and they do fulfill the first condition. I am not sure how to go about doing this.

• Looking for matrices for this is the wrong method, as matrices directly correspond to linear transformations (provided you've chosen a basis). – Mark Mar 9 '17 at 23:35

$$T(x_1,x_2) = \big((x_1^2x_2)^{1/3},x_1+x_2\big)$$

Based on Doug M's strategy, here's a symmetrical one:

$$T(x,y) = \left((x^3+y^3)^\frac{1}{3},(x^3+y^3)^\frac{1}{3}\right)$$

Or, with the same idea, an even simpler one:

$$T(x,y) = \left((x^3+y^3)^\frac{1}{3}\,,\,0\right)$$

Still with the same idea, but dispensing with symmetry, here's a bijective one:

$$T(x,y) = \left((x^3+y^3)^\frac{1}{3},(x^3-y^3)^\frac{1}{3}\right)$$

• +1 for the bijective one. More generally to produce a bijection one may consider any invertible matrix $M$, any odd positive integer $k$, and define $T(x)= (Mx^k)^{1/k}$ to obtain a bijection (where the powers are taken component wise). – Surb Mar 10 '17 at 1:17

$f(x,y) = \begin{cases} (0,0) & x \ne 0 \\ (x,y) & x = 0\end{cases}\\$

Conceptually, the given condition requires that if we restrict $f$ to any line through the origin, it is linear. Since each of those lines only overlaps at the origin, we can construct $f$ easily by constructing its value on each of those lines one by one.

In this case, we choose $f$ to be zero on every line except on the y axis. On the y axis, it just looks like a the identity function. The function constructed this way is not linear because its value on the y axis doesn't "agree" with its value everywhere else. If we try to check whether the law of linearity applies, we check two points $p_1, p_2$, and verify that

$f(p_1 + p_2) = f(p_1) + f(p_2)$

Now, if both $p_1$ and $p_2$ are on the y axis, or both of them are not on the y axis, this equality will hold. However, if one of them is on the y axis and the other isn't, you'll run into problems. For example:

$p_1 = (1,0)$ $p_2 = (1,1)$ $f(p_1+p_2) = f(2,1) = (0,0)$ $f(p_1) + f(p_2) = (1,0) + (0,0) = (1,0)$

In general, you can see that there are infinitely many ways that we could choose different linear functions on all the lines through the origin and "piece together" a novel f that violates the constraints, even a continuous or differentiable one, as other people have demonstrated.

$f(x,y) = \begin{cases} \frac {x^2}{y} & y\ne 0 \\ 0 & y = 0\end{cases}\\ f(kx,ky) = kf(x,y)$

Here we have a non-linear map from $\mathbb R^2 \to \mathbb R$

now find a $g(x,y)$ along similar lines (or even a linear $g(x)$ at this point.)

and $h(x,y) = f(x,y),g(x,y)$ will do what you need it to do.

• So quick with the -1's. No good reason for that. Make the comment, then give the poster a chance to fix it, or delete it. – quasi Mar 9 '17 at 23:42
• @quasi - At least had the courtesy to comment. The -1 with no explanation irritate me far more. – Doug M Mar 9 '17 at 23:44
• Now the map fails to be homogeneous for $k<0$. – Erick Wong Mar 9 '17 at 23:44
• @DougM I did downvote without initially commenting, but I thought the explanation was apparent (you did indeed know it was incomplete when posting it), and I tend to hold people with obvious experience on the site to high standards. – Erick Wong Mar 9 '17 at 23:50
• @Surb: Yes, of course -- I misread it as a product instead of an $n$-tuple. Then as in my deleted comment, $\alpha=0$ works nicely. – quasi Mar 10 '17 at 0:10

Generally, any function of the form $$f(x) = \left(x_1 g\left(\frac{x_2}{x_1}\right),x_1 h\left(\frac{x_2}{x_1}\right)\right)$$ This works because $kx_2/kx_1=x_2/x_1$

Note that the value at $x=(0,a)$ should be taken as the limit as $x_1\to 0$ where possible.

• The value at $x=(0,0)$ must always be taken as $(0,0)$ regardless of the limit. There is no requirement that $f$ be continuous. – Erick Wong Mar 10 '17 at 4:24
• @ErickWong - True, but I was referring to the expression as provided. Then again, it occurs to me that the value when $x_1=0$ is the concern for "dealing with special cases", so I'll adjust the wording. – Glen O Mar 10 '17 at 4:28
• Ah that makes more sense, however it's still true that there is no requirement that $f$ be continuous. Since this is meant to be a general solution, the phrasing "should be taken" seems odd. Generally, one can choose $f(0,a)$ to be $a$ times an arbitrary constant vector, independent of $g$ and $h$. – Erick Wong Mar 10 '17 at 14:40
• @ErickWong - yes, but then it wouldn't match the form I gave, whereas it is common practice to convert$x_1\frac{x_2}{x_1}$, for example, into $x_2$ and interpret it as satisfying the given form. – Glen O Mar 10 '17 at 15:45
• It makes sense at the level of the OP. I don't think of it as not matching your form, it's just that the domain of $g$ "wants" to be the real projective line rather than $\mathbb R$. And the nice case is where $g$ is continuous, but in general $g(\infty)$ isn't obliged to agree with its neighborhood any more than $g(2)$ is :). – Erick Wong Mar 10 '17 at 18:38

Inspired by Surd, I slightly generalize his result as follows:

$$T(x_1,x_2) = (\sum_{i=1}^N a_ix_1^{n_{1,i}/n_i}x_2^{n_{2,i}/n_i}, \sum_{j=1}^M b_jx_1^{m_{1,j}/m_j}x_2^{m_{2,j}/m_j}),$$ where $a_i, b_j \in {\mathbb R}$ [$(\prod_{i = 1}^{N} a_i) (\prod_{j = 1}^{M} b_j) \neq 0$], $M,N \in \overline{\mathbb Z}_+$ ($MN \neq 0$), $n_{1,i},n_{2,i},m_{1,j},m_{2,j} \in {\mathbb Z}^+$, $n_{1,i} + n_{2,i} = n_i$, $m_{1,j}+m_{2,j} = m_j$, and $n_i,m_j$ are odd.

• @Surb Thanks very much, and I will modify it. – Ryan Mar 10 '17 at 0:08