What formula can make a non-linear transformation into a matrix transformation?
This problem showed up on an exam,
I was really shocked that I never heard one of them...
From my understanding, this question asks me is there any formula which can make $T(x) = x^2$, a non-linear transformation into a matrix transformation ? What am I supposed to answer to this question?
Please save me from full of confusion, thank you!
I think that the question is rather poor. There is no linear transformation $S$ such that $S(x) = x^2$, so there is no linear transformation that agrees with $T$ everywhere.
On the other hand, there is a "natural" way to obtain a linear transformation $S$ from $T(x) = x^2$, and that would be to take the derivative. The derivative $S(x) = T'(x) = 2x$ is the linear map which scales vectors by $2x$.
More generally, given a "non-linear" (but still differentiable) map $T\colon\mathbb R^n\to\mathbb R^m$, written in the standard bases as $$ T(x_1,\dots,x_n) = \big(T_1(x_1,\dots,x_n),\dots,T_m(x_1,\dots,x_n)\big), $$ the derivative $DT_p$ at any point $p = (p_1,\dots,p_n)$ is a linear map $\mathbb R^n\to\mathbb R^m$. Moreover, in the standard bases, $DT_p$ can be identified with the $m\times n$ Jacobian matrix of partial derivatives: $$ DT_p = \begin{pmatrix} {\partial T_1 \over \partial x_1}(p) & {\partial T_1 \over \partial x_2}(p) & \dots & {\partial T_1 \over \partial x_n}(p) \\ {\partial T_2 \over \partial x_1}(p) & {\partial T_2 \over \partial x_2}(p) & \dots & {\partial T_2 \over \partial x_n}(p) \\ \vdots & \vdots & \ddots & \vdots \\ {\partial T_m \over \partial x_1}(p) & {\partial T_m \over \partial x_2}(p) & \dots & {\partial T_m \over \partial x_n}(p) \end{pmatrix}. $$ The upshot is that as long as we look at points $q = (q_1,\dots,q_n)$ that are "very small," then the affine transformation $q\mapsto T(p) + DT_p(q)$ closely approximates the (in general nonlinear) map $q\mapsto T(p+q)$.
So, for the example $T\colon \mathbb R\to\mathbb R$ defined by $T(x) = x^2$, we get a linear map at every point $p$ defined as $DT_p$, which we typically just write as $T'(p)$, which is really just a $1\times 1$ matrix $T'(p) = (2p)$. As a linear map acting on $1\times 1$ column vectors $(q)$, we have $$ DT_p(q) = T'(p)(q) = (2p)(q) = (2pq). $$ Hence the end result is that "vectors" $(q)$ are mapped linearly to $(2pq)$ by the derivative, and if we identify $1\times 1$ matrices with numbers, \begin{align*} T(p+q) = (p+q)^2 &= p^2 + 2pq + q^2 \\ &\approx p^2 + 2pq \qquad\qquad\text{(small numbers like $q$ are "negligible" if we square them)}\\ &= T(p) + 2pq \\ &= T(p) + T'(p)q, \end{align*} so long as $q$ is sufficiently small. You may recognize that $T(p) + T'(p)q$ is really the first-order Taylor polynomial of $T$ at $p$ (i.e. the best affine approximation near $p$).
