What is the analogue of linear algebra for the quadratic case Let $m,n$ be positive integers.  $f:\mathbb{R}^m\rightarrow \mathbb{R}^n$ is said to be a quadratic transformation (a term I just made) iff:
1) $f(0)=0$  
2) For every $x,y,z\in\mathbb{R}^m$ , we have:
$$[f(x+z)-f(x)]-[f(x)-f(x-z)]=[f(y+z)-f(y)]-[f(y)-f(y-z)]$$
Now what mathematical theory is concerned with such transformations ? Does there exist versions of the determinant for such transformtions ?
Notes: The following can be shown easily
1) Every linear transformation is a quadratic transformation
2) $f$ is a linear transformation iff $f(0)=0$,and for every  $x,y,z\in\mathbb{R}^m$ , we have:
$$f(x+z)-f(x)=f(y+z)-f(y)$$
Thank you

 A: Let $V$ and $W$ be finite-dimensional vector spaces over a field $K$ with $\operatorname{char} K \neq 2$. Let's first recall the following facts about linear tranformations:


*

*One has a natural isomorphism $W \otimes V^\ast \cong L(V,W)$, given by $(w \otimes f) \mapsto (v \mapsto f(v)w)$. 

*In particular, recall that $L(V) := L(V,V) \cong V \otimes V^\ast$. Then, the trace $\operatorname{Tr} : L(V) \to K$ corresponds to the canonical map $\tau : V \otimes V^\ast \to K$ given by $\tau(v \otimes f) := f(v)$.

*Recall that for $n := \dim V$, $\wedge^n V$ is $1$-dimensional, so that $L\left(\wedge^n V\right)$ is also $1$-dimensional, with $L\left(\wedge^n V\right) = K \operatorname{Id}_{\wedge^n V}$. Then, for $T \in L(V)$, $\det(T) \in K$ is defined by $$\det(T) \operatorname{id}_{\wedge^n V} := \wedge^n T : v_1 \wedge \cdots \wedge v_n \mapsto Tv_1 \wedge \cdots \wedge Tv_n.$$


Let's now turn to quadratic forms. One defines a $W$-valued quadratic form on $V$ to be a map $q : V \to W$ such that $$b_q(v_1,v_2) := \tfrac{1}{2}(q(v_1+v_2)-q(v_1)-q(v_2))$$ defines a bilinear map $V \times V \to W$, which, by construction, is symmetric. Indeed, $q \mapsto b_q$ defines a canonical isomorphism from the space of all $W$-valued quadratic forms on $V$ to the space of all bilinear maps $V \times V \to W$, with inverse $b \mapsto (v \mapsto b(v,v))$. Hence, it really suffices to think about bilinear maps $V \times V \to W$, particularly the symmetric forms.


*

*One has a natural isomorphism from $W \otimes (V^\ast)^{\otimes 2}$ to the space of all bilinear maps $V \times V \to W$, given by $$w \otimes f_1 \otimes f_2 \mapsto \left((v_1,v_2) \mapsto f_1(v_1)f_2(v_2)w\right).$$ In particular, the symmetric maps (which correspond bijectively with the quadratic forms) correspond to $W \otimes S^2 V^\ast$.

*Conventionally, one observes that if $W = K$, then a bilinear form $b : V \times V \to K$ corresponds unambiguously to a linear transformation $B : V \to V^\ast$, $B(v) := b(\cdot,v) = b(v,\cdot)$. More abstractly, $K \otimes (V^\ast)^{\otimes 2} = V^\ast \otimes V^\ast \cong L(V,V^\ast)$ by the natural isomorphism $W \otimes V^\ast \cong L(V,W)$. Then, for any isomorphism $\phi : V^\ast \to V$ (e.g., from a choice of basis on $V$), $\phi \circ B \in L(V)$, so that one could define $\operatorname{Tr}_\phi(b) := \operatorname{Tr}(\phi \circ B)$, $\det_\phi(b) := \det(\phi \circ B)$. Observe, however, the dependence on $\phi$. In many applications, however, one works with $V = K^n$, which has a standard ordered basis, and hence a canonical isomorphism $\phi : (K^n)^\ast \cong K^n$, which is essentially the transpose map taking row vectors to column vectors.


In any event, this is all the standard machinery. Of course, a bilinear form $V \times V \to V \otimes V$ naturally corresponds to an element of $(V \otimes V) \otimes (V^\ast \otimes V^\ast)$, which corresponds to a linear transformation $V \otimes V \to V \otimes V$ thanks to the natural isomorphism $V\ast \otimes V^\ast \cong (V \otimes V)^\ast$, in which case one could immediately define the trace and determinant of such a bilinear form to simply be the trace and determinant, respectively, of the corresponding element of $L(V \otimes V)$. However, I'm not sure this is anywhere close to being as useful or interesting as the usual constructions in terms of bilinear forms $V \times V \to K$.
