Is there such a thing as "quadratic independence" (and higher generalizations of linear independence)?

Solution 1:

There is "algebraic independence" - if there is a non-zero polynomial $f$ of $n$ variables, with coefficients in the field $K$, such that $$f(\alpha_1,\ldots,\alpha_n)=0\ ,$$ then $\alpha_1,\ldots,\alpha_n$ are said to be algebraically dependent over $K$; if there is no such polynomial then they are algebraically independent over $K$.

For example, $\pi$ and $\sqrt{2\pi}$ are algebraically dependent over $\Bbb Q$ because if $f(z_1,z_2)=2z_1-z_2^2$ then $f(\pi,\sqrt{2\pi})=0$.

It would seem totally plausible that $e$ and $\pi$ are algebraically independent, but as far as I know this is still an unsolved problem - see here.

For a small number of examples and general theorems on algebraically independent numbers look here.

There is also a general concept of dependence relations which includes linear dependence and algebraic dependence as special cases. Haven't searched online but you can find it in N. Jacobson, Basic Algebra vol.II, section 3.6.

Solution 2:

I would say that $x$ and $y$ are quadratically dependent if there exist constants $a,$ $b,$ $\ldots,$ $f$ not all equal to $0$ such that $$ax^2 + bxy + cy^2 + dx + ey + f = 0$$

Here, $x$ and $y$ can be real numbers (or complex numbers), or real-valued functions (or complex-valued functions), or other contexts where this makes sense.

I've actually used this idea in classes (I used the phrase quadratically related) as an intuitive explanation of why compositions like $\sin(\arccos x)$ wind up being relatively simple functions in the sense that only squares and square roots are involved, and no transcendental functions. The explanation is that each trig function is quadratically related to every other trig function, so while the sine function doesn't entirely un-do the arc-cosine function, it comes "quadratically close" to doing so because the sine and cosine functions are quadratically related.

Solution 3:

If I understand you correctly, the question is:

Is "linear independence" a special case of a more general notion of independence?

Yes, it is.

Definition. Let $\mathbf{C}$ denote a concrete category whose forgetful functor $U_\mathbf{C} : \mathbf{C} \rightarrow \mathbf{Set}$ has a left-adjoint $F_\mathbf{C} : \mathbf{Set} \rightarrow \mathbf{C}$. Consider an object $X$ and function $f : I \rightarrow U_\mathbf{C}(X)$ from some index set $I$. Then $f$ is independent/generating/basislike iff the corresponding morphism $F_\mathbf{C}(I) \rightarrow X$ is injective/surjective/bijective.

Examples:

• If $\mathbf{C}$ is the category of $R$-modules for some commutative ring $R$, then $F_\mathbf{C}(I)$ equals $R^I_{\mathrm{fin}}$ and these terms mean what they usually mean. For example, $f : I \rightarrow U_\mathbf{C}(X)$ is independent in the above sense iff it is linearly independent in the usual sense iff the corresponding linear transform $R^I_{\mathrm{fin}} \rightarrow X$ is injective.
• Consider a field extension $E/F$. Functions into $E$ from some index set are algebraically independent over $F$ in the usual sense iff they're the independent in the above sense when $E$ is viewed as a commutative $F$-algebra.
• No non-empty subset of a finite group is ever independent, since all non-trivial free groups are infinite.

Solution 4:

As a result of my search for answers to this question, here is the answer I encountered:

For a set $\{x^1,\ldots x^k\}\subseteq F^s$ the set is quadratically dependent if there are coefficients $a_1,\ldots ,a_k$ not all zero such that: $$\forall i_1,i_2\in[s]:\ \sum_{j=1}^{k}a_j(x^j_{i_1})(x^j_{i_2})=0$$

Similarly, one can define $n$-degree dependence by considering monomials of degree n, instead of 2.

My answer is based on this paper by H. Tatcher Jr.