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The notion of linear independence is very well-known and well-understood.

However, is there a way to generalize the definition to other types of independence -- such as perhaps "quadratic independence", "polynomial independence", "harmonic independence", etc.?

(Sorry if isn't a good tag; I couldn't think of a better one.)

user541686
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  • Linear independence comes from linear combination. How would one generalize this to quadratic combinations ?? – daw Apr 28 '14 at 12:46
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    @daw I think that's the essence of the question – Chris Brooks Apr 30 '14 at 00:28
  • Aside from algebraic independence, the other thing that comes to mind is the concept of "matroid", which is the sort of abstract general structure linear independent sets of vectors have. – Mark S. Jun 19 '15 at 11:32

4 Answers4

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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.

Community
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David
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  • +1 fantastic answer. I'll keep the question open for a bit to see if there are other ones coming, but will probably accept this haha. – user541686 Apr 30 '14 at 08:36
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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.
goblin GONE
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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.

Dave L. Renfro
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    Isn't there always an $f$ that makes this true in the case of numbers? – user541686 Apr 30 '14 at 18:41
  • @Mehrdad: Ah, good catch! Should be rational coefficients for numbers. And for functions it seems that two potentially useful classifications arise: using real numbers as coefficients or using rational functions (with real number coefficients) as coefficients. – Dave L. Renfro Apr 30 '14 at 19:11
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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.

NL1992
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  • Thanks!! I'm a little bit confused by the notation though. If you fix $i_1$ and $i_2$ then what is $j$ doing there? The elements of the set are just x[1] ... x[k], right? Why are there 2 subscripts (/superscripts) instead of just 1? – user541686 Jul 29 '21 at 15:25
  • This is a definition of quadratic dependence for a set of $k$ vectors in $F^s$, that’s what the j’s stand for. $i_1,i_2$ are the coordinates (between $1\to s$).

    For example, if $s=3$, and $k=4$, with $x_1=(1,0,0),x_2=(0,1,0),x_3=(1,1,0),x_4=(0,0,0)$, we have the following 6 equations (corresponding to choosing 2 elements of a 3 element set with replacement):

     – NL1992 Jul 31 '21 at 16:46
  • $a_1(1)^2+a_2(0)^2+a_3(1)^2+a_4(0)^2=0$, $a_1(1)(0)+a_2(0)(1)+a_3(1)(1)+a_4(0)(0)=0$, $a_1(1)(0)+a_2(0)(0)+a_3(1)(0)+a_4(0)(0)=0$,
    $a_1(0)^2+a_2(1)^2+a_3(1)^2+a_4(0)^2=0$, $a_1(0)(0)+a_2(1)(0)+a_3(1)(0)+a_4(0)(0)=0$, $a_1(0)^2+a_2(0)^2+a_3(0)^2+a_4(0)^2=0$     – NL1992 Jul 31 '21 at 16:51
  • Given by the monomials: $x_1^2,x_1x_2,x_1x_3,x_2^2,x_2x_3,x_3^2$ – NL1992 Jul 31 '21 at 16:53
  • If there are $a_1,\ldots, a_4$ such that these equations are all satisfied, then the elements may be called quadratically dependent, and otherwise, quadratically independent. – NL1992 Jul 31 '21 at 16:57
  • +1 Interesting, thanks. But does this mean the linear version of this is actually different than our definition of linear independence in linear algebra? Since in this case we're defining an inner product over nth-degree products, whereas linear dependence just looks at 1st-degree products (i.e. the vectors themselves) and linearly combines those without involving an inner product. – user541686 Aug 02 '21 at 00:42
  • Actually, in the linear version you get precisely the same definition as your usual linear dependence. Indeed, in our running example, these are the equations corresponding to $x_1,x_2,x_3$, so we get the equations: $b_1(1)+b_2(0)+b_3(1)+b_4(0)=0$, $b_1(0)+b_2(1)+b_3(1)+b_4(0)=0$, $b_1(0)+b_2(0)+b_3(0)+b_4(0)=0$, which is solved with the same coefficients as the vector equation $b_1(1,0,0)+b_2(0,1,0)+b_3(1,1,0)+b_4(0,0,0)=(0,0,0)$. – NL1992 Aug 02 '21 at 07:34
  • We do not define an inner product here - we solve the equations $\sum_{x\in{x^1,\ldots , x^k}}a_x\cdot x_i x_j$ for every two coordinates $i$ and $j$, where $a_x$ is uniform over choices of a pair of coordinates (but not necessarily over the different choices for $x$). – NL1992 Aug 02 '21 at 07:37
  • Ohh I see! I misunderstood what was going on. Very nice, thanks! – user541686 Aug 03 '21 at 10:26