A $4\times 4$ symmetric matrix
$$
\left(
\begin{array}{cccc}
 a\_{11} & a\_{12} & a\_{13} & a\_{14} \\
 a\_{12} & a\_{22} & a\_{23} & a\_{24} \\
 a\_{13} & a\_{23} & a\_{33} & a\_{34} \\
 a\_{14} & a\_{24} & a\_{34} & a\_{44} \\
\end{array}
\right)
$$
contains exactly 21 minors of order 2, but there is a linear combination of them, namely
$$
-(a\_{13} a\_{24}-a\_{14} a\_{23})+(a\_{12} a\_{34}-a\_{14} a\_{23})-(a\_{12} a\_{34}-a\_{13} a\_{24})\, ,
$$
which vanishes, as opposed as to the case of symmetric $3\times 3$ symmetric matrices, whose 6 minors of order two are linearly independent.



> 
> **BIG QUESTION**: Why is that? I mean, what is the theory behind such a phenomenon?
> 
> 
> PHILOSOPHICAL QUESTION: What is the "true number" of minors (from the example above), 20 or 21?
> 
> 
> 


I already gave myself an explanation, but I still cannot see the big picture. I would be grateful if anyone pointed out a reference, sparing me the efforts of reinventing the wheel.


If an $n\times n$ matrix $A$ is regarded as an element of $V\otimes\_{\mathbb{K}} V^\ast$, with $V\equiv \mathbb{K}^n$, then there is an obvious way to extend $A$ to a $\mathbb{K}$-linear map
$$
A^{(k)}:\bigwedge^kV\longrightarrow\bigwedge^kV^\ast\, .
$$
If $A$ is symmetric, then so is $A^{(k)}$, i.e., there is a (polynomial of degree $k$) map
$$
S^2(V^\ast)\ni A\stackrel{p}{\longmapsto} A^{(k)}\in W^k:=S^2\left( \bigwedge^kV ^\ast \right)\, .
$$
Observe that $W\_k$ has dimension $\frac{{n\choose k}\left({n\choose k}+1\right)}{2}$, which is 21 for $n=4$ and $k=2$. So, I was led to identify $W^k$ with the space of "formal minors" of order $k$ of a $n\times n$ matrix (indeed, the entries of $A^{(k)}$ are precisely such minors).


How to explain now the dependency of three of them?


There is a linear map
$$
S^2\left( \bigwedge^2V ^\ast \right)\ni\rho\odot\eta\stackrel{\epsilon}{\longmapsto}\rho\wedge\eta\in \bigwedge^4V ^\ast\equiv\mathbb{K}\, ,
$$
and it can be proved that $\epsilon\circ p=0$, i.e., that $p$ takes its values in the subspace
$$
W^2\_0:=\ker\epsilon\, ,
$$
which has dimension 20 (vanishing of the above linear combination corresponds precisely to the equation $\epsilon(p(A))=0$).



> 
> SIDE QUESTION: How to define this $W\_0^k$ for arbitrary $n$ and $k$? It should be the "linear envelope" of the image of $p$: but how to exhibit it explicitly?
> 
> 
> 


Probably, the theory of representations may answer this: $W^k$ is not always an irreducible $\mathfrak{gl}(n,\mathbb{K})$-module, and $W\_0^k$ is the only irreducible component which contains the image of $p$. Provided this guess is true, it doesn't help me finding the expression of $W\_0^k$ in terms of tensors on $V$.



Why there is a relation among the second-order minors of a symmetric $4\times 4$ matrix?
This is about your specific question. For any vector space $V$ of dimension $n$, one has canonical decompositions
$$
S^2(S^2V^\*) \simeq S^4(V^\*)\oplus K(V^\*)
$$
and 
$$
S^2(\Lambda^2V^\*) \simeq \Lambda^4(V^\*)\oplus K(V^\*),
$$
where $K(V^\*)$ is an irreducible $\mathrm{GL}(V)$-module of dimension $n^2(n^2{-}1)/12$. It follows that, up to a constant multiple, there is only one nontrivial $\mathrm{GL}(V)$-equivariant linear mapping
$$
S^2(S^2V^\*) \longrightarrow S^2(\Lambda^2V^\*),
$$
and hence that the number of linearly independent minors of order $2$ of an $n$-by-$n$ symmetric matrix is $n^2(n^2{-}1)/12$.


You can answer similar questions by comparing $S^k(S^2V^\*)$ with $S^2(\Lambda^kV^\*)$ for higher values of $k$. This kind of question goes by the name 'plethysm' in representation theory. For example, when $k=3$, we find that $S^3(S^2V^\*)$ and $S^2(\Lambda^3V^\*)$ share only one irreducible $\mathrm{GL}(V)$-constituent, which has dimension
$$
\frac{(n{-}2)(n{-}1)^2n^2(n{+}1)}{144},
$$
so this is the number of linearly independent $3$-by-$3$ minors of an $n$-by-$n$ symmetric matrix.


**Update:** Here is a conjectural answer to the big question: Recall that, for a vector space $V$ of dimension $n$, a fundamental system of irreducible $\mathrm{SL}(V)$ modules is given by $\Lambda^kV^\*$ for $0<k<n$. If, as is traditional, one assigns highest weights so that $\Lambda^kV^\*$ has highest weight $\alpha\_k$, then any irreducible representation $W$ of $\mathrm{SL}(V)$ is determined by its highest weight $w = m\_1\,\alpha\_1 + \cdots + m\_{n-1}\,\alpha\_{n-1}$, where the $m\_i$ are nonnegative integers. 


Using this nomenclature, $S^2(V^\*)$, which is irreducible, has highest weight $2\,\alpha\_1$. 


The following Conjecture is supported by explicit computations for $0<k< n\le 9$. (Presumably, this conjecture (if true) would be easy to prove for an expert combinatorialist or representation theorist, but I am neither.)


**Conjecture:** The $\mathrm{SL}(V)$-modules $S^k(S^2V^\*)$ and $S^2(\Lambda^kV^\*)$ have only one common irreducible $\mathrm{SL}(V)$-constituent, namely, the irreducible $\mathrm{SL}(V)$-module $W$ of highest weight $2\,\alpha\_k$.


Since, by the Weyl Character formula, the dimension of the $W$ with highest weight $2\,\alpha\_k$ is
$$
\frac1{k{+}1}{n\choose k}{{n+1}\choose k},
$$
it would follow from the Conjecture that this is the number of linearly independent $k$-by-$k$ minors of a symmetric $n$-by-$n$ matrix.


