1D interpolation: finding a polynomial satisfying $\forall_i\ p(x_i)=y_i$ can be written as a system of linear equations, having well known Vandermonde determinant: $\det=\prod_{i<j} (x_i-x_j)$. Hence, the interpolation problem is well defined as long as the system of equations is determined ($\det\neq 0$), that is equivalent with condition of having no two repeating $x$-s.

I need something analogous for quadratic form in $n$ dimensions: we would like to find symmetric matrix $A$ satisfying $\forall_k\ f(x^k)=y_k$, where $f(x)=x^T A x$, this time $x^k$ are vectors.

We get a system of linear equations: $$ \forall_k\ \sum_i A_{ii} (x^k_i)^2 + 2\sum_{i<j} A_{ij} x_i^k x_j^k =y_k$$ for $D=n(n+1)/2$ coefficients of symmetric $A$.

In analogy to interpolation problem, having values in $D$ points, we would like to find $A$. However, it requires that $\det\neq 0$ for the above set of linear equations.

Is there known a compact form for this determinant? (in analogy to Vandermonde)

If not, are there some known conditions ensuring it is nonzero - making fitting quadratic form well defined? These conditions need to contain e.g. that no two points are in one line $(x^k=a\cdot x^l)$. In what I need we can assume that all points lie on a sphere.

Specifically, my motivation is that looking at eigenspaces of adjacency matrix, we can convert the graph isomorphism problem into a question if two sets of points differ only by rotation (page 9-11 here. For strongly regular graphs these points are on a sphere and form a very regular polyhedron. Hence, I wanted to use an affine space of quadratic forms defining "wobbling" ellipsoids, such that they intersect only in our set - then we could use characteristic polynomial to test if they differ only by rotation. The crucial question is if e.g. $\{x: x^T A x=1 \textrm{ for all }A=A_0 + a\cdot A_1\}$ doesn't add too many extra points to the description. Geometrically: if "wobbling" ellipsoids with fixed some points, what extra fixed points would their intersection have?

Here is example of 2D situation: describing two points as intersection of ellipses/hyperbolas. Intersection only adds symmetric ($-x$) points, the question is when it is true, also in higher dimensions: enter image description here

  • $\begingroup$ I have browsed your paper about P=NP. Very interesting. About the fact that two set of points differ by a certain rotation, do you the Procustes algorithm (testing with matrix $UV^T$ coming from a certain SVD decomposition ? Besides, you work on spectral theory of graphs on adjacency matrices ; sometimes laplacians of graphs (diagonal matrix of degrees minus adjacency matrix) can bring more information : you can find many many things about this theory in the very rich book "An introduction to the theory of Graph Spectra" by Cvetkovic, Rowlinson and Simic, Cambridge Ed. 2010. $\endgroup$
    – Jean Marie
    Dec 3, 2017 at 8:30
  • $\begingroup$ Thanks, it is far from proving P=NP (yet?), only tries to gather approaches which are not likely to be found by people usually working on it. E.g. Grassmann variables known mostly by solid state physicists - P!=NP says their representation needs exponential cost. Graph isomorphism is simpler, I will have to look at Procrustes ... I had some experience with rotational invariants using spherical harmonics - see slide 5 here. Quadratics seem promising, but might add too many extra points? $\endgroup$
    – Jarek Duda
    Dec 3, 2017 at 10:00
  • $\begingroup$ ps. regarding using laplacian of graph instead, for regular there is no difference of eigenspaces, and I have to admit that my background is rather in MERW ( en.wikipedia.org/wiki/Maximal_entropy_random_walk ), which uses eigendecomposition of just adjacency matrix - it is more natural for me. Laplacian is close to heat kernel, which can be also done with MERW - nice paper: citeseerx.ist.psu.edu/viewdoc/… $\endgroup$
    – Jarek Duda
    Dec 3, 2017 at 23:15
  • 1
    $\begingroup$ Related: mathoverflow.net/questions/188746/… $\endgroup$
    – Dap
    Dec 8, 2017 at 14:22

1 Answer 1


I am not certain that I fully answer your question. Tell me if such is the case.

Let me introduce the following notation, different from yours : let us set

$$Y_{ij}:=X_i^TAX_j \in \mathbb{R}.$$

With this notation, do we agree that the system of constraints can be written :


where $X=[X_1|X_2|\cdots|X_n]$ ? (the columns of $X$ are the $X_i$s).

But (1) can be written, using Kronecker (=tensor) product $\otimes$ and operator "vec" (which means a conversion of a $n \times n$ matrix into a $n^2 \times 1$ column vector). See for that "Matrix equations" in (https://en.wikipedia.org/wiki/Kronecker_product):

$$\tag{2}\underbrace{(X^T \otimes X^T)}_{\text{known}}\underbrace{vec(A)}_{\text{unknown}}=\underbrace{vec(Y)}_{\text{known}}$$

where "the compact expression" $X^T \otimes X^T$ is a $n^2 \times n^2$ matrix .

Thus we have transformed the issue into the resolution of a linear system.


$$\det(X^T \otimes X^T)= \det(X^T)^{2n}, $$ (Determinant of the Kronecker Product of Two Matrices)

the condition for this system to have a unique solution is still $\det(X^T) \ne 0 \iff \det(X) \ne 0$.

A remark: one can wonder if the constraint of symmetry is taken into account: this is the case because in (2), $vec(A)$ and $vec(Y)$ correspond to symmetrical unknowns (for $A$) and symmetrical data (for $Y$).

  • $\begingroup$ Any comment ?.. $\endgroup$
    – Jean Marie
    Dec 3, 2017 at 8:31
  • $\begingroup$ Thank you, I am just starting today - I have to look closer, but at the first look there is a dimensionality problem: the number of A coefficients and so points is D=n(n+1)/2, so your N matrix is n x D, should I understand it that any n x n minor has det != 0? $\endgroup$
    – Jarek Duda
    Dec 3, 2017 at 9:44
  • $\begingroup$ I understand your argument. I am going to take (maybe this evening) a simple example in order to see (first for me) how it works. $\endgroup$
    – Jean Marie
    Dec 3, 2017 at 9:49
  • $\begingroup$ And we don't know the entire Y, only its diagonal ... maybe seeing A in diagonal form could help: A = O^T D O ... $\endgroup$
    – Jarek Duda
    Dec 3, 2017 at 11:21
  • $\begingroup$ About procrustes analysis, I like this overview by Higham, a prominent numerical analyst (www.maths.manchester.ac.uk/~higham/talks/procrust94.ps.gz) $\endgroup$
    – Jean Marie
    Dec 3, 2017 at 11:39

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