# Discriminant of characteristic polynomial as sum of squares

The characteristic polynomial of a real symmetric $$n\times n$$ matrix $$H$$ has $$n$$ real roots, counted with multiplicity. Therefore the discriminant $$D(H)$$ of this polynomial is zero or positive. It is zero if and only if there is a degenerate eigenvalue.

Thus $$D(H)$$ is a non-negative (homogeneous) polynomial in the $$\frac12n(n+1)$$ entries of $$H$$. Some non-negative polynomials can be written as a sum of squares and I am interested in whether $$D(H)$$ can. There is a concrete question at the end, but any insights into the general case are also welcome.

The size of the problem grows very quickly with dimension, so I will only look at $$n=2$$ (which I do understand) and $$n=3$$ (which I am yet to understand).

## 2D

In two dimensions it is pretty easy to write down the polynomial and its discriminant and see by eye that $$D(H) = (h_{11}-h_{22})^2 + 4h_{12}^2,$$ which is indeed a sum of two squares.

Having a degenerate eigenvalue is a polynomial condition: it happens if and only if $$D(H)=0$$. The discriminant is a second order polynomial, but writing it as a sum of squares leads to far simpler algebraic condition: $$h_{11}-h_{22}=0$$ and $$h_{12}=0$$. Simple algebraic conditions for degeneracy are the goal here, but I thought the question would be of some interest in itself.

## 3D

In three dimensions the discriminant is pretty big: $$D(H) = h_{22}^2h_{33}^4-2h_{11}h_{22}h_{33}^4+4h_{12}^2h_{33}^4+h_{11}^2h_{33}^4-2h_{22}h_{23}^2h_{33}^3+2h_{11}h_{23}^2h_{33}^3-8h_{12}h_{13}h_{23}h_{33}^3-2h_{22}^3h_{33}^3+2h_{11}h_{22}^2h_{33}^3+2h_{13}^2h_{22}h_{33}^3-8h_{12}^2h_{22}h_{33}^3+2h_{11}^2h_{22}h_{33}^3-2h_{11}h_{13}^2h_{33}^3-8h_{11}h_{12}^2h_{33}^3-2h_{11}^3h_{33}^3+h_{23}^4h_{33}^2+8h_{22}^2h_{23}^2h_{33}^2-10h_{11}h_{22}h_{23}^2h_{33}^2+2h_{13}^2h_{23}^2h_{33}^2+20h_{12}^2h_{23}^2h_{33}^2+2h_{11}^2h_{23}^2h_{33}^2+12h_{12}h_{13}h_{22}h_{23}h_{33}^2+12h_{11}h_{12}h_{13}h_{23}h_{33}^2+h_{22}^4h_{33}^2+2h_{11}h_{22}^3h_{33}^2+2h_{13}^2h_{22}^2h_{33}^2+2h_{12}^2h_{22}^2h_{33}^2-6h_{11}^2h_{22}^2h_{33}^2-10h_{11}h_{13}^2h_{22}h_{33}^2+20h_{11}h_{12}^2h_{22}h_{33}^2+2h_{11}^3h_{22}h_{33}^2+h_{13}^4h_{33}^2+20h_{12}^2h_{13}^2h_{33}^2+8h_{11}^2h_{13}^2h_{33}^2-8h_{12}^4h_{33}^2+2h_{11}^2h_{12}^2h_{33}^2+h_{11}^4h_{33}^2-10h_{22}h_{23}^4h_{33}+8h_{11}h_{23}^4h_{33}-36h_{12}h_{13}h_{23}^3h_{33}-2h_{22}^3h_{23}^2h_{33}-10h_{11}h_{22}^2h_{23}^2h_{33}-2h_{13}^2h_{22}h_{23}^2h_{33}-2h_{12}^2h_{22}h_{23}^2h_{33}+20h_{11}^2h_{22}h_{23}^2h_{33}-2h_{11}h_{13}^2h_{23}^2h_{33}-38h_{11}h_{12}^2h_{23}^2h_{33}-8h_{11}^3h_{23}^2h_{33}+12h_{12}h_{13}h_{22}^2h_{23}h_{33}-48h_{11}h_{12}h_{13}h_{22}h_{23}h_{33}-36h_{12}h_{13}^3h_{23}h_{33}+72h_{12}^3h_{13}h_{23}h_{33}+12h_{11}^2h_{12}h_{13}h_{23}h_{33}-2h_{11}h_{22}^4h_{33}-8h_{13}^2h_{22}^3h_{33}+2h_{12}^2h_{22}^3h_{33}+2h_{11}^2h_{22}^3h_{33}+20h_{11}h_{13}^2h_{22}^2h_{33}-10h_{11}h_{12}^2h_{22}^2h_{33}+2h_{11}^3h_{22}^2h_{33}+8h_{13}^4h_{22}h_{33}-38h_{12}^2h_{13}^2h_{22}h_{33}-10h_{11}^2h_{13}^2h_{22}h_{33}+8h_{12}^4h_{22}h_{33}-10h_{11}^2h_{12}^2h_{22}h_{33}-2h_{11}^4h_{22}h_{33}-10h_{11}h_{13}^4h_{33}-2h_{11}h_{12}^2h_{13}^2h_{33}-2h_{11}^3h_{13}^2h_{33}+8h_{11}h_{12}^4h_{33}+2h_{11}^3h_{12}^2h_{33}+4h_{23}^6+h_{22}^2h_{23}^4+8h_{11}h_{22}h_{23}^4+12h_{13}^2h_{23}^4+12h_{12}^2h_{23}^4-8h_{11}^2h_{23}^4-36h_{12}h_{13}h_{22}h_{23}^3+72h_{11}h_{12}h_{13}h_{23}^3+2h_{11}h_{22}^3h_{23}^2+20h_{13}^2h_{22}^2h_{23}^2+2h_{12}^2h_{22}^2h_{23}^2+2h_{11}^2h_{22}^2h_{23}^2-38h_{11}h_{13}^2h_{22}h_{23}^2-2h_{11}h_{12}^2h_{22}h_{23}^2-8h_{11}^3h_{22}h_{23}^2+12h_{13}^4h_{23}^2-84h_{12}^2h_{13}^2h_{23}^2+20h_{11}^2h_{13}^2h_{23}^2+12h_{12}^4h_{23}^2+20h_{11}^2h_{12}^2h_{23}^2+4h_{11}^4h_{23}^2-8h_{12}h_{13}h_{22}^3h_{23}+12h_{11}h_{12}h_{13}h_{22}^2h_{23}+72h_{12}h_{13}^3h_{22}h_{23}-36h_{12}^3h_{13}h_{22}h_{23}+12h_{11}^2h_{12}h_{13}h_{22}h_{23}-36h_{11}h_{12}h_{13}^3h_{23}-36h_{11}h_{12}^3h_{13}h_{23}-8h_{11}^3h_{12}h_{13}h_{23}+4h_{13}^2h_{22}^4+h_{11}^2h_{22}^4-8h_{11}h_{13}^2h_{22}^3-2h_{11}h_{12}^2h_{22}^3-2h_{11}^3h_{22}^3-8h_{13}^4h_{22}^2+20h_{12}^2h_{13}^2h_{22}^2+2h_{11}^2h_{13}^2h_{22}^2+h_{12}^4h_{22}^2+8h_{11}^2h_{12}^2h_{22}^2+h_{11}^4h_{22}^2+8h_{11}h_{13}^4h_{22}-2h_{11}h_{12}^2h_{13}^2h_{22}+2h_{11}^3h_{13}^2h_{22}-10h_{11}h_{12}^4h_{22}-2h_{11}^3h_{12}^2h_{22}+4h_{13}^6+12h_{12}^2h_{13}^4+h_{11}^2h_{13}^4+12h_{12}^4h_{13}^2+2h_{11}^2h_{12}^2h_{13}^2+4h_{12}^6+h_{11}^2h_{12}^4 .$$ (I got this by Maxima.) This is indeed a non-negative homogeneous polynomial of degree six in six variables, but it is too big for me to see any structure by eye and I cannot tell whether it is a sum of squares.

In the diagonal case $$h_{12}=h_{13}=h_{23}=0$$ the discriminant has a simpler expression: $$D(H) = (h_{11}-h_{22})^2 (h_{22}-h_{33})^2 (h_{11}-h_{33})^2.$$ This form is not at all surprising, as it should be a sixth degree polynomial vanishing if and only if two diagonal entries coincide.

My concrete question is: Is this $$D(H)$$ of the case $$n=3$$ a sum of squares (without assuming it is diagonal)? If yes, what are the squared polynomials and how unique are they?

I have understood that there are computational tools for finding a sum of squares decomposition, but I have yet to find one that I could run with the software I have. And I assume this particular polynomial has structure which simplifies matters: for example, the polynomial is invariant under orthogonal changes of basis and the non-negativity has a geometric meaning. One can indeed diagonalize the matrix, but I cannot see a way to use this to understand what the polynomial is in terms of the original basis. The 2D case and the diagonal 3D case suggest that being a sum of squares is a reasonable guess.

The answer for a general $$n$$ is positive: the discriminant is a sum of squares of polynomials in the entries of $$H$$. The first formula was given by Ilyushechkin and involves $$n!$$ squares. This number was improved by Domokos into $$\binom{2n-1}{n-1}-\binom{2n-3}{n-1}.$$ See Exercise #113 on my page.

Details of Ilyushechkin's solution. Consider the scalar product $$\langle A,B\rangle={\rm Tr}(AB)$$ over $${\bf Sym}_n({\mathbb R})$$. It extends as a scalar product over the exterior algebra. Then the discriminant equals $$\|I_n\wedge H\wedge\cdots\wedge H^{n-1}\|^2,$$ which is a sum of squares of polynomials.

The answer is Yes in any dimension by a result of Ilyushechkin in Mat. Zametki, 51, 16-23, 1992.

See my previous MO answer

real symmetric matrix has real eigenvalues - elementary proof

We know that $$H$$ is symmetric, and therefore, diagonalizable, as $$H = Q^TDQ$$ for some orthogonal matrix $$Q$$. Moreover, $$D$$ and $$Q$$ have the same eigenvalues, and thus the same characteristic polynomials. Perhaps this can be used?

In any case, this reference by Domokos mentions the other answers and references as well. It gives some explicit expressions in the 3x3 case, both in five squares (theorem 7.3) and in seven squares (theorem 7.4), showing that the decomposition is not unique.