Motivated by this question I propose a simplification:

Question Let matrices $A,B,C\in M_{2}(\mathbb{C})$ be Hermitian and positive definite, such that:$$A+B+C=I_2$$ Show that: $$\det\left(6(A^3+B^3+C^3)+I_{2}\right)\ge 5^2\det(A^2+B^2+C^2)$$ where $I_{2}$ is the identity matrix.

The original question is of unknown origin and I am hoping for a substantial simplification in the $2 \times 2$ case where the following expansion holds:

$$\left[ \begin{array}{cc} x_1 & y + iz \\ y - iz & x_2\end{array}\right] = \left[ \begin{array}{cc} x_1 & 0 \\ 0 & x_2\end{array}\right] + y\left[ \begin{array}{cc} 0 & 1 \\ 1 & 0\end{array}\right] + z\left[ \begin{array}{rc} 0 & i \\ - i & 0\end{array}\right]$$

maybe with expansion to Pauli spin matrices this is solvable.

There is a nice article by Knutson and Tao on Honeycombs that might be of assistnce:

  • $\begingroup$ In some of the (upvoted) comments accompanying the following question numerical evidence is presented by user1551, confirming that the statement is true for $10000$ of such random $2\times 2$ matrices: How to construct symmetric and positive definite $A,B,C$ such that $A+B+C=I$? $\endgroup$ Jan 12, 2017 at 20:25
  • $\begingroup$ Nice (+1) attempt to settle the problem, though! I'm breathlessly awaiting, really. $\endgroup$ Jan 12, 2017 at 20:28
  • $\begingroup$ @HandeBruijn contest problems and university exams can be tricky, it seems that discussion has gone down the wrong track and never recovered. I am hoping for a solution that emaphasizes general principles. I think this problem could be solved using convexity or matrix norms, for example. $\endgroup$
    – cactus314
    Jan 12, 2017 at 21:08
  • $\begingroup$ Considering the difficulties with proving such inequalities for real numbers, I doubt that someone would come up with a solution even for symmetric instead of Hermitian $2\times 2$ matrices. Thus it seems that the "power of mathematics" is somewhat unevenly distributed among its areas of interest. $\endgroup$ Jan 17, 2017 at 11:40
  • $\begingroup$ For the question here, the 2 by 2 case, a hermitian matrix can be thought of as an element of Minkowski 4-dimensional spacetime, with the determinant giving the Minkowski product of the vector with itself. The conditions that the matrix is positive definite amounts to the condition that the vector be timelike and future-pointing. $\endgroup$
    – Malkoun
    May 7, 2017 at 0:11

1 Answer 1


Let $L=\{(A,B,C)\in M_2(\mathbb{C})^3; A,B,C\text{ are hermitian }\geq 0,A+B+C=I_2\}$ and

$res:L\rightarrow \det(6(A^3 +B^3+C^3)+I_2)-25\det(A^2 +B^2+C^2)$.

Since we seek $\min_L res$, we may assume that $A$ is a diagonal matrix and that $res$ is defined on the set $K$ parameterized by

$A=diag(u,v),B=\begin{pmatrix}p&r+is\\r-is&q\end{pmatrix},C=\begin{pmatrix}1-u-p&-r-is\\-r+is&1-v-q\end{pmatrix}$, where

$u,v,p,q,1-u-p,1-v-q,pq-r^2-s^2,(1-u-p)(1-v-q)-r^2-s^2\geq 0$.

Note that $K$ is a compact convex subset of $\mathbb{R}^6$. The edge of $K$, $\partial K$ is the set of points of $K$ s.t. at least one of the $8$ terms above is zero and $K\setminus \partial K$ is the interior of $K$.

Of course, $\min_K res$ is reached in at least one point of $K$. We know that $\min_K res\leq 0$ and that, if $AB=BA$, then $res(A,B,C)\geq 0$.

We want to show

$\textbf{Theorem}.$ One has $\min_K res=0$ and this bound is reached only by commuting triplets of matrices.

$\bullet$ In the sequel, we assume that $AB\not= BA$.

An essential tool will be the Grobner basis theory on $\mathbb{C}$ and on $\mathbb{R}$. Since the theory over $\mathbb{R}$ is much less powerful, the idea is to reduce our problem to a polynomial system on $\mathbb{C}$ constituted by polynomials simpler than $res$ and to show that this system has no real solutions satisfying the $8$ conditions above.

$\textbf{Lemma}.$ We may assume $v>u\geq 0$ and $p,q,1-u-p,1-v-q>0$.

$\textbf{Proof}$. For example, if $pq=0$, then $r=s=0$ and $AB=BA$. $\square$

$\textbf{Proposition 1.}$ If $u>0,pq>r^2+s^2,(1-u-p)(1-v-q)>r^2+s^2$ (that is, $\det(A),\det(B),\det(C)>0$), then $(A,B,C)$ does not realize $\min_K res$.

$\textbf{Proof}$. Here $(A,B,C)\in K\setminus\partial K$; if it reaches $\min_K res$, then it must cancel the partial derivatives of $res$.

Step 1. We use the Maple library FGb (working over $\mathbb{C}$) to study the $3$ systems

$\dfrac{\partial res}{\partial u}=\dfrac{\partial res}{\partial v}=\dfrac{\partial res}{\partial p}=\dfrac{\partial res}{\partial q}=\dfrac{\partial res}{\partial r}=\dfrac{\partial res}{\partial s}=0,u-v\not= 0$, AND

$[(rs\not=0)$ OR $(r\not=0,s=0)$ OR $(r=0,s\not=0)]$,

asking (among other polynomials) for relations between $u,v$.

We obtain, in each case, a Grobner basis containing (in particular) $3$ polynomials $P,Q,R$ which only depend on $u,v$.

Step 2. We use the Maple library RAG (working over $\mathbb{R}$) to study the system


We obtain that there are no solutions in each of the $3$ cases. $\square$

$\textbf{Proposition 2}$. If $\det(A)$ or $\det(B)$ or $\det(C)=0$, then $(A,B,C)$ does not realize $\min_K res$.

$\textbf{Proof}$. We may assume that $\det(A)=0$, that is, $u=0$. If we put $a=r^2+s^2$ (with $a>0$ because $AB\not= BA$), then (Maple form)


We use the Maple library RAG (working over $\mathbb{R}$) to study the system

$f=0,v > 0, p > 0, q > 0, 1-p > 0, 1-v-q > 0, a > 0, a < 1$.

We obtain that there are no solutions.

Remark that $U=\{(v,p,q,a);v > 0, p > 0, q > 0, 1-p > 0, 1-v-q > 0, a > 0, a < 1\}$

is an open convex subset of $\mathbb{R}^4$ and that $f(0.1,0.3,0.4,0.1)\approx 5.6>0$.

We conclude that $f>0$ on $U$ and that $\min_K res$ is not reached on $U$. $\square$

$\textbf{Proof of Theorem.}$ According to Propositions 1 and 2, $\min_K res$ is reached only by triplets $(A,B,C)$ s.t. $AB=BA$; therefore $\min_k res=0$ and is reached only (up to simultaneous similarity) by the triplets

i) $diag(1/2,1/2),diag(1/2,0),diag(0,1/2)$

ii) $A=B=C=1/3.I_2$

iii) $A=B=0.5.I_2,C=0_2$

iv) $A=diag(1/3,1/2),B=diag(1/3,0),C=diag(1/3,1/2)$.


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