# Finding the minimal value of a $4\times 4$ determinant

The question.

Let $$\xi=(\xi_1,\xi_2,\xi_3,\xi_4)\in\mathbb R^4$$ be a vector with irrational coordinates.

I am interested in finding the minimal value $$\mu_\xi$$ of

$$\left\vert \det \begin{pmatrix} a_1 & a_2 & 0 & 1 \\ b_1 & b_2 & 1 & 0 \\ c_1 & c_2 & \xi_1 & \xi_3 \\ d_1 & d_2 & \xi_2 & \xi_4 \end{pmatrix} \right\vert,$$

for $$a_1,b_1,c_1,d_1,a_2,b_2,c_2,d_2\in\mathbb Z$$, in terms of the area of the parallelepiped formed by the two vectors $$X_i:=(a_i,b_i,c_i,d_i)$$, $$i=1,2$$ (which I assume linearly independent), in $$\mathbb R^4$$.

Let's call this area $$D(X_1,X_2)$$. I know we have

\begin{align*} D(X_1,X_2)^2 &= \Vert X_1\Vert^2\Vert X_2\Vert^2-(X_1\cdot X_2)^2 \end{align*},

but I have no clue on how to proceed from here.

The conjecture.

My hope (which would help the construction of another proof a lot) would be that if we chose the $$\xi_i$$ properly, we can show that the minimal value verifies

$$$$\mu_\xi\geqslant \frac c{D(X_1,X_2)^2}\qquad\qquad (1)$$$$

where $$c$$ is a constant (it may depends on $$\xi$$).

Final remarks.

Despite the fact that I strongly believe that $$(1)$$ is true, any proof that would show that

$$\mu_\xi\geqslant \frac c{D(X_1,X_2)^\gamma}$$

for a $$\gamma<4$$ would be of great interest.

• By the definition of the determinant, your minimun depends well on $\xi$ vector. If you take the two vectors the area formula $D(X_1,X_2)$ isn't right if it is the triangle area formed by $X_1,X_2$. So either you compute the expression and minimize if possible, either i guess taking particular values of $\xi$ and see a geometric proof. – Toni Mhax Sep 26 '18 at 12:56
• @ToniMhax I edited to specify what I mean by the area formed by $X_1$ and $X_2$. Yes, the minimum depends on $\xi$, but I am interested in a particular value of $\xi$ that would solve the conjecture. – E. Joseph Sep 26 '18 at 13:04
• I think $D(X_1,X_2)^2=||X_1||^2||X_2||^2-(X_1•X_2)^2$ – Empy2 Sep 26 '18 at 13:38
• @Empy2 You're right, thank you, I edited. – E. Joseph Sep 26 '18 at 13:39
• – Toni Mhax Sep 26 '18 at 14:14

The answer is wide so we need notations, for any $$(X_1,X_2)$$ in $${\mathbb{Z}}^2$$ distinct, there is $$A$$ such that $$\dfrac{1}{D(X_1,X_2)}\le \dfrac{1}{A}$$. $$C:=area(X_1,X_2)\ge a$$ (parallelogram) also $$a\neq 0$$ exist. The determinant is the volume of the four column vectors of the given matrix [which we suppose invertible] and equals $$\mu_{\xi}=C.\alpha_{\xi}\ge \frac{c_{\xi}}{A}\ge \frac{c_{\xi}}{D(X_1,X_2)}$$ for some $$\alpha_{\xi}$$ where $$c_{\xi}= a.A.\alpha_{\xi}$$.
Edit $$D(X_1,X_2)$$ is the volume of $$(X_1,X_2, X_1\times X_2)$$
Finding the minimum value $$a$$ explicitly is the same as finding the minimum value of $$A$$.
• It shows for any $\xi$ the existence of $c_{\xi}$ as in the bound. – Toni Mhax Sep 26 '18 at 17:10
• But your $c_\xi$ depends on $X_1,X_2$, which is not allowed unfortunately. – E. Joseph Sep 26 '18 at 17:29
• Nop. Anyway we hit a wall of supposing the matrix invertible. $a$ and $A$ can be computed, (hard now) when the vectors $X_1,X_2$ are in ${\mathbb{Z}}^4$ – Toni Mhax Sep 26 '18 at 17:44
• I could not follow any of your computations. How can you get a sharp bound when your proof basically starts saying that $D(X_1,X_2)$ is non-zero? – C. Falcon Sep 26 '18 at 19:12