Kernel of a Vandermonde like matrix

I am wondering how to show that the following matrix has trivial kernel: $$\begin{bmatrix} 1&1&1&1&1&1 \\ s_1&s_2&s_3&s_4&s_5&s_6 \\ s_1^2&s_2^2&s_3^2&s_4^2&s_5^2&s_6^2 \\ s_1^3&s_2^3&s_3^3&s_4^3&s_5^3&s_6^3 \\ e^{s_1}s_1^3&e^{s_2}s_2^3&e^{s_3}s_3^3&e^{s_4}s_4^3&e^{s_5}s_5^3&e^{s_6}s_6^3 \\ e^{s_1}s_1^4&e^{s_2}s_2^4&e^{s_3}s_3^4&e^{s_4}s_4^4&e^{s_5}s_5^4&e^{s_6}s_6^4 \\ e^{s_1}s_1^5&e^{s_2}s_2^5&e^{s_3}s_3^5&e^{s_4}s_4^5&e^{s_5}s_5^5&e^{s_6}s_6^5 \end{bmatrix}$$ where $\zeta,\lambda\in\mathbb{R}\setminus\{0\}$, and $s_i \in \mathbb{C}$ are the roots of the polynomial $\zeta ^2 s^6-s^4+\lambda ^2=0$, which are all non-zero and distinct.

• I think you can use linear independence of monomials. $x^0,x^1,x^2,\cdots$ are all linearly independent. – mathreadler May 18 '16 at 7:40
• "The roots are all non-zero" ? But $s=0$ is a root of $\zeta ^2 s^6-s^4+\lambda ^2s=0$, is it not ? Your question is unclear – Ewan Delanoy May 21 '16 at 7:03
• Sorry, in the polynomial expression there was a typo. corrected – Sara Winslet May 21 '16 at 15:20
• Is there anything to say about $\zeta$ and $\lambda$? Are they complex? – Alex M. May 21 '16 at 17:33
• @Alex M. No, they have real non-zero values – Sara Winslet May 21 '16 at 22:08

Assuming that $\zeta,\lambda\in\color{red}{\mathbb{Q}}$, it is a consequence of the Lindemann-Weierstrass theorem.

The existence of a non-zero vector $(a_1,a_2,a_3,a_4,a_5,a_6,a_7)$ in the null-space of $M^T$ would imply the vanishing of $$f(z)=a_1+a_2 z+a_3 z^2+ a_4 z^3+ z^3 \color{red}{e^{z}}\left(a_5+a_6 z+a_7 z^2\right)$$ at every root of $\zeta^2 z^6-z^4 +\lambda^2$, i.e. over the set $\{s_1,s_2,s_3,s_4,s_5,s_6\}$. However, assuming that $s_1,s_2,s_3$ are linearly independent, $e^{s_1},e^{s_2},e^{s_3}$ are algebraically independent, hence that cannot happen and $M$ has to be an invertible matrix.

We may also replace the initial assumption $\zeta,\lambda\in{\mathbb{Q}}$ with "$\zeta,\lambda$ are two algebraic numbers over $\mathbb{Q}$" and the argument proving $\det(M)\neq 0$ is more or less the same. Moreover, $\det(M)$ is a continuous function of $\{s_1,s_2,s_3,s_4,s_5,s_6\}$ and every $s_i$ is a continuous function of $\zeta,\lambda$, so $\det(M)$ is a continuous function of $(\zeta,\lambda)\in\mathbb{R}^2$, non-vanishing over $\mathbb{Q}^2$. However, that is not enough for stating $\det(M)\neq 0$ for any $(\zeta,\lambda)\in\mathbb{R}^2$, since such a determinant may vanish over a set of isolated points $(\zeta,\lambda)$ having a trascendental coordinate. In the general case, I fear that some lower bound on $\left|\det M\right|$ (or, probably better to handle, $\det(MM^T)$) is required, but luckily enough it is not out of reach, since we may write every $s_i$ as an explicit function of $\zeta$ and $\lambda$ through the cubic formula.

• First of all, thank you for answering this question. I read through your answer a couple of times, I would like to point out some comments on your answer. Firstly, I was confused about taking determinant and talking about invertibility of a non-square matrix denoted by $M$. Secondly, coefficients $(a_1,a_2,a_3,a_4,a_5,a_6,a_7)$ are not necessarily algebraic numbers; we have the flexibility to pick them up from anywhere on real line. Hence, I am afraid that Lindemann-Weierstrass theorem is not applied properly. – Sara Winslet May 27 '16 at 3:19
• Thirdly, is it really a reasonable assumption that $s_1, s_2, s_3$ are all linearly independent over $\mathbb{Q}$, I am wondering how to say that the roots of a polynomial with rational coefficients are linearly independent? what if some roots are rationals distinct numbers. – Sara Winslet May 27 '16 at 3:19

I think that the problem is ill-posed.

In the sequel, we assume that the $(s_i)$ are non-zero real numbers. Let $A$ be the considered matrix and $B$ be the matrix deduced from $A$ by removing its first row. If there are $i\not=j$ s.t. $s_i=s_j$, then $rank(A)<6$. If the $(s_j)$ are distinct, then is $rank(A)=6$ true ? I think that the answer is yes. It suffices to show that if the $(s_j)$ are distinct, then $\det(B)\not= 0$.

Numerical experiments seems (at least to me) to "show" that the previous result is true. Proceed as follows: give random distinct values to $(s_i)_{i\geq 2}$. Then plot the graph of the function $f$, where $\det(B)=f(s_1)$. I obtain always only the $5$ solutions $s_1=s_2,\cdots,s_6$.

EDIT. I did not see that the $(s_i)$ are in fact complex numbers. I don't know if the proposed result could stand on complex numbers!

• Thanks for your answer. This is exactly what I started with, every numerical computation agrees that the kernel of the matrix is trivial; however, it should be proved analytically. There is also an interesting fact, I proved that the matrix formed by deleting the fourth row has vanishing determinant for some $\lambda$'s belong to a specific infinite but countable set of numbers. – Sara Winslet May 27 '16 at 3:28
• @ Sara Winslet , I think that your problem is difficult and interesting. Can you tell me the origin of this problem ? On the other hand, if you delete the fourth (or the fifth row), the valuations of the considered functions are distinct and the problem is less difficult; how have you shown your result above: "the matrix formed by deleting the fourth row has vanishing determinant for some λ's belong to a specific infinite but countable set of numbers." ? – loup blanc Jun 28 '16 at 10:23