Linear dependence of determinants over $\operatorname{Mat}^{\mathbb{C}}_{ n}$

I try to solve task 2b from here.

Let us have $$\operatorname{Mat}^{\mathbb{C}}_k$$ and a set of functions $$\operatorname{Mat}^{\mathbb{C}}_{k} \to \mathbb{C}$$: $$\det(X-mE), \det(X-(m-1)E), ... ,\det(X), \det(X+E), \det(X+2E), ..., \det(X+mE).$$ Prove that there exists such sufficiently large $$m$$ that these functions are linearly dependent over $$\mathbb{C}$$.

The equivalent statement is that there exists $$m'$$ s.t. $$\det(X), ..., \det(X + m'E)$$ is linearly dependent.

Part $$2a$$ stated that for $$k=3$$, and $$m=1$$ this set of functions is linearly independent. It can be proved, if we take $$X = 0, E$$ and $$\begin{pmatrix} 0 & 0 & 1\\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{pmatrix}$$ and get $$\begin{pmatrix} -1 & 0 & 1\\ 0 & 1 & 8 \\ 0 & -1 & 0 \end{pmatrix}\begin{pmatrix} A \\ B \\C \end{pmatrix} = \begin{pmatrix} 0 \\0 \\ 0\end{pmatrix}$$ where $$A,B,C$$ are linear combination coefficients. Since matrix is non-degenerate, it has only trivial solution.

I see that if $$k$$ is even then the matrix $$\begin{pmatrix} 0 & 1^n & 2^n & \dots & (2k)^n \\ 1^n & 0 & 1^n & \dots & (2k-1)^n \\ \dots & \dots & \dots & \dots & \dots \\ (2k)^n & (2k-1)^n & (2k-2)^n & \dots & 0\end{pmatrix},$$ which we can get substituting $$X$$ by $$-kE , ... , 0 , ... , kE$$ is non-degenerate, but if $$k$$ is odd, then we get a skew-symmetric matrix with zero determinant. This problem for $$k=3$$ was solved with a trick, but I don't know how to prove it in general case.

• Hint: Theorem 1 in math.stackexchange.com/questions/2520370/… . More precisely, not Theorem 1 but the slight generalization you get when you replace $\det\left(\sum\limits_{i\in I} A_i\right)$ but $\det\left(X + \sum\limits_{i\in I} A_i\right)$ for some given square matrix $X$. (The proof is the same, except that you apply Corollary not to $m = 0$ but to $m = X$.) This shows that you can take the $m'$ in your equivalent statement to be $k + 1$. Commented May 1, 2019 at 21:16

The functions $$\det(X - mE)$$ are multilinear functions of the elements of $$X$$: for example, when $$k=2$$ (the matrices are $$2\times 2$$) and $$m = 1$$, we have $$\det(X - mE) = (x_{11}-1)(x_{22}-1) - x_{12}x_{21} = x_{11}x_{22} - x_{12}x_{21} - x_{11} - x_{12} + 1.$$ There are finitely many terms possible in such a multilinear function. No entry of $$X$$ is ever multiplied by itself, so either it appears in a term with power $$1$$ or it does not, giving us $$2^{k^2}$$ possible terms. In other words, the functions $$\det(X - mE)$$ live in the vector space of dimension $$2^{k^2}$$ spanned by $$\{1, x_{11}, x_{12}, \dots, x_{kk}, x_{11}x_{12}, x_{11}x_{13}, \dots\},$$ that is, all the possible products of some of the entries.
We could do better. For example, an $$x_{11}x_{12}$$ term can never appear in $$\det(X - mE)$$ for any $$E$$. But if we don't care about the implied constant in "sufficiently large", we don't have to be careful about this sort of thing.
In particular, if we take $$2^{k^2}+1$$ functions of this form, such as $$\det(X), \det(X+E), \det(X+2E), \dots, \det(X + 2^{k^2}E),$$ then they are $$2^{k^2}+1$$ elements of a vector space of dimension $$2^{k^2}$$, so they are linearly dependent.