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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.

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  • $\begingroup$ 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 $k + 1$. $\endgroup$ Commented May 1, 2019 at 21:16

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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.

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