Let $G$ be a collection of $n\times n$ invertible matrices, satifying that $A,B\in G, A\neq B, |A+B|=0$. Show $G$ is a finite set.

Clearly, if $n=1$, $G=\{a,-a\}, a\neq 0$. I think it is impossible to use induction on $n$.

Try another approach. Can we make some linear combination of the first rows of matrices in $G$, so that the sum is zero...Oh.

Let $A=(\alpha_1,\cdots,\alpha_n)\in G$, then we can set $G=\{A, (-\alpha_1,\alpha_2,\cdots, \alpha_n)\}$ as an example. Can this help?

Any finite subset of $\{det(X+A); A\in G, X\in \Bbb P^{n\times n}\}$ is linearly independent. Is this helps?

  • $\begingroup$ A potentially helpful idea: for each $A \in G$, we have $G \subset S_A:= \{X: |A + X| = 0\}$, which is the zero-set of a polynomial. By adding elements to $G$, you reduce the dimension of the variety $\bigcap_{A \in G}S_A$ until you reach the point where $G \subset \bigcap_{A \in G}S_A$ ensures that $G$ is a finite set. $\endgroup$ Jan 28 at 5:31

1 Answer 1


Suppose that $G$ is infinite. For any $N$ distinct elements $A_1,\ldots,A_N$, consider the following multivariable polynomials in variables $x_{11}, x_{12}, \ldots, x_{nn}$ (entries of $X$ as variables): $$ \det (A_1+X), \det (A_2+X), \ldots, \det (A_N+X). $$

Suppose that scalars $c_1,\ldots, c_N$ with $$ c_1\det (A_1+X) + c_2\det (A_2+X)+ \cdots + c_N \det (A_N+X)=0. $$ Then for each $i=1,\ldots, N$, by the property of $G$, $$ c_i\det(A_i+A_i)=0 $$ That is, $c_i 2^n \det (A_i)=0$. This gives $c_i=0$. Thus, those $N$ multivariable polynomials are linearly independent.

We take $N$ at least one plus the number of nonconstant terms of $\det (A+X)$. Then the nonconstant terms of $\det (A_i+X)$'s are linearly dependent. There are scalars $c_1,\ldots, c_N$ not all zeros such that $$ c_1 \det (A_1+X)+c_2\det (A_2+X)+\cdots +c_N \det (A_N+X)$$ has all nonconstant terms vanish.

By the linear independence of those polynomials, the constant term of the above cannot vanish. Then we have $c\neq 0$ with $$ c_1 \det (A_1+X)+c_2\det (A_2+X)+\cdots +c_N \det (A_N+X)=c.$$

Since we assume $G$ is infinite, there is $A_{N+1}$ in $G$, different from each $A_i$, $i=1,\ldots, N$. Then putting $X=A_{N+1}$ gives $0=c$, contradiction.


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