# 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.

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?

• 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. Jan 28 at 5:31

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.