Matrix Group under multiplication

Question

Suppose the collection $\{A_1, A_2,...,A_k\}$ forms a Group under matrix multiplication, where each $A_i$ is an $n \times n$ real matrix. Let $A = \sum_{i=1}^{k} A_i$

• Show that $A^2 = kA$
• If the trace of $A$ is zero, then show that $A$ is the zero matrix.

Context

I am new to Group Theory and while browsing the web, I recently stumbled upon this question. I have tried to prove the first part but have no idea how to proceed with the next.

Attempt:

Part 1

Let us construct a Cayley Table, T

$\times$	$A_1$	$A_2$	$...$	$A_k$
$A_1$	$A_1 A_1$	$A_1 A_2$	$...$	$A_1 A_k$
$A_2$	$A_2 A_1$	$A_2 A_2$	$...$	$A_2 A_k$
$...$	$...$	$...$	$...$	$...$
$A_k$	$A_k A_1$	$A_k A_2$	$...$	$A_k A_k$

$$ A^2 = \{A_1 + A_2 + ... + A_k \}\{A_1 + A_2 + ... + A_k \}\\ A^2 = \{A_1 A_1 + A_1 A_2 + ... + A_1 A_k + A_2 A_1 + A_2 A_2 + ... + A_2 A_k +...A_k A_1 + A_k A_2 + ... + A_k A_k\}\\ A^2 = \sum_{i=0}^k \sum_{j=0}^k a_{ij}, \forall a_{ij} \in T\\ $$

Since, each row of a Cayley table consists of unique elements of the Group, hence the sum of all elements in each row should be equal to $A$. As there are k such rows, thus, the sum of all the elements of all the rows must be $kA$.

Therefore, $$ A^2 = kA $$

Part 2 If,

$$ A^2 = kA \\ \implies A^2 - kA = 0 \\ $$ As eigenvalues of $A$ must satisfy the characteristic equation, hence

$$ \lambda^2 - k\lambda = 0 $$ Comparing this with the general format of a quadratic characteristic equation $$ \lambda^2 - (trace (A))\lambda + det(A)=0 $$ We get, $k=trace(A)=0$, as per the given data. This gives as $\lambda=0,0$. Hence, all the eigen values are 0.

Question:

1. Is there any more elegant proof of the first part than this?
2. How to proceed in the second part? I am completely lost.

Answer 1

If $G$ is a finite group, then for each $g\in G$ the map $x\mapsto gx$ is a bijection.

In your case, when you do $$ A_i(A_1+A_2+\dots+A_k) $$ you're simply doing $A_1+A_2+\dots+A_k$, just in a different order. Therefore $$ A^2=\sum_{i=1}^k A_i(A_1+A_2+\dots+A_k)=\sum_{i=1}^k(A_1+A_2+\dots+A_k)=kA $$

Therefore, if $\lambda$ is an eigenvalue of $A$ (in the complex numbers), you have $\lambda^2=k\lambda$, so either $\lambda=0$ or $\lambda=k$. Thus, if one eigenvalue is nonzero, the trace would be nonzero as well.

On the other hand, the minimal polynomial of $A$ divides $x(x-k)$, so it has distinct roots and therefore $A$ is diagonalizable. Hence $A=0$.

Answer 2

I do not think that your argument for part 2 is correct.

We have that $A^2=kA$.

Hence the minimal polynomial of $A$ divides $X(X-k)$

Suppose that $A$ is not equal to $0$; hence the minimal polynomial of $A$ cannot be $X$, and so is either $X-k$ or $X(X-k)$. In either case the characteristic polynomial of $A$ is $X^{n-s}(X-k)^s$ for some $s\ne 0$. The trace is the sum of the eigenvalues, and so the coefficient of $-X^{n-1}$ in the characteristic polynomial. That is the trace is $sk$. As $s\ne 0$ this forces $k=0$, and this contradicts the fact that $k$ (the order of a group) is at least $1$.