Matrix Group under multiplication 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:

*

*Is there any more elegant proof of the first part than this?

*How to proceed in the second part? I am completely lost.

 A: 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 $A=0$.
A: 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 $1$.
