Matrices such that $(A+B)^n = A^n + B^n$ How can I prove that $(A+B)^n = A^n + B^n$ for all integers $n \geq 1$ ?
I have been thinking about induction? Start for example with basecase 2 and then assume it's true for n = k, which would imply it's true for n = k + 1. But exactly how would this look, could someone help me please? I guess something like $(A+B)^{k+1} = (A+B) \cdot (A+B)^k$
Oh yes...and A is a matrix which looks like $ \begin{bmatrix}
                                                1 & 1 & 1 \\
                                                1 & 1 & 1 \\
                                                1 & 1 & 1 \\
\end{bmatrix}$
B looks like $ \begin{bmatrix}
                -2 & 1 & 1 \\
                1 & -2 & 1 \\
                1 & 1 & -2 \\
\end{bmatrix}$

Once you establish $AB = BA = 0$ the induction is easy:
We will take as our base case $n = 2$ (so we may assume later $n$ is at least $2$ -the case $n = 1$ is trivial).
$(A + B)^2 = (A + B)(A + B) = A^2 + AB + BA + B^2 = A^2 + 0 + 0 + B^2 = A^2 + B^2$.
Suppose $(A + B)^{n-1} = A^{n-1} + B^{n-1}$.
Then $(A + B)^n = (A + B)(A + B)^{n-1} = A(A + B)^{n-1} + B(A + B)^{n-1}$
$= A(A^{n-1} + B^{n-1}) + B(A^{n-1} + B^{n-1})$ (by our induction hypothesis)
$= A^n + AB^{n-1} + BA^{n-1} + B^n$ (the distributive law for matrices)
$= A^n + AB(B^{n-2}) + BA(A^{n-2}) + B^n$ (perhaps now you see why we want $n \geq 2$)
$= A^n + 0(B^{n-2}) + 0(A^{n-2}) + B^n = A^n + B^n$.
There is no need to use the binomial theorem, although this is indeed a "special case" of when it applies (namely, when $A$ and $B$ commute).
