Matrix P to the power of 4, i.e $P^4$, is this the same as $P^2 \cdot P^2$? Basically, what it says in the title. 
I have a $5 \times 5$ matrix and I need to work out $P^4$, is it possible to just do $P^2$ and multiply this with itself?
 A: 
I have a $5 \times 5$ matrix and I need to work out $P^4$, is it possible to just do $P^2$ and multiply this with itself?

Yes, Why?: Because A square matrix always commutes with itself, under matrix multiplication. $\quad\;\;$WHY? Because matrix multiplication (when well-defined) is associative. $$P\cdot (P \cdot (P\cdot P)) = P\cdot ((P \cdot P)\cdot P) = (P\cdot (P\cdot P)) \cdot P $$ $$ =  ((P\cdot P) \cdot P) \cdot P = (P\cdot P) \cdot (P \cdot P) = P^2\cdot P^2 = P^4.$$ 

But be careful. Generally speaking, for square matrices $A, B$ such that $P = AB$, $$P^4 = (AB)^4 = (AB)^2\cdot (AB)^2 \ne A^2 \cdot A^2\cdot B^2 \cdot B^2 = A^4\cdot B^4  $$ since here, we cannot assume that $A$ and $B$ commute.
A: $P^4=P^2\times P^2$ for a matrix $P$. Multiplication of matrices is associative, so you can do that.
A: What does $P^4$ mean to you? It might mean $$P\cdot P\cdot P\cdot P$$ but this is not exactly defined until you agree that matrix multiplication is asociative. Otherwise $P^4$ should mean something like $$((P\cdot P)\cdot P)\cdot P$$ Now if you believe that the multiplication is associative, then this is the same as $$(P\cdot P)\cdot (P\cdot P)$$ which is almost certainly what you would mean by $P^2\cdot P^2$.
