If the matrices $A^3 = 0$, $B^3=0$ and $AB=BA$ then show this: The question: If $A$ and $B$ are square matrices of the same type such that $A^3=0$, $B^3=0$ and $AB=BA$. Show that
$$\left(I+A+\frac{A^2}{2!}\right)\left(I+B+\frac{B^2}{2!}\right)=I+(A+B)+\frac{(A+B)^2}{2!}+\frac{(A+B)^3}{3!}+\frac{(A+B)^4}{4!}$$
This is how I started. 
$$\left(I+A+\frac{A^2}{2!}\right)\left(I+B+\frac{B^2}{2!}\right)=I+(A+B)+\frac{A^2+B^2}{2!}+AB+\frac{AB^2}{2!}+\frac{A^2B}{2!}+\frac{A^2B^2}{2!2!}$$ 
I tried to get 
$$\frac{A^2+B^2}{2!}+AB+\frac{AB^2}{2!}+\frac{A^2B}{2!}+\frac{A^2B^2}{2!2!}$$
to be equal to $$\frac{(A+B)^2}{2!}+\frac{(A+B)^3}{3!}+\frac{(A+B)^4}{4!}$$ But I'm not getting there. 
 A: I would do it without brute force:


*

*Since $A^3=0$ we have
$$
e^A=I+A+\frac{A^2}{2!}.
$$
Similar for $B$.

*The LHS of your expression is then $e^Ae^B$. For commuting matrices it is true that
$$
e^Ae^B=e^{A+B}.
$$

*Finally notice that if $A^3=B^3=0$ and $AB=BA$ then $(A+B)^5=0$ (e.g. do the binomial expansion and see that you get at least $A^3$ or $B^3$ in all terms). Thus, the RHS is precisely $e^{A+B}$.

A: Try to expand the right-hand side instead. You have
\begin{align}
\frac{1}{2}(A + B)^2 &= \frac{1}{2}A^2 + AB + \frac{1}{2}B^2\\[1mm]
\frac{1}{6}(A + B)^3 &= \frac{1}{2}A^2B + \frac{1}{2}AB^2\\[1mm]
\frac{1}{24}(A + B)^4 &= \frac{1}{4}A^2B^2,
\end{align}
where I have used that $A$ and $B$ commute as well as $A^k = B^k = 0$ for all integers $k \geq 3$. Summing everything up gives you the desired result.
A: You can simply match the terms of second order, third order and fourth order to the corresponding terms in the answer:
For second order terms: by multiplication $(A+B)^2=A^2+AB+BA+B^2$ and using $AB=BA$ we see that 
$$
\frac{(A+B)^2}{2!} = \frac{A^2+B^2}{2} + AB
$$
For third order terms we have by multiplication $(A+B)^3 = A^3 + (A^2B+ABA+B^2A) + (AB^2+BAB+B^2A) + B^3$.  By $A^3=B^3=0$ this becomes
$(A+B)^3=(A^2B+ABA+B^2A) + (AB^2+BAB+B^2A)$ and using $AB=BA$ gives 
$(A+B)^3=3(A^2B+AB^2)$.  The commutativity property simply allows us to reorder the terms in the multiplications.  Therefore
$$
\frac{(A+B)^3}{3!}=\frac{A^2B+B^2A}{2!}
$$.
Finally, for the fourth order term we note that by commutativity we have
$$
(A+B)^4 = A^4+4A^3B+6A^2B^2+4AB^3+B^4
$$
Since $A^3=B^3=0$ only the middle term remains.  Hence
$$
\frac{(A+B)^4}{4!} = \frac{A^2B^2}{2!2!}.
$$
