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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.

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    $\begingroup$ Well, have you tried expanding $\frac{(A+B)^2}{2!}+\frac{(A+B)^3}{3!}+\frac{(A+B)^4}{4!}$? $\endgroup$ – Eric Wofsey Nov 13 '16 at 21:44
  • $\begingroup$ My bad, E is supposed to be the identity matrix I. Updated it now. $\endgroup$ – E.Bob Nov 13 '16 at 21:45
  • $\begingroup$ @EricWofsey I haven't. I thought I wasn't suppose to do that. $\endgroup$ – E.Bob Nov 13 '16 at 21:48
  • $\begingroup$ @EricWofsey Alright figured it out, thank's! $\endgroup$ – E.Bob Nov 13 '16 at 22:10
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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.

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I would do it without brute force:

  1. Since $A^3=0$ we have $$ e^A=I+A+\frac{A^2}{2!}. $$ Similar for $B$.
  2. The LHS of your expression is then $e^Ae^B$. For commuting matrices it is true that $$ e^Ae^B=e^{A+B}. $$
  3. 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 $e^{A+B}$.
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  • $\begingroup$ Nice answer! I like the methodology a lot. $\endgroup$ – Peder Nov 13 '16 at 22:10
  • $\begingroup$ I was about to do it this way but I forgot to do it when I thought of a different way(made everything more complicated). But thank's, you made it more clear. I figured it out thank's. $\endgroup$ – E.Bob Nov 13 '16 at 22:16
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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 $$ \frac{(A+B)^4}{4!} = \frac{A^2B^2}{2!2!}. $$

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