1
$\begingroup$

Prove that if $A$ is an $n \times n$ matrix such that $A^{4}$ = 0 then: $$(I_n - A)^{-1}=I_n+A+A^2+A^3$$

My proof is as follows: $$(I_n - A)(I_n - A)^{-1}=I_n$$ $$(I_n - A)^{-1}=I_n/(I_n - A)$$ $$I_n/(I_n - A)=I_n+A+A^2+A^3$$ $$I_n=(I_n - A)(I_n+A+A^2+A^3)$$ $$I_n=I_n+A+A^2+A^3-A-A^2-A^3-A^4$$ $$I_n=I_n-A^4$$ because we know that: $$A^4=0$$ therefore: $$I_n=I_n$$

Is this an acceptable justification or have I made an error in my logic?

*I apologize for any poor formatting

$\endgroup$
  • 5
    $\begingroup$ First of all, you shouldn't ever 'divide' by matrices. Only multiply with their inverse $\endgroup$ – Lukas Rollier Oct 5 '20 at 14:41
  • $\begingroup$ @LukasRollier Right, thank you, I will go back and work through it again. $\endgroup$ – Nish Oct 5 '20 at 14:43
4
$\begingroup$

The essence of your proof is correct, but the first few lines are very confusing. Why not something more clear like:

In order to show $(I_n-A)^{-1}=I_n + A +A^2 +A^3$, it suffices to show $$ (I_n-A)(I_n + A +A^2 +A^3)=(I_n + A +A^2 +A^3)(I_n-A)=I_n. $$ It's easy to show that the left equals the middle. Then by multiplying through and using the fact that $A^4=0$, it's easy to show that both of them equal the right.

$\endgroup$
  • $\begingroup$ Thank you, that is much clearer $\endgroup$ – Nish Oct 5 '20 at 15:10
1
$\begingroup$

Hint: Let $B=I+A+A^2+A^3$. Compute $AB$.

$\endgroup$
-1
$\begingroup$

To prove $a$ is inverse of $b$. Just find $ab$ and $ba$. If $b$ is inverse of $a$ then $ab=I=ba$. Thus Take $B=I_n+A+A^2+A^3$ and solve $AB$ and $BA$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.