I need to show that if $A^3=0$, then $I^3−A$ is invertible and its inverse is equal to $I^3+A+A^2$. I have been at this question for almost an hour and do not know how to approach it, any help would be appreciated :)
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2Literally all you need to do is show that $(I_3  A)(I_3 + A + A^2) = I_3$. – angryavian Nov 12 '19 at 02:57
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Hint:
Compute $(I^3A)(I^3+A+A^2)$, from there, you are able to answer the two questions.
Siong Thye Goh
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Doing this I arrive at $I^2+A^2IA^2A^3$, where do I go from here? – Altaaf Ackbar Nov 12 '19 at 03:07

your expression can be simplify to $I+A^2A^2O$ right, further simplify it and see what you get and think from the definition of inverse. – Siong Thye Goh Nov 12 '19 at 03:08

Sorry, this is a dumb question, but how did you get to $I+A^2A^2O$ from my expression? I know that $A^3$ is just $0$, but how did you simplify the rest? – Altaaf Ackbar Nov 12 '19 at 03:22

$I$ is the identity matrix, hence $A^2I = A^2$ and also $I^3=I$. – Siong Thye Goh Nov 12 '19 at 03:25