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This question already has an answer here:

A square matrix $A$ is idempotent if $A^2 = A.$

Prove that if $A$ is an $n\times n$ matrix that is idempotent and invertible, then $A$ is the identity.

How do i prove this?

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marked as duplicate by rschwieb, Leucippus, Pragabhava, Chris Godsil, user228113 Jan 27 '16 at 0:18

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    $\begingroup$ what have you tried? You have the equation $A^2=A$, you know that $A^{-1}$ exists, how can use that? $\endgroup$ – Ofir Schnabel Aug 20 '15 at 10:31
  • $\begingroup$ You could always multiply both sides of the equation by $A^{-1}$ $\endgroup$ – Omnomnomnom Aug 20 '15 at 13:22
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We have $$I = A A^{-1} \overset{(*)}{=} A^2 A^{-1} = A (A A^{-1})=A I =A,$$ where we used in $(*)$ that we have $A=A^2$.

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  • $\begingroup$ is this the final answer? or do we have to use values of In to prove? $\endgroup$ – shive Aug 20 '15 at 10:47
  • $\begingroup$ @shive Why not? $\endgroup$ – principal-ideal-domain Aug 20 '15 at 10:48
  • $\begingroup$ @ principal-ideal-domain ohh is it? $\endgroup$ – shive Aug 20 '15 at 10:50
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    $\begingroup$ @shive The question was to show $A=I$. I showed $I=A$ using multiple times the transitivity of the equality relation. The last step which I want you to do is using the symmetry of the equality relation to conclude $A=I$. $\endgroup$ – principal-ideal-domain Aug 20 '15 at 10:53

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