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Given the identity matrix $I_n$ and a matrix $J_n$ with only 1's as every element.

I must prove that if $n>0$ then $(I_n - J_n)^{-1} = I_n - \frac{1}{n-1} J_n$

I'm trying to understand if there is some sort of theorem that states that $(I_n - J_n)^{-1} = I_n^{-1} - J_n^{-1}$ or something like that?


marked as duplicate by draks ..., Git Gud, Davide Giraudo, Martin Sleziak, Claude Leibovici Mar 6 '14 at 11:19

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  • 1
    $\begingroup$ $J_n^{-1}$ doesn’t exist as $J_n$ is not invertible. $\endgroup$ – k.stm Mar 6 '14 at 10:52
  • $\begingroup$ Then I need some hints :( $\endgroup$ – Paze Mar 6 '14 at 10:52
  • $\begingroup$ I've got: $(I - c \cdot J)(I-J) = I$ Where C is the constant $\frac{1}{n-1}$ How can I proceed? $\endgroup$ – Paze Mar 6 '14 at 11:11
  • $\begingroup$ Can someone explain the last step in the algebra proof math.stackexchange.com/questions/441505/… $\endgroup$ – Paze Mar 6 '14 at 11:22
  • $\begingroup$ Why is $A^2 = nA$? $\endgroup$ – Paze Mar 6 '14 at 11:22

Just multiply out the things that are supposed to be inverses of each other, and check that it gives $I_n$.

  • $\begingroup$ Would that be a proof? $\endgroup$ – Paze Mar 6 '14 at 10:56
  • $\begingroup$ If you prove that $AB = I$, then you have proven that $B=A^{-1}$. By definition. $\endgroup$ – 5xum Mar 6 '14 at 10:59
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    $\begingroup$ ...and by uniqueness of the inverse. $\endgroup$ – DonAntonio Mar 6 '14 at 11:00
  • $\begingroup$ Please see my new comment above. $\endgroup$ – Paze Mar 6 '14 at 11:12
  • $\begingroup$ I need an explanation of the steps in this answer math.stackexchange.com/questions/441505/… The one with 7 votes. How does he just plug an $n$ in and how does he get rid of the $-A$? $\endgroup$ – Paze Mar 6 '14 at 11:26

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