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So my friend and I are working on this and here is what we have so far.

We want to show that $\exists \, B$ s.t. $(I+A)B = I$. We considered the fact that $I - A^k = I$ for some positive $k$. Now, if $B = (I-A+A^2-A^3+ \cdots -A^{k-1})$, then $(I+A)B = I-A^k = I$. My question is: in matrix $B$, why is the sign for $A^{k-1}$ negative? Couldn't it be positive, in which case we'd get $(I+A)B = I + A^k$?

Thank you.

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  • $\begingroup$ Duplicate of math.stackexchange.com/q/140348/264 and/or math.stackexchange.com/q/119904/264 $\endgroup$ Commented Mar 9, 2013 at 7:23
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    $\begingroup$ But $A^k=0$, what do you care ? $(I+A)B=I+A^k$ would still be $I$. You just need to make expression for $B$ correct $\endgroup$ Commented Mar 9, 2013 at 7:25
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    $\begingroup$ Yes, it can be negative, but there are ways around it (see the answers), and you don't particularly care :-) BTW, I want to commend you on your way of asking homework questions. Very constructive approach to first show what you think! $\endgroup$ Commented Mar 9, 2013 at 7:52
  • $\begingroup$ I have voted to reopen this question, because the OP is actually not asking for a proof, but an explanation on how his/her particular proof can be modified to work. And as Jyrki said, the OP's way of asking homework question is very constructive. $\endgroup$ Commented Mar 9, 2013 at 11:54
  • $\begingroup$ @user1551: I agree with the question being a good example on how to ask a homework question, but it is still an abstract duplicate, and the accepted answer to the other question uses the exact same technique and determines the sign correctly. $\endgroup$ Commented Mar 9, 2013 at 13:08

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It's the usual polynomial identity $$ 1 - x^{k} = (1 - x)(1 + x + x^{2} + \dots + x^{k-1}), $$ where you are substituting $x = -A$.

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  • $\begingroup$ This only works if k is odd - see Andre' Nicolas's note. $\endgroup$ Commented Mar 9, 2013 at 8:08
  • $\begingroup$ @StevenStadnicki, sorry, but I do not see your point. As noted in the answer by Julian Kuelshammer, when you substitute $x = - A$, you will get $(-1)^{k-1} A^{k-1}$ as the last term in the second term of RHS. $\endgroup$ Commented Mar 9, 2013 at 8:11
  • $\begingroup$ Ahh, right; mea culpa - I was looking at the LHS, but as noted in a comment it doesn't actually matter whether you wind up with $I+A^k$ or $I-A^k$, either is just $I$ by the nilpotency. $\endgroup$ Commented Mar 9, 2013 at 8:14
  • $\begingroup$ @StevenStadnicki, ok, no problem! $\endgroup$ Commented Mar 9, 2013 at 8:15
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By incrementing if necessary by $1$, we can make sure that $k$ is *odd. Then we have the familiar identity (for odd $k$) $1+x^k=(1+x)(1-x+x^2-x^3+\cdots +x^{k-1})$.

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As you can see already from your formula for $B=I-A\mathbf{+} A^2-\dots$ the sign can be $+$ for example if $k=2$. In fact the sign that occurs is $(-1)^{k-1}$.

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