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I have the following question:

Suppose A is a linear transformation. Show that if $A^2 - A + I = O$, then A is invertible.

To show $A$ is invertible, I can show if $Ax = 0$ then $x = 0$.

But, if $Ax = 0$, then $(Ax)^2 + (Ax) + 1 = 0 + 0 + 1 \neq 0$ so the equation in the question is not satisfied. What gives?

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    $\begingroup$ You surely mean $A^2-A+I=0$, not "A^2 - A + 1 = 0". From $A^2-A+I=0$ you should be able to just write down an inverse for $A$. $\endgroup$ – Lord Shark the Unknown Sep 19 '17 at 17:43
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    $\begingroup$ Rewrite $A^2-A+I=O$ in the form $AB=I$, then you have $B=A^{-1}$ $\endgroup$ – StubbornAtom Sep 19 '17 at 17:44
  • $\begingroup$ You have to start knowing that $A^2-A+I=O$ and then prove that $A$ is invertible $\endgroup$ – Raffaele Sep 19 '17 at 17:44
  • $\begingroup$ That's what I mean Lord Shark. Forgive my lack of typesetting $\endgroup$ – pellucidcoder Sep 19 '17 at 17:47
  • $\begingroup$ It's not just the typesetting, it's also the confusion of $I$ with $1$. $\endgroup$ – Lord Shark the Unknown Sep 19 '17 at 17:55
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From $A^2-A+I=O$

You have $A-A^2=I$ which means that $A(I-A)=I$ then $A$ has an inverse which is $I-A$

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From the equation $A^2-A+I=0$ we have that the minimal polynomial $m(x)|x^2-x+1$ so $m(0)\not=0 \Rightarrow 0$ is not an eigenvalue of $A \Rightarrow A$ is invertible

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