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I'm preparing for a test in linear algebra and I've come across a problem I'm having trouble with for some reason:

Given a square matrix A, $A^2=2I$, prove that $A-I$ is invertible.

I know this is pretty simple but I can't seem to play with the equations to get it so that for some $B$, $B(A-I)=I$

It's pretty easy to see that $A^{-1}=\frac{1}{2}A$, but beyond that I haven't been able to get very far.

Can anyone help with this?

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5 Answers 5

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From your original equation subtract $\;I\;$ from both sides, and you'll get at once what you want:

$$A^2=2I\implies A^2-I=2I-I=I\implies (A-I)(A+I)=I\;\;\color{green}\checkmark$$

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No “guess and check” is needed. Set $B=A-I$, so $A=B+I$; then $$ (B+I)^2=2I $$ that becomes $$ B^2+2B+I=2I $$ or $$ I=B^2+2B=B(B+2I) $$ Thus $$ (A-I)^{-1}=B^{-1}=B+2I=A+I $$

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    $\begingroup$ I would say that setting $B=A-I$ counts as a guess-and-check, how do you know it will help? $\endgroup$
    – David
    Feb 20, 2016 at 8:15
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    $\begingroup$ @David No guess: it's the matrix we want to know about. By using $A=B+I$ and substituting, we get rid of $A$. Whether the method works is not known in advance, of course, but it's an attempt to be made. $\endgroup$
    – egreg
    Feb 20, 2016 at 9:24
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    $\begingroup$ @David What on Earth would you consider not to be "guess-and-check" in this case? You have to think at least a tiny bit to do mathematics! $\endgroup$ Feb 20, 2016 at 13:31
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    $\begingroup$ @David I'm not saying this method is better or worse than “guess-and-check”. Of course, observing that $(A-I)(A+I)=I$ (in this particular case) is much faster. A “safe-though-slower” method should always be available and this was the purpose of my answer. $\endgroup$
    – egreg
    Feb 20, 2016 at 14:05
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Note the if $A-I$ is not invertible then exist a vector $v$ such that $(A-I)v=0$ and $Av=v$. Therefore $v$ is an eigenvector with eigenvalue $1$ but the possible eigenvalues of $A$ are $\pm \sqrt 2$. Indeed if $v$ is an eigenvector for $A$ with eigenvalue $\lambda$ then $$\lambda ^2v-2v=0$$ $$v(\lambda ^2-2)=0$$ from here $\lambda =\pm \sqrt 2$

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    $\begingroup$ This argument can be made somewhat more elementary by eschewing the concept of "eigenvalue" and simply noting that since $Av=v$ we have $2v=A^2v=Av=v$ so $v=0$, contradiction. $\endgroup$ Feb 20, 2016 at 18:05
  • $\begingroup$ @MeniRosenfeld right $\endgroup$ Feb 20, 2016 at 18:07
  • $\begingroup$ But the concept of eigenvalue is beautiful $\endgroup$ Feb 20, 2016 at 18:08
  • $\begingroup$ Of course it is. But there's merit in not bringing into a solution heavier machinery than is necessary. $\endgroup$ Feb 20, 2016 at 18:33
  • $\begingroup$ Not to mention, there's a reasonable chance the OP hasn't yet gotten as far as eigenvalues in their studies. $\endgroup$ Feb 20, 2016 at 18:58
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We have $$(A + I)(A - I) = A^2 - I^2 = 2I - I = I.$$ So $(A - I)^{-1} = A + I$.

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  • $\begingroup$ Cool. Was that kind of a "guess and check" situation or did you know that if you'd multiply $(A+I)(A-I)$ you'd get 0 as the coefficient of $A$ in the resultant polynomial and that that might solve the problem? $\endgroup$ Feb 19, 2016 at 22:11
  • $\begingroup$ I knew that because of the difference of two squares formula. The answer below shows a more deductive way of finding the inverse. $\endgroup$ Feb 19, 2016 at 22:16
  • $\begingroup$ Difference-of-squares is useful in many areas of math. The generalized difference-of-powers is also useful in places you might not expect: proofwiki.org/wiki/Difference_of_Two_Powers $\endgroup$ Feb 20, 2016 at 6:22
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Suppose that $A-I$ is not invertible. Then $\lambda = 1$ is an eigenvalue of $A$, but then $1$ is an eigenvalue of $A^{2}$. So that $v = A^{2}v = 2v \implies 1 = 2$ a contradiction.

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