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Suppose a square matrix $A$ such that $A^2+cA+cI=0$ for all $c \in \mathbb{Z}$. How can I show that $A+(c-1)I$ is invertible and find its inverse?

I started off this way: $A+(c-1)I = A+cI-I$

Then $(A+cI-I)(d_1A+d_2I)=I$, where $d_1, d_2 \in \mathbb{Z}$.

Expand it and it becomes: $d_1A^2+d_1cA-d_1A+d_2A+(c-1)d_2I=I$

$\Rightarrow (c-1)d_2=1 \;\; and \; \; d_1A^2+d_1cA-d_1A+d_2A=0$

$\Rightarrow d_2=\frac{1}{(c-1)}$

Continue from $d_1A^2+d_1cA-d_1A+d_2A=0$, after some manipulation, I got $d_1(A^2+cA+cI)-d_1cI-d_1A+d_2A=0$.

Since given that $A^2+cA+cI=0$,

$d_1(A^2+cA+cI)-d_1cI-d_1A+d_2A=0$

$\Rightarrow -d_1cI-d_1A+d_2A=0$

$\Rightarrow -d_1cI-d_1A+\frac{1}{c-1}A=0$

$\Rightarrow -d_1cI-d_1A=-\frac{1}{c-1}A$

$\Rightarrow d_1(cI+A)=\frac{1}{c-1}A$

At this point, since I don't know if $A$ is invertible yet, I cannot do it as $d_1=\frac{1}{c-1}\frac{A}{(cI+A)}$.

Even if I did this, I still cannot find a value for $d_1$ and $d_2$ to find the inverse of $A+(c-1)I$. How should I continue from here?

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Did you really mean for all $c \in \mathbb{Z}$? Wouldn't then $A^2 = A+I = 0$, hence $I^2 = 0$? –  Niels Diepeveen Sep 25 '11 at 18:23
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Instead of all this, rewrite your starting equation as $A^2+(c-1)A+A+(c-1)I=-I$. Then factor the left-hand-side...

But if you favor a more systematic approach you can also proceed as you do until $$d_1A^2+d_1cA-d_1A+d_2A+(c-1)d_2I=I$$ At this point you decide to set $(c-1)d_2=1$, but that is too early! First get rid of the $A^2$ using $A^2=-c(A+I)$, to get $$-d_1c(A+I)+d_1cA - d_1A + d_2A + (c-1)d_2I = I$$ which simplifies to $$(d_1c-d_1+d_2-d_1c)A + ((c-1)d_2-d_1c-1)I = 0$$ $$(d_2-d_1)A + ((c-1)d_2-cd_1-1)I$$ Now set both coefficients to 0. This gives immediately $d_1=d_2$, and we then need to solve $$(c-1)d-cd-1=0$$ in which the $cd$'s cancel and give us $d=-1$. So the sought inverse is $-A-I$.

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hmm...What happens if I couldn't see this and didn't start from this equation? I remember I had another problem to find inverses and I used a similar method and worked. I had the problem posted here too at math.stackexchange.com/questions/58841/… –  xenon Sep 25 '11 at 15:23
    
I've added a more systematic approach. You have to start from the given relation, though -- otherwise you have no way to know there is even an inverse anywhere. –  Henning Makholm Sep 25 '11 at 15:36
    
Thanks! But how's $A^2=-(A+I)$ and not $A^2=-(cA+cI)$? –  xenon Sep 25 '11 at 15:57
    
At one part, you set both the coefficients to be $0$. Is there a reason for doing this? Because I was thinking why wouldn't it be $((c-1)d_2-cd_1-1)=1$ and $(d_2-d_1)A=-I$? Of course, what you have done by having the coefficients zero makes the whole thing simpler but I was just thinking what if I set it something else other than zero. –  xenon Sep 25 '11 at 16:21
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It follows from some experience with quotients of polynomial rings. Your original relation can be used to prove that any linear combination of powers of $A$ can be rewritten into the form $pA+qI$, and if that is all we know about $A$, then there is only one such form. So if we have $pA+qI=0$ then either it must be because $p=q=0$ or else $A$ happened to satisfy a nicer property than the one we can depend on. (In your example the nicer property is "is a multiple of $I$"). It's always worth it to try a general method first and try more complex tactices if that fails. –  Henning Makholm Sep 25 '11 at 16:31
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NEW ANSWER. In the "long" division $$X^2+cX+c=(X+c-1)(X+1)+1,$$ replace $X$ by $A$: $$0=A^2+cA+cI=\Big(A+(c-1)I\Big)\Big(A+I\Big)+I,$$ $$\Big(A+(c-1)I\Big)^{-1}=-A-I.$$

[EDIT. I'm realizing that this answer is the same as user1551's, who posted it before. Sorry...]

OLD ANSWER. Let $A$ be an $n$ by $n$ matrix with coefficients in a field $K$, let $f\in K[X]$ be a polynomial annihilating $A$, and let $g\in K[X]$ be any polynomial.

If $g$ is prime to $f$, then $g(A)$ is invertible, and the inverse of $g(A)$ is given by $h(A)$ where $h\in K[X]$ is an inverse of $g$ mod $f$.

Moreover, there is a closed formula for such an $h$ (see this answer).

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$$ \begin{eqnarray*} x^2+cx+c = \left[x+(c-1)\right](x+1)+1\\ \therefore 0 = A^2+cA+cI = \left[A+(c-1)I\right](A+I)+I\\ \therefore \left[A+(c-1)I\right](A+I)=-I\\ \therefore \left[A+(c-1)I\right]^{-1} = -(A+I). \end{eqnarray*} $$

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Perhaps a more transparent way: "change of variables". Let $B = A + (c-1) I$, or $A = B - (c-1) I$. Then $0 = A^2 + c A + c I = (B - (c-1) I)^2 + c (B - (c-1) I) + c I= B^2 + (2-c) B + I$. Multiply by $B^{-1}$ to get $B + (2-c) I + B^{-1} = 0$, i.e. $B^{-1} = -B + (c-2) I$ or $(A + (c-1) I)^{-1} = -A - (c-1) I + (c-2) I = -A - I$.

You may find it a bit dodgy to multiply by $B^{-1}$ before you know that $B^{-1}$ exists, but once you have the result $-A-I$ it's easy to verify that this works by multiplying it by $A + (c-1) I$.

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