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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 $I^2 = 0$?2011-09-25

4 Answers 4

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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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    Thank you very much $f$or the help!2011-09-25
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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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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*} $