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I need to proof that if $A^2 + A + I_n = 0$ then matrix $A$ is invertible.

I can see why $A^2 + A$ is invertible, but can't find a way to proof it on $A$.

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    $\begingroup$ The usual terminology is "A is invertible" (and not reversable). $\endgroup$ – P Vanchinathan Mar 11 at 9:29
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Let $Ax=0$, then $0=(A^2+A+I_n)x=I_nx=x$, hence $ker(A)=\{0\}$.

Conclusion ?

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  • $\begingroup$ I would add general conclusion that if only $p(A)=0$ with non-zero part $I_n$ then $A$ must be invertible. $\endgroup$ – Widawensen Mar 11 at 10:48
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    $\begingroup$ Yes, if $ p(x)=x^n+a_{n-1}x^{n-1}+...+a_1x$ and $p(A)+I_n=0$, then $A$ is invertible. $\endgroup$ – Fred Mar 11 at 10:58
  • $\begingroup$ Why the downvote? ? $\endgroup$ – Fred Mar 11 at 19:09
  • $\begingroup$ I upvoted both answers, they seem to be very good. I'm also very interested what weaknesses someone noticed in the reasoning about kernel .. $\endgroup$ – Widawensen Mar 12 at 7:09
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$I_n=-A(A+I_n)$, hence $1= \det(I_n)= \det(-A)\det(A+I_n)=(-1)^n \det(A) \det(A+I_n).$

This gives $ \det(A) \ne 0$.

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    $\begingroup$ And more explicitly, $A^{-1}=-A-I_n$. $\endgroup$ – Bernard Mar 11 at 9:39
  • $\begingroup$ @Dietrich: $A=-I_n$ doe not satisfy the equation $A^2 + A + I_n = 0.$ $\endgroup$ – Fred Mar 11 at 9:55
  • $\begingroup$ @Fred Good point! $\endgroup$ – Dietrich Burde Mar 11 at 9:56
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The inverse is $-(A+I)$, since by direct calculation

$$-(A+I)A=-A^2-A=I.$$

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  • $\begingroup$ This is the best answer here. I have allowed myself to reformat. Undo if you want. $\endgroup$ – Yves Daoust Mar 11 at 20:25
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To show that any square matrix A is invertible, you need to show that there exists a square matrix $A^{-1}$ (called the inverse of A) such that $AA^{-1} = A^{-1}A = I_n $. That's the definition of a matrix being invertible.

Now, we can use the equation $A^2 + A + I_n = O_n$ to find a potential candidate for the inverse $A^{-1}$ such that the equation $AA^{-1} = A^{-1}A = I_n $ holds. Let us solve for $I_n$:

$A^2 + A + I_n = O_n$$I_n = -A^2 - A$$I_n = A(-A - I_n)$

Here, we have our candidate inverse matrix, which is just $A^{-1} = (-A - I_n)$. Now, you must show that it satisfies the equation $AA^{-1} = A^{-1}A = I_n$.

Let $A^{-1} = -A - I_n$. Then $AA^{-1} = A(-A - I_n) = -A^2 - AI_n = -A^2 - A = -(A^2 + A) = I_n$

Similarly, $A^{-1}A = (-A - I_n)A = -A^2 - I_{n}A = -A^2 - A = I_n$.

By the definition, A is invertible.

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