# $A^2 + A + I_n = 0 \implies$ matrix $A$ is invertible

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 $$A$$.

• The usual terminology is "A is invertible" (and not reversable). – P Vanchinathan Mar 11 at 9:29

## 4 Answers

Let $$Ax=0$$, then $$0=(A^2+A+I_n)x=I_nx=x$$, hence $$ker(A)=\{0\}$$.

Conclusion ?

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

$$I_n=-A(A+I_n)$$, hence $$1= \det(I_n)= \det(-A)\det(A+I_n)=(-1)^n \det(A) \det(A+I_n).$$

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

• And more explicitly, $A^{-1}=-A-I_n$. – Bernard Mar 11 at 9:39
• @Dietrich: $A=-I_n$ doe not satisfy the equation $A^2 + A + I_n = 0.$ – Fred Mar 11 at 9:55
• @Fred Good point! – Dietrich Burde Mar 11 at 9:56

The inverse is $$-(A+I)$$, since by direct calculation

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

• This is the best answer here. I have allowed myself to reformat. Undo if you want. – Yves Daoust Mar 11 at 20:25

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.