Let $A$ be $10 \times 10$ matrix such that $A^2 + A +I = 0$ then the matrix is invertible.

$A^2 + A = –I$

$A(–A–I) = I$ So, – $A$$I$ is inverse of A which is clearly $10 \times 10$ matrix, since $A$ and $I$ are $10 \times 10$ matrices.

Is this explanation enough to prove the statement true?

  • 1
    $\begingroup$ Yes, it is fine. As an alternative, you may assume that $A$ is not invertible, hence there is some $v\neq 0$ such that $Av=0$. In such a case, however, $v=(A^2+A+I)v = (0)v = 0$ leads to a contradition. In general, $p(A)=0$ implies that the eigenvalues of $A$ belong to the set of roots of $p$. Zero is not a root of $x^2+x+1$, hence all the matrices fulfilling $A^2+A+I=0$ are invertible (i.e. $0\not\in\text{Spec}A$). $\endgroup$ – Jack D'Aurizio Dec 9 '18 at 17:27

I don't think you need to stress that the sizes of the matrices in sight are still $10\times 10$, as that is clear from context. The factorization looks fine, but you should probably quote some definition of invertibility or some part of an Invertible Matrix Theorem if you are looking for lots of rigor.


Short Answer: Yes.

Long Answer (explaining what you have used): Since you are only working over a matrix ring $R^{10 \times 10}$, you have additive inverses which allows you to conclude $A^2+A = -I$ and since you have the distributive law also $A (A+I) = -I$. Then you use the properties that $-(-X) = X$ and $-(XY) = X(-Y)$ for all $X,Y \in R^{10\times 10}$ and get $A (- (A+I)) = I$. So by definition of invertibility, $A$ is invertible.


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