Q:

If $\,A^3-A+I=0,\,$ then $A$ is invertible

Prove or disprove. If $A$ is a square matrix and $A^3-A+I=0,$ then $A$ is invertible.

Is it possible to say the characteristic polynomial of $A$ is $\,p(t)=t^3-t+1$, and $A$ is invertible since $0$ is not an eigenvalue of the characteristic polynomial? 

A:

Hint.
$$
A^3-A+I=0\quad\Longrightarrow\quad A(I-A^2)=I.
$$

A:

Call $B= (-1)(A^2 - I)$, now $(-1)(A^2 - I)A=I$ so you have $B$ as one side inverse, on the other side also $A(-1)(A^2-I) =I$.

A:

Check this out:
Since
$A^3 - A+ I = 0, \tag{1}$
we have
$A(I - A^2) = (I - A^2)A = I.  \tag{2}$
(2) shows that
$A^{-1} = I - A^2, \tag{3}$
so $A$ is invertible.  
It is not in general true that (1) implies the characteristic polynomial of $A$ is $t^3 - t + 1$; if $\text{size}(A) \ne 3$, for example, it cannot be the case, since the degree of the characteristic polynomial is equal to the size of the matrix.
What we can say, however, is that every eigenvalue $\lambda$ of $A$ satisfies
$\lambda^3 - \lambda + 1 = 0, \tag{4}$
for
$Av = \lambda v \tag{5}$
yields
$A^3 v = A^2(Av) = A^2(\lambda v) = \lambda A(Av)$
$= \lambda^2 Av = \lambda^3 v, \tag{6}$
so that
$(\lambda^3 - \lambda + 1)v = (A^3 - A + I)v = 0, \tag{7}$
and since $v \ne 0$ (it is an eigenvector), (4) binds.  Thus no eigenvalue can vanish; again, $A$ is invertible.

