How can I prove that a square matrix is invertible if it satisfies this polynomial equation? For a 3x3 matrix $C$, it is given that
$$C^3+I=3C^2-C$$
I am then required to prove that $C$ is invertible.
I have attempted a proof, below, but I am not sure it is valid or if there is a better solution.
Attempted proof
$$C^3 + I = 3C^2 - C$$
$$I = - C^3 +3C^2-C$$
If it is assumed that $C^{-1}$ exists then
$$I = C^{-1}(-C^4+3C^3-C^2)$$
If $C^{-1}$ is defined, then $I=C^{-1}C$; therefore test whether
$$C \stackrel{!}{=} -C^4 + 3C^3 - C^2$$
$$ 0 = -C^4 + 3C^3 - C^2 - C$$
$$ C = 0, 1, 1\pm\sqrt{2}$$
Is this at all in the right direction?

If $C$ is not invertible, it has a non-trivial null space and hence a non-trivial eigenvector $v$ with eigenvalue $\lambda=0$.
We have $(C^3+I)v=(3C^2-C)v$ whence $(\lambda^3+1)v=(3\lambda^2-\lambda)v$ and setting $\lambda=0$ we see that $v=0$, which is a contradiction.
Easiest (and in fact more general) to use the explicit inverse in the other answers, but this is a way to use the eigenvalues as suggested in your answer. 
