Real matrix satisfying $A^3=4I_n-3A$

Question

Let $A\in M_n(\mathbb{R}) $ so that $A^3=4I_n-3A$. Prove that $\det(A+I_n) =2^n$.

My work : $A$ 's eigenvalues are the roots of $x^3+3x-4=0$, so one of the eigenvalues is $1$ and the others are $\lambda_1$ and $\lambda_2$, the roots of $x^2+x+4=0$.

Hence,

$$\det(A+I_n) =2(\lambda_1+1)(\lambda_2+1)=2(\lambda_1 \lambda_2 +\lambda_1+\lambda_2+1) =2(4-1+1)=8$$ What is my mistake?

Answer 1

Careful, you only know that the spectrum of $A$ is contained in the set of zeroes of $x^3+3x-4$.

Since $A$ is a real matrix, its minimal polynomial has real coefficients and divides $x^3+3x-4$, so it is one of $$x-1, x^2+x+4, (x-1)(x^2+x+4)$$

Hence the characteristic polynomial is of the form $\chi_A(x) = (x-1)^{n-2r}(x^2+x+4)^{r}$ for some $0 \le r \le \frac{n}2$.

Therefore if we denote $\lambda_1, \lambda_2$ the zeroes of $x^2+x+4$, we have $$\det(A+I_n) = (1+1)^{n-2r}(\lambda_1+1)^r(\lambda_2+1)^r = 2^{n-2r}(\lambda_1\lambda_2+\lambda_1+\lambda_2+1)^r = 2^{n-2r}4^r = 2^n$$

Answer 2

There are a couple of mistakes here: You don’t know that $x^3+3x-4$ is the characteristic polynomial of $A$ (indeed, it can’t be for $n\ne3$), so you can only conclude that its eigenvalues are also roots of this polynomial, not that the all of the roots of the polynomial are eigenvalues. Moreover, you don’t know their algebraic multiplicities, either.