Determine the number of possible values for $\det(A)$, given that $A$ is an $n \times n$ matrix with real entries such that $A^3 - A^2 -3A +2I=0$.

Question

Determine the number of possible values for $\det(A)$, given that $A$ is an $n \times n$ matrix with real entries such that $A^3 - A^2 -3A +2I=0$.

here is the source of the problem. In the last comment, I don't understand why we can say for sure that $A$ has only three distinct eigenvalues. Can anyone explain to me?

Remark: I understand that the roots of the equation $x^3-x^2-3x+2=0$ are eigenvalues of $A$. My problem is why the matrix $A$ cannot have another eigenvalue?

Answer 1

Hint : The minimal polynomial of the matrix $A$ must be a divisor of the polynomial

$$x^3-x^2-3x+2=(x-2)(x^2+x-1)$$

Answer 2

Suppose $v$ is an eigenvector of $A$, with eigenvalue $\lambda$. Then $\lambda$ must satisfy the same cubic equation that $A$ satisfies: $$(A^3-A^2-3A+2I)v=0v=0\\ (A^3-A^2-3A+2I)v=AAAv-AAv-3Av+2v\\ =AA\lambda v-A\lambda v-3\lambda v+2v\\ =\lambda A(Av)-\lambda(Av)-3\lambda v+2v\\ =\lambda A(\lambda v)-\lambda^2 v-3\lambda v+2v\\ =\lambda^2(Av)-\lambda^2v-3\lambda v+2v =(\lambda^3-\lambda^2-3\lambda+2)v$$ From the first line, the answer is zero, but $v$ is not the zero vector, so $\lambda^3-\lambda^2-3\lambda+2=0$

Answer 3

The eigenvalues of $A$ are the roots of its minimal polynomial, and $\bigl\{P(x)\in \mathbf R[x] \mid P(A)=0\bigr\}$ is the set of multiples of the minimal polynomial of $A$. Hence if $P(A)=0$, the eigenvalues of $A$ are among the roots of $P(x)$.