If $A$ is an $n\times{n}$ matrix such that $A^3-A^2-3A+2I=0$, how many possible values for $\det(A)$ exists?

Question

The exercise is to find an expression in the variable $n$ for the number of possible values for the determinant of the matrix $A$ of real entries.

I tried to sum $I$ on both sides, getting $(A^2-3I)(A-I)=I$, what means that $(A^2-3I)=(A-I)^{-1}$, but I could't develop this idea.

I also tried to factorize the expression, getting $(A-2I)[A-\frac{-1+√5}{2} I][A-\frac{-1-√5}{2} I]=0$.

I know $2$, $(-1+√5)/2$ and $(-1-√5)/2$ are eigenvalues of $A$, but aren't there others?

Answer 1

HINT

Recall that

• $p(A)=0 \implies p(\lambda)=0$

and

• $\det(A)=\prod \lambda_i$

Answer 2

These are all the eigenvalues; any eigenvalue $x$ of $A$ must solve the characteristic polynomial $x^3-x^2-3x+2=0$. Writing the eigenvalues as $2,\,\lambda,\,-1/\lambda$, how many products of powers summing to $n$ can you get?

Answer 3

The condition indicates that $ p(x) =x^3 -x^2 -3x + 2$ nullifies $\boldsymbol A$, hence the minimal polynomial of $\boldsymbol A$ is a factor of $p$. As the OP indicated, $$ p(x) = (x-2 ) \left(x - \frac {-1+\sqrt 5}2\right) \left(x - \frac {-1-\sqrt 5}2\right), $$ thus the minimal polynomial $m(x)$ of $\boldsymbol A$ has $2^3 - 1 =7$ [if $m(x) = 1$, then $\boldsymbol A = \boldsymbol O$, which is not nullified by $p$] possibilities, each of which is a product of $\deg = 1$ monic polynomials with power not exceeding $1$. Since the minimal polynomial and the characteristic polynomial have the exact same roots [ignore the multiplicities], $\boldsymbol A$ can only have these numbers as eigenvalues.

If $\deg(m)=1$, then $\boldsymbol A$ has only 1 eigenvalue, hence there are 3 possible determinants.

If $\deg(m)=2$, then $\det(\boldsymbol A) = c_1^k c_2^{n-k}$, using star and bars we know that $\det (\boldsymbol A )$ has $n-1$ possibilities.

Similarly for $\deg(m) =3$, there are $\binom {n-1} 2$ possibilities.

In total $\det (\boldsymbol A)$ has $$ 3 + 3(n-1) + \binom {n-1} 2 = 3n + \frac {(n-2)(n-1)}2 = \frac {(n+1)(n+2)}2 $$ possibilities [if no coincidences].