The exercise is the following: Assume that $A$ is a $n \times n$ matrix satisfying the identity $A^3-cA=I_n$ ($c>0$).

a) Prove that $A^{-1}=A^2-cI_n$.

b) Compute the possible values of $\det(A)$.

The proof of a) is obvious from the definition. However the proof of b) seems to be cumbersome, if we try to compute it based on the characteristic polynomial $p(\lambda)=\lambda^3-c\lambda-1$.

However, I think if we apply the Cayley-Hamilton theorem, it would be possible to compute $\det(A)$ based on the binomial expansion identity

$$ A^{-n}=(A^2-cI_n)^n=\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor-1}\left(\begin{array}{lll} n \\ k \end{array}\right)(-c)^{n-k}A^{2k}+A^{2\lfloor \frac{n}{2} \rfloor}\sum_{k=0}^{n-\lfloor \frac{n}{2} \rfloor}\left(\begin{array}{ccc} n \\ \lfloor \frac{n}{2} \rfloor+k \end{array}\right)(-c)^{n-k}A^{2k}.$$

Up to my knowledge, one needs to find the coefficients of order $0$ on both sides of the polynomial equation.

It is clear from the Cayley-Hamilton theorem that $A^{-n}$ equals to a polynomial of degree $n-1$, whose coefficient of order $0$ equals to $$-(-1)^n\det(A^{-1})=\dfrac{(-1)^{n+1}}{\det(A)}.$$

On the right-hand side one finds that the coefficient of order $0$ equals to $$(-c)^n-(-1)^n\det(A)\left(\begin{array}{ccc} n \\ \lfloor \frac{n}{2} \rfloor \end{array}\right)(-c)^n,$$ if $n$ is even and $(-c)^n$ if $n$ is odd.

Again, I've used the Cayley-Hamilton theorem to compute the coefficient of order $0$ underlying to the polynomial expansion of the matrix $A^{2\lfloor \frac{n}{2} \rfloor}=A^n$ [equals to $-(-1)^n\det(A)$].

Thereby, for $n$ odd $\det(A)=-c^{-n}$. For $n$ even, the possible values for $\det(A)$ may be found as solutions of the quadratic equation

$$\left(\begin{array}{ccc} n \\ \lfloor \frac{n}{2} \rfloor \end{array}\right)c^n\det(A)^2-c^n\det(A)-1=0.$$

Is this procedure right?

  • 2
    $\begingroup$ The characteristic polynomial of $A$ had degree $n$. At best you can say that it and $p(\lambda)$ are both multiples of the minimal polynomial. $\endgroup$
    – amd
    May 27, 2018 at 21:14
  • $\begingroup$ That's a standard result from matrix theory. And what about the computation of $\det(A)$? Can you do it explicitly through the computation of the minimal polynomial without lenghty computations? $\endgroup$ May 27, 2018 at 21:25

1 Answer 1


For simplicity, I assume that $A$ is allowed to be a matrix with complex entries, but I have added a quick note on the bottom addressing the real case.

We note that for all values of $c > 0$ with $c \neq \frac{3}{\sqrt[3]{4}}$, the roots of $p(x) = x^2 - cx - 1$ are distinct (this may be quickly seen by calculating the discriminant of the polynomial, which is $\Delta = 4c^3 - 27$). In this case, let $\lambda_1,\lambda_2,\lambda_3$ denote the roots of this polynomial.

By standard results on the minimal polynomial of a matrix, a matrix $A$ will satisfy $p(A) = 0$ if and only if it is diagonalizable with eigenvalues taken from the set $\{\lambda_1,\lambda_2,\lambda_3\}$. Thus, $\det(A)$ may have any value of the form $$ \det(A) = \lambda_1^{k_1} \cdot \lambda_2^{k_2} \cdot \lambda_3^{k_3} $$ where $k_1 + k_2 + k_3 = n$.

In the case of repeated roots the same result holds for the determinant, but $A$ is not necessarily diagonalizable. For $c = \frac{3}{\sqrt[3]{4}}$, we have $-1/\sqrt[3]{2}$ as a double root and $x = \sqrt[3]{4}$.

If $A$ is necessarily a real valued matrix, then another consideration is required in the above. In particular, we note that a diagonalizable matrix is similar to a real matrix if and only if its eigenvalues come in conjugate pairs.

In this case: if $p(x)$ has complex roots $\lambda,\bar \lambda$ and real root $\mu > 0$, we find that $\det(A)$ will necessarily have the form $$ \det(A) = \lambda^{k_1} \cdot \bar \lambda^{k_1} \cdot \mu^{k_2} = |\lambda|^{2k_1} \cdot \mu^{k_2} $$ where $2k_1 + k_2 = n$.

For $c = 2$, we may argue that there are at least $2n+1$ values for $\det(A)$ as follows: We find the possible eigenvalues are $-1$ and $\frac {1 \pm \sqrt{5}}2$. The magnitudes of these eigenvalues are $1, \phi^{\pm 1}$, where $\phi = \frac{1 + \sqrt{5}}{2}$. We see that $|\det(A)|$ can be $\phi^{k}$ for any $k \in \{-n,\dots,n-1,n\}$.

  • $\begingroup$ That's indeed right, but you consider a polynomial of degree $2$, instead of a polynomial of degree $3$ [$x^3-cx-1$]. $\endgroup$ May 27, 2018 at 21:56
  • $\begingroup$ @NelsonFaustino I'm in the process of fixing some errors $\endgroup$ May 27, 2018 at 21:57
  • $\begingroup$ @NelsonFaustino see the latest edit $\endgroup$ May 27, 2018 at 21:59
  • $\begingroup$ Alright. Anyway, recall than for any eigenpair $(\lambda,v)$ of $A$ we then have $$ \lambda^3=c\lambda+1.$$ Thus $p(\lambda)=\lambda^3-c\lambda-1$ is the natural candidate to be minimal polynomial. $\endgroup$ May 27, 2018 at 22:04
  • $\begingroup$ @NelsonFaustino sure. However, I still find it difficult to believe that, for arbitrary odd $n$, $\det(A)$ can take on only two values $\endgroup$ May 27, 2018 at 22:06

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