A question about eigenvalues of $5\times5$ matrix

$A\in M_5({R})$ and $A^2-4A-I=0$ If $a_1,a_2,a_3,a_4,a_5$ are eigenvalues of $A$ what is value of $$\left(a_1-\dfrac{1}{a_1}\right)+\left(a_2-\dfrac{1}{a_2}\right)+\cdots+\left(a_5-\dfrac{1}{a_5}\right)$$ I tried to work with $A^2-4A-I=0$ but get nothing because $A_{5\times5}$ characteristic polynomial is something like $A^5+k_1A^4+...$
Any hint will be appreciated .

If $\lambda$ is an eigenvalue of $A$ then:

$$\lambda^2-4\lambda-1=0\to \lambda=2\pm\sqrt{5}\to \frac{1}{\lambda}=-2\pm\sqrt{5}$$

which means that

$$\lambda-\frac{1}{\lambda}=2\pm\sqrt{5}-(-2\pm\sqrt{5})=4$$

so,

$$\left(a_1-\dfrac{1}{a_1}\right)+\left(a_2-\dfrac{1}{a_2}\right)+\cdots+\left(a_5-\dfrac{1}{a_5}\right)=4+4+4+4+4=5\cdot4 =20$$

EDITED:

If $a_i$ are the eigenvalues of $A$, $a_i - 1/a_i$ are the eigenvalues of $A - A^{-1}$. Since $A^2 - 4 A - I = 0$, $A^{-1} = A^{-1}(A^2 - 4 A) = A - 4 I$ and $A - A^{-1} = 4 I$. Thus the result is $\text{trace}(4I) = 20$.