Find $(a_1-\frac{1}{a_1})+(a_2-\frac{1}{a_2})+⋯+(a_5-\frac{1}{a_5})$ if $a_i$'s are eigenvalues of $A$

Question

Please help to reach answer for the following problem:

$A∈M_5(R)$ is a matrix and in relation to $A^2-4A-I=0$. if $a_1,a_2,a_3,a_4,a_5$ is eigenvalues for A; solve the following problem:

$(a_1-\frac{1}{a_1})+(a_2-\frac{1}{a_2})+⋯+(a_5-\frac{1}{a5})$

Thanks a lot.

Answer 1

$$(a_1-\frac{1}{a_1})+(a_2-\frac{1}{a_2})+⋯+(a_5-\frac{1}{a_5}) = tr{(A-A^{-1})}$$

$$A^2-4A-I = 0 \Rightarrow A(A-4I)= I \Rightarrow A^{-1}=A-4I \Rightarrow A-A^{-1} = A-(A-4I) = 4I$$

$$tr(A-A^{-1}) = 4\cdot 5 = 20$$

Answer 2

The polynomial $x^2-4x-1$ annihilates the matrix $A$ so $$\sigma(A) \subseteq \{\text{ zeros of } x^2-4x-1\} = \{2-\sqrt5, 2+\sqrt5\}$$

Assume that $2-\sqrt{5}$ is the eigenvalue of $A$ with mulyiplicity $k \ge 0$ and $2+\sqrt5$ is the eigenvalue with multiplicity $5-k$.

Then we have

\begin{align} \sum_{i=1}^5 \left(a_i - \frac1{a_i}\right) &= k\underbrace{\left(2-\sqrt{5} - \frac1{2-\sqrt{5}}\right)}_{=4} + (5-k)\underbrace{\left(2+\sqrt{5} - \frac1{2+\sqrt{5}}\right)}_{=4} \\ &= 4k + 4(5-k) \\ &= 4\cdot 5 \\ &= 20 \end{align}

We see that the result is independent of $k$.