Solution: We will use the following facts:

• If $\lambda$ is an eigenvalue of $A$ with an eigenvector $v$, then $\lambda ^k $ is an eigenvalue of $A^k$ with teh same eigenvector $v$.
• Sum of eigenvalue of $A$ is equal to $\operatorname{trace}\left( A \right) $.
• Trace is linear, that is, $\operatorname{trace}\left( A + B \right) = \operatorname{trace}\left( A \right) + \operatorname{trace}\left( B \right) $.

We need to determine $\operatorname{trace}\left( A +B \right) $. Note that \begin{align*} \operatorname{trace}\left( A + B \right) & = \operatorname{trace}\left( A \right) + \operatorname{trace}\left( B \right) \\ & = \operatorname{trace}\left( A \right) + \operatorname{trace}\left( A^4 - 5A^2 + 5I \right) \\ & = \operatorname{trace}\left( A \right) + \operatorname{trace}\left( A^4 \right) - 5 \operatorname{trace}\left( A^2 \right) + 5 \operatorname{trace}\left( I \right). \end{align*} Observe that \begin{align*} & \operatorname{trace}\left( A \right) = \text{ sum of eigenvalue of } A = -1 \\ & \operatorname{trace}\left( A^4 \right) = \text{ sum of eigenvalue of } A^4 = 19 \\ & \operatorname{trace}\left( A^2 \right) = \text{ sum of eigenvalue of } A^2 = 7 \\ & \operatorname{trace}\left( I \right) = \text{ sum of eigenvalue of } I = 4. \end{align*} Therefore, \begin{align*} \operatorname{trace}\left( A + B \right) & = \operatorname{trace}\left( A \right) + \operatorname{trace}\left( A^4 \right) - 5 \operatorname{trace}\left( A^2 \right) + 5 \operatorname{trace}\left( I \right) \\ & = -1 + 19 - 5 \times 7 + 5 \times 4 \\ & = 3. \end{align*}