1. **Stating the problem:** Find the eigenvalues of the matrix $$A = \begin{pmatrix}4 & 6 & 6 \\ 1 & 3 & 2 \\ -1 & -5 & -2 \end{pmatrix}$$ given that two eigenvalues are equal and each is double the third eigenvalue. Then find the eigenvalues of $$A^2$$. 2. **Understanding eigenvalues relationship:** Let the eigenvalues be $$\lambda_1, \lambda_2, \lambda_3$$. Given two eigenvalues are equal, say $$\lambda_1 = \lambda_2$$. Given these two equal eigenvalues are double the third eigenvalue: $$\lambda_1 = \lambda_2 = 2 \lambda_3$$. 3. **Sum of eigenvalues:** Sum of eigenvalues equals trace of matrix: $$\lambda_1 + \lambda_2 + \lambda_3 = \text{trace}(A) = 4 + 3 + (-2) = 5$$ Substitute $$\lambda_1 = \lambda_2 = 2 \lambda_3$$: $$2\lambda_1 + \lambda_3 = 5$$ But since $$\lambda_1 = 2 \lambda_3$$, $$2(2\lambda_3) + \lambda_3 = 5 \implies 4\lambda_3 + \lambda_3 = 5 \implies 5\lambda_3 = 5 \implies \lambda_3 = 1$$ 4. **Finding other eigenvalues:** $$\lambda_1 = \lambda_2 = 2 \times 1 = 2$$ 5. **Eigenvalues of $$A^2$$:** Eigenvalues of $$A^2$$ are squares of eigenvalues of $$A$$: $$\lambda_1^2 = 2^2 = 4, \quad \lambda_2^2 = 2^2 = 4, \quad \lambda_3^2 = 1^2 = 1$$ **Final answer:** Eigenvalues of $$A$$ are $$2, 2, 1$$. Eigenvalues of $$A^2$$ are $$4, 4, 1$$.