The eigenvalues and corresponding eigenvectors of the matrix $A = \begin{pmatrix} 8 & -6 & 2 \\ -6 & 7 & -4 \\ 2 & -4 & 3 \end{pmatrix}$ are as follows:

Eigenvalues:

1. $\lambda_1 = 15$
2. $\lambda_2 = 3$
3. $\lambda_3 \approx 0$ (very close to zero)

Eigenvectors (corresponding to each eigenvalue):

1. For $\lambda_1 = 15$, the eigenvector is: $v_1 = \begin{pmatrix} -\frac{2}{3} \\ \frac{2}{3} \\ -\frac{1}{3} \end{pmatrix}$
2. For $\lambda_2 = 3$, the eigenvector is: $v_2 = \begin{pmatrix} \frac{2}{3} \\ \frac{1}{3} \\ -\frac{2}{3} \end{pmatrix}$
3. For $\lambda_3 \approx 0$, the eigenvector is: $v_3 = \begin{pmatrix} \frac{1}{3} \\ \frac{2}{3} \\ \frac{2}{3} \end{pmatrix}$

Would you like further explanations or details on the calculation process?

Related questions:

1. How are eigenvalues and eigenvectors used in matrix diagonalization?
2. What does it mean if an eigenvalue is zero?
3. How can eigenvectors be used in physical systems, such as in quantum mechanics?
4. What is the geometric interpretation of eigenvectors and eigenvalues?
5. How can we compute the eigenvectors manually for a matrix like this?

Tip:

Eigenvectors corresponding to distinct eigenvalues are linearly independent, which means they can form a basis for the space.