Matrix \(D\):
\[ D = \begin{bmatrix} -2 & 1 & -2 & -4 \\ 12 & 1 & 4 & 9 \\ 6 & 5 & -2 & -4 \\ 3 & -4 & 5 & 10 \end{bmatrix} \]
Solve the characteristic equation \(\text{det}(D - \lambda I) = 0\), where \(\lambda\) is the eigenvalue and \(I\) is the identity matrix.
The equation is:
\[ \text{det}(D - \lambda I) = \text{det}\left(\begin{bmatrix} -2-\lambda & 1 & -2 & -4 \\ 12 & 1-\lambda & 4 & 9 \\ 6 & 5 & -2-\lambda & -4 \\ 3 & -4 & 5 & 10-\lambda \end{bmatrix}\right) = 0 \]
Find the eigenvectors for each eigenvalue by solving \((D - \lambda I)\mathbf{v} = \mathbf{0}\).
\[ (D - \lambda I)\mathbf{v} = \begin{bmatrix} -1 & 1 & -2 & -4 \\ 12 & 2 & 4 & 9 \\ 6 & 5 & -3 & -4 \\ 3 & -4 & 5 & 11 \end{bmatrix}\mathbf{v} = \mathbf{0} \]
The solution gives the eigenvector \(\mathbf{v}_1 = \begin{bmatrix} 1 \\ -4 \\ -1 \\ 1 \end{bmatrix}\).
\[ (D - \lambda I)\mathbf{v} = \begin{bmatrix} -5 & 1 & -2 & -4 \\ 12 & -2 & 4 & 9 \\ 6 & 5 & -5 & -4 \\ 3 & -4 & 5 & 7 \end{bmatrix}\mathbf{v} = \mathbf{0} \]
The solution gives the eigenvector \(\mathbf{v}_2 = \begin{bmatrix} 1 \\ 2 \\ 1 \\ 1 \end{bmatrix}\).
\[ (D - \lambda I)\mathbf{v} = \begin{bmatrix} -4 & 1 & -2 & -4 \\ 12 & -1 & 4 & 9 \\ 6 & 5 & -4 & -4 \\ 3 & -4 & 5 & 8 \end{bmatrix}\mathbf{v} = \mathbf{0} \]
The solution gives the eigenvector \(\mathbf{v}_3 = \begin{bmatrix} 1 \\ 1 \\ 0 \\ 1 \end{bmatrix}\).
\[ (D - \lambda I)\mathbf{v} = \begin{bmatrix} -8 & 1 & -2 & -4 \\ 12 & -5 & 4 & 9 \\ 6 & 5 & -8 & -4 \\ 3 & -4 & 5 & 4 \end{bmatrix}\mathbf{v} = \mathbf{0} \]
The solution gives the eigenvector \(\mathbf{v}_4 = \begin{bmatrix} 1 \\ -3 \\ 1 \\ 1 \end{bmatrix}\).
The eigenvalues are \(\text{-1, 3, 2, 6}\), and the corresponding eigenvectors are:
\[ \mathbf{v}_1 = \begin{bmatrix} 1 \\ -4 \\ -1 \\ 1 \end{bmatrix} \]
\[ \mathbf{v}_2 = \begin{bmatrix} 1 \\ 2 \\ 1 \\ 1 \end{bmatrix} \]
\[ \mathbf{v}_3 = \begin{bmatrix} 1 \\ 1 \\ 0 \\ 1 \end{bmatrix} \]
\[ \mathbf{v}_4 = \begin{bmatrix} 1 \\ -3 \\ 1 \\ 1 \end{bmatrix} \].