91Ó°ÊÓ

Find the eigenvalues of the matrix \( \mathbf{A} \) by solving the characteristic equation \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \). For our matrix, \( \mathbf{I} \) is the identity matrix and \( \lambda \) is the eigenvalue. \[\det\left(\begin{array}{ccc} 1 - \lambda & 1 & 0 \ 0 & 2 - \lambda & 0 \ 0 & 0 & 3 - \lambda \end{array}\right) = 0\]The determinant can be simplified as:\[(1-\lambda)((2-\lambda)(3-\lambda)) = 0\]The eigenvalues are \( \lambda = 1, 2, 3 \).

A matrix is diagonalizable if, for each eigenvalue, its algebraic multiplicity equals its geometric multiplicity. Here, we have three distinct eigenvalues \( \lambda = 1, 2, 3 \), each repeated once (i.e., algebraic multiplicity is 1 for each). Therefore, we only need to find a linearly independent set of eigenvectors corresponding to each eigenvalue.

For each eigenvalue \( \lambda_i \), solve \( (\mathbf{A} - \lambda_i \mathbf{I}) \mathbf{x} = \mathbf{0} \) to find the eigenvectors.**Eigenvalue \( \lambda = 1 \):**Solve \((\mathbf{A} - \mathbf{I}) \mathbf{x} = \mathbf{0}\), which results in:\[\begin{pmatrix} 0 & 1 & 0 \ 0 & 1 & 0 \ 0 & 0 & 2 \end{pmatrix} \mathbf{x} = \mathbf{0}\]Eigenvector: \( \mathbf{x}_1 = \begin{pmatrix} 1 \ 0 \ 0 \end{pmatrix} \).**Eigenvalue \( \lambda = 2 \):**\[\begin{pmatrix} -1 & 1 & 0 \ 0 & 0 & 0 \ 0 & 0 & 1 \end{pmatrix} \mathbf{x} = \mathbf{0}\]Eigenvector: \( \mathbf{x}_2 = \begin{pmatrix} 1 \ 1 \ 0 \end{pmatrix} \).**Eigenvalue \( \lambda = 3 \):**\[\begin{pmatrix} -2 & 1 & 0 \ 0 & -1 & 0 \ 0 & 0 & 0 \end{pmatrix} \mathbf{x} = \mathbf{0}\]Eigenvector: \( \mathbf{x}_3 = \begin{pmatrix} 0 \ 0 \ 1 \end{pmatrix} \).

Compose matrix \( \mathbf{P} \) using the eigenvectors as columns. So for our case:\[\mathbf{P} = \begin{pmatrix} 1 & 1 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{pmatrix}\]

Matrix \( \mathbf{D} \) is simply the diagonal matrix composed of the eigenvalues in the same order as their corresponding eigenvectors in \( \mathbf{P} \):\[\mathbf{D} = \begin{pmatrix} 1 & 0 & 0 \ 0 & 2 & 0 \ 0 & 0 & 3 \end{pmatrix}\]

Check that \( \mathbf{AP} = \mathbf{PD} \). If true, the matrix is correctly diagonalized. Calculate:\[\mathbf{AP} = \begin{pmatrix} 1 & 1 & 0 \ 0 & 2 & 0 \ 0 & 0 & 3 \end{pmatrix} \begin{pmatrix} 1 & 1 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 2 & 0 \ 0 & 2 & 0 \ 0 & 0 & 3 \end{pmatrix}\]Since \( \mathbf{PD} \) also equals:\[\begin{pmatrix} 1 & 1 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \ 0 & 2 & 0 \ 0 & 0 & 3 \end{pmatrix} = \begin{pmatrix} 1 & 2 & 0 \ 0 & 2 & 0 \ 0 & 0 & 3 \end{pmatrix}\]Both equals verify the result.