91Ó°ÊÓ

A block diagonal matrix is a special type of square matrix that is organized into smaller matrices, called blocks, along its diagonal. These blocks are bounded by zeros on all off-diagonal elements. In essence, each block behaves like a smaller matrix within the larger matrix, making computations more manageable. In our problem, the matrix \( A \) is set up as two \( 2 \times 2 \) blocks:

• \( \begin{bmatrix} -7 & 24 \ 24 & 7 \end{bmatrix} \)
• \( \begin{bmatrix} -7 & 24 \ 24 & 7 \end{bmatrix} \)

These repeating blocks suggest that similar matrix properties hold, aiding in simplifications like finding eigenvalues and eigenvectors. Hence, treating each block independently when diagonalizing is both efficient and effective.

Eigenvalues and eigenvectors are fundamental in understanding the behavior of linear transformations represented by matrices. An eigenvalue is a scalar that indicates how a transformation stretches or shrinks vectors. The eigenvector, associated with this eigenvalue, essentially points in a direction that remains unchanged under transformation.

To find eigenvalues for a matrix block, one solves the characteristic equation \( \det(\lambda I - B) = 0 \), where \( B \) is the block in consideration. For instance, in our problem, calculating for the block \( \begin{bmatrix} -7 & 24 \ 24 & 7 \end{bmatrix} \), we find the eigenvalues \( \lambda = \pm 25 \).

Once eigenvalues are determined, the next step is to find corresponding eigenvectors by solving \( (B - \lambda I)\vec{v} = \vec{0} \). For each eigenvalue, a specific eigenvector emerges, such as \( \begin{bmatrix} 4 \ 3 \end{bmatrix} \) for \( \lambda = 25 \), helping us build the orthogonal matrix later.

An orthogonal matrix, denoted as \( P \), has a unique property: its columns (and rows) are mutually orthogonal unit vectors. This means each vector is perpendicular and has a magnitude of one.

For our case, we construct the orthogonal matrix from the normalized eigenvectors of the blocks of matrix \( A \). The result is:

• \( \begin{bmatrix} (4/5, 3/5, 0, 0) \ (-3/5, 4/5, 0, 0) \ (0, 0, 4/5, 3/5) \ (0, 0, -3/5, 4/5) \end{bmatrix} \)

To check if a matrix is truly orthogonal, verify that \( P^T P = I \), where \( I \) is the identity matrix. Observing this holds, confirms \( P \) is orthogonal. This property facilitates computations like defining the diagonal matrix \( D \) = \( P^{-1} A P = P^T A P \), simplifying to \( D \) as diagonal with eigenvalues on its diagonal.