\( \left.\begin{array}{l}\text { Given the covariance matrix } S=\left[\begin{array}{cc}4.0 & 15.0 \\ 15.0 & 57.0\end{array}\right] \text { with eigenvalues } \lambda_{1}=60.95 \text { and } \lambda_{2}=0.05 \text { and } \\ \text { eigenvectors } \mathbf{u}_{1}=\left[\begin{array}{c}-0.97 \\ 0.25\end{array}\right] \text { and } \mathbf{u}_{2}=\left[\begin{array}{l}0.25 \\ 0.97\end{array}\right] \text {, what is the diagonal matrix } D \text { such that } D=P^{T} S P \\ D=[\text { Ex: } 1.23 \\ \square\end{array}\right] \)

To find the diagonal matrix \( D \) where \( D = P^{T} S P \), we first construct the matrix \( P \) using the eigenvectors as columns. Thus, we have: \[ P = \begin{bmatrix} -0.97 & 0.25 \\ 0.25 & 0.97 \end{bmatrix} \] The diagonal matrix \( D \) will contain the eigenvalues on its diagonal, as follows: \[ D = \begin{bmatrix} 60.95 & 0 \\ 0 & 0.05 \end{bmatrix} \] Thus, the complete diagonal matrix \( D \) is: \[ D = \begin{bmatrix} 60.95 \\ 0.05 \end{bmatrix} \] This confirms that the process to derive \( D \) from \( S \) through \( P \) is in alignment with the properties of eigenvalues and eigenvectors in the context of covariance matrices!