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

Given the covariance matrix \[ S=\begin{pmatrix} 4.0 & 15.0 \\ 15.0 & 57.0 \end{pmatrix}, \] its eigenvalues are given by \[ \lambda_{1}=60.95, \quad \lambda_{2}=0.05, \] and the corresponding eigenvectors are \[ \mathbf{u}_{1}=\begin{pmatrix}-0.97 \\ 0.25\end{pmatrix}, \quad \mathbf{u}_{2}=\begin{pmatrix}0.25 \\ 0.97\end{pmatrix}. \] If we form the matrix \( P \) by placing the eigenvectors as columns \[ P = \begin{pmatrix} -0.97 & 0.25 \\ 0.25 & 0.97 \end{pmatrix}, \] then the diagonalization of \( S \) is written as \[ S = PDP^{T}, \] where \( D \) is the diagonal matrix containing the eigenvalues. Equivalently, by premultiplying by \( P^{T} \) and postmultiplying by \( P \) we have \[ D = P^{T}SP. \] Since the eigenvectors are arranged in the same order as the eigenvalues (first column with \( \lambda_1 \) and second with \( \lambda_2 \)), the diagonal matrix \( D \) is \[ D = \begin{pmatrix} 60.95 & 0 \\ 0 & 0.05 \end{pmatrix}. \]