91Ó°ÊÓ

In order to show that \(D\) is idempotent, we need to prove that \(D^2 = D\). Recall that \(D\) is a diagonal matrix with its diagonal elements being 0 or 1. Let's multiply \(D\) with itself: \[ D^2 = DD = \begin{pmatrix} d_1 & 0 & \cdots & 0 \\ 0 & d_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & d_n \end{pmatrix} \begin{pmatrix} d_1 & 0 & \cdots & 0 \\ 0 & d_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & d_n \end{pmatrix}, \] where \(d_i \in \{0, 1\}\) for all \(i \in \{1, 2, \ldots, n\}\). Now, let's evaluate the product by multiplying the elements: \[ D^2 = \begin{pmatrix} d_1^2 & 0 & \cdots & 0 \\ 0 & d_2^2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & d_n^2 \end{pmatrix}. \] Since \(d_i^2 = d_i\) for all \(i\) (because \(d_i\) are either 0 or 1), we have \(D^2 = D\). Therefore, \(D\) is idempotent.

Next, let's prove that if \(A = XDX^{-1}\) for a nonsingular matrix \(X\), then \(A\) is also idempotent. First, we'll calculate \(A^2\): \[ A^2 = (XDX^{-1})(XDX^{-1}) \] Now, let's multiply the matrices by applying the associative property: \[ A^2 = X(DX^{-1})(XD)(X^{-1}) \] Since \(D\) is idempotent, as we showed in Step 1, we can insert on \(DX^{-1}XD\): \[ A^2 = X(DX^{-1}X)D(X^{-1}) \] Now, notice that \((X^{-1}X) = I\) where \(I\) is the identity matrix. So, \[ A^2 = X(DI)D(X^{-1}) \] Since multiplying any matrix by the identity matrix will result in the same matrix, we get: \[ A^2 = XDD(X^{-1}) = XDX^{-1} \] This result shows that \(A^2 = A\). Therefore, \(A\) is idempotent. This completes our proof.