91Ó°ÊÓ

First, we'll show that any diagonal matrix A is congruent to the diagonal matrix D with diagonal entries \(a_1k_1^2, \ldots, a_nk_n^2\). Two matrices A and D are congruent if there exists an invertible matrix P such that \(D = P^TAP\). Consider the matrix P with diagonal entries \(k_1, \ldots, k_n\) and all other entries being 0. As all its diagonal entries are nonzero, P is invertible. Let's compute \(P^TAP\): \(P^TAP = (P^TA)P = P^T(diag(a_1, \ldots, a_n))P\) Notice the product of P with the diagonal matrix A is another diagonal matrix with entries \(a_1k_1, \ldots, a_nk_n\) on the diagonal. Therefore, \(P^TAP = P^T(diag(a_1k_1, \ldots, a_nk_n))\) Now, the product of the transpose of P with the diagonal matrix we found above is yet another diagonal matrix, with entries \(a_1k_1^2, \ldots, a_nk_n^2\). So we have, \(P^TAP = diag(a_1k_1^2, \ldots, a_nk_n^2) = D\) This proves that A is congruent to D.

Now, we will show that if \(K = \mathbf{C}\), we can choose D such that its diagonal entries are only 1's and 0's. If an entry on the diagonal of A is 0, we keep it as such in D. For a non-zero diagonal entry \(a_i\), choose \(k_i = \frac{1}{\sqrt{a_i}}\). Then for the corresponding diagonal entry in D, we have \(a_ik_i^2 = a_i \frac{1}{a_i} = 1\). Therefore, by appropriately choosing the values of \(k_1, \ldots, k_n\), we can obtain a diagonal matrix D with diagonal entries being only 1's and 0's if \(K = \mathbf{C}\).

Finally, we will show that if \(K = \mathbf{R}\), we can choose D such that its diagonal entries are only 1's, -1's, and 0's. Similar to part (a), if an entry on the diagonal of A is 0, we keep it as such in D. If a non-zero diagonal entry of A is positive, we can choose the corresponding \(k_i = \frac{1}{\sqrt{a_i}}\), resulting in a diagonal entry in D being 1. If a non-zero diagonal entry of A is negative, we can choose the corresponding \(k_i = \frac{-1}{\sqrt{a_i}}\), and the diagonal entry in D will be -1. Therefore, by appropriately choosing the values of \(k_1, \ldots, k_n\), we can obtain a diagonal matrix D with diagonal entries being only 1's, -1's, and 0's if \(K = \mathbf{R}\). This concludes the proof for part (b) and the entire exercise.