91Ó°ÊÓ

To show that H is symmetric, we need to prove that H^T = H. Using the definition of H, we calculate H^T: \( H^T = (I - 2\mathbf{uu}^T)^T \) Distribute the transpose operation across the subtraction: \( H^T = I^T - (2\mathbf{uu}^T)^T \) Since \(I^T = I\), and taking the transpose of the scalar multiple is the same as taking the transpose and then scaling: \( H^T = I - 2(\mathbf{uu}^T)^T \) Now, use the property that \((AB)^T = B^TA^T\) to find the transpose of \(\mathbf{uu}^T\): \( H^T = I - 2\mathbf{u}^T\mathbf{u} \) Since \(H = I - 2\mathbf{uu}^T\) and \(H^T = I - 2\mathbf{u}^T\mathbf{u} = I - 2\mathbf{uu}^T\), we can conclude that H is symmetric.

To show that H is orthogonal, we need to prove that H^T * H = I. Since H is symmetric, we know that H^T = H. Therefore, we need to find H * H: \(H * H = (I - 2\mathbf{uu}^T)(I - 2\mathbf{uu}^T)\) Now, use the distributive law to expand the product: \(H * H = I^2 - 2\mathbf{uu}^TI - 2\mathbf{uu}^TI + 4\mathbf{uu}^T\mathbf{uu}^T\) Since \(I^2 = I\), this simplifies to: \(H * H = I - 4\mathbf{uu}^T + 4\mathbf{uu}^T\mathbf{uu}^T\) Now, let's focus on the product \(\mathbf{uu}^T\mathbf{uu}^T\). Since \(\mathbf{u}\) is a unit vector, \(\mathbf{u}^T\mathbf{u} = ||\mathbf{u}||^2 = 1\). Therefore, \(\mathbf{uu}^T\mathbf{uu}^T = \mathbf{uu}^T\). Substituting this back, we get: \(H * H = I - 4\mathbf{uu}^T + 4\mathbf{uu}^T = I\) Therefore, H is orthogonal.

Since H is orthogonal (i.e., H^T * H = I), and H is symmetric (i.e., H^T = H), it must be true that H * H = I. This directly implies that H is its own inverse, which completes the proof.