91Ó°ÊÓ

We need to show that \[ \|\mathbf{x}-\mathbf{y}\|_{2}^{2}=2(\mathbf{x}-\mathbf{y})^{T} \mathbf{x}. \] First, let's compute the squared norm of the vector difference \(\mathbf{x}-\mathbf{y}\): \[ \|\mathbf{x}-\mathbf{y}\|_{2}^{2}=(\mathbf{x}-\mathbf{y})^{T} (\mathbf{x}-\mathbf{y}). \] Now, we perform the matrix multiplication and simplify: \begin{align*} (\mathbf{x}-\mathbf{y})^{T} (\mathbf{x}-\mathbf{y}) &= (\mathbf{x}^T - \mathbf{y}^T)(\mathbf{x}-\mathbf{y}) \\ &= \mathbf{x}^T\mathbf{x} - \mathbf{x}^T\mathbf{y} - \mathbf{y}^T\mathbf{x} + \mathbf{y}^T\mathbf{y} \\ \end{align*} Since \(\|\mathbf{x}\|_{2}=\|\mathbf{y}\|_{2}\), we have that \[ \mathbf{x}^T\mathbf{x} = \mathbf{y}^T\mathbf{y}. \] Plugging this into the previous equation, we have: \[ (\mathbf{x}-\mathbf{y})^{T} (\mathbf{x}-\mathbf{y}) = \mathbf{x}^T\mathbf{x} - \mathbf{x}^T\mathbf{y} - \mathbf{y}^T\mathbf{x} + \mathbf{x}^T\mathbf{x}. \] Notice that since \((\mathbf{x}-\mathbf{y})^T \mathbf{x} = \mathbf{x}^T\mathbf{x} - \mathbf{y}^T\mathbf{x}\), the expression in the right-hand side \((\mathbf{x}-\mathbf{y})^{T} \mathbf{x}\) can be rewritten as: \[ 2(\mathbf{x}-\mathbf{y})^T \mathbf{x} = 2(\mathbf{x}^T\mathbf{x}-\mathbf{y}^T\mathbf{x}). \] Thus, \[ \|\mathbf{x}-\mathbf{y}\|_{2}^{2} = 2(\mathbf{x}-\mathbf{y})^T \mathbf{x}, \] as was desired to be shown.

Recall that \(Q = I - 2\mathbf{u}\mathbf{u}^T\) where \(\mathbf{u}=\frac{1}{\|\mathbf{x}-\mathbf{y}\|_{2}}(\mathbf{x}-\mathbf{y})\). To prove that \(Q \mathbf{x} = \mathbf{y}\), let's compute the product \(Q\mathbf{x}\) and try to simplify it: \begin{align*} Q \mathbf{x} &= (\mathbf{I} - 2\mathbf{u}\mathbf{u}^T) \mathbf{x} \\ &= \mathbf{x}- 2\mathbf{u}\mathbf{u}^T \mathbf{x} \\ \end{align*} Now we need to compute the term \(\mathbf{u}\mathbf{u}^T \mathbf{x}\): \[ \mathbf{u}\mathbf{u}^T \mathbf{x} = \frac{1}{\|\mathbf{x}-\mathbf{y}\|_{2}^2}(\mathbf{x}-\mathbf{y})(\mathbf{x} - \mathbf{y})^T \mathbf{x}. \] From part (a), we know that \(\|\mathbf{x}-\mathbf{y}\|_{2}^{2} = 2(\mathbf{x}-\mathbf{y})^T\mathbf{x}\). Substituting this expression into the above equation, we get: \[ \mathbf{u}\mathbf{u}^T \mathbf{x} = \frac{1}{2(\mathbf{x}-\mathbf{y})^T\mathbf{x}}(\mathbf{x}-\mathbf{y})(\mathbf{x} - \mathbf{y})^T \mathbf{x}. \] Notice that the terms \((\mathbf{x}-\mathbf{y})^T\mathbf{x}\) in the numerator and denominator will cancel out. Thus, we have: \[ \mathbf{u}\mathbf{u}^T \mathbf{x} = \frac{1}{2}(\mathbf{x}-\mathbf{y})(\mathbf{x} - \mathbf{y})^T \mathbf{x}. \] Now, we can plug this result back into the expression for \(Q\mathbf{x}\): \begin{align*} Q \mathbf{x} &= \mathbf{x} - 2\mathbf{u}\mathbf{u}^T\mathbf{x} \\ &= \mathbf{x} - 2\left(\frac{1}{2}(\mathbf{x}-\mathbf{y})(\mathbf{x}-\mathbf{y})^T \mathbf{x}\right) \\ &= \mathbf{x} - (\mathbf{x}-\mathbf{y})(\mathbf{x}-\mathbf{y})^T \mathbf{x} \\ \end{align*} From part (a), we have: \[ (\mathbf{x}-\mathbf{y})^T \mathbf{x} = \frac{1}{2} \|\mathbf{x} - \mathbf{y}\|_{2}^2. \] We can plug this expression back into the equation for \(Q\mathbf{x}\) and get: \[ Q \mathbf{x} = \mathbf{x} - (\mathbf{x}-\mathbf{y})\left(\frac{1}{2} \|\mathbf{x} - \mathbf{y}\|_{2}^2 \right). \] This equation simplifies to: \[ Q \mathbf{x} = \mathbf{x} - \frac{1}{2}(\mathbf{x}-\mathbf{y})\|\mathbf{x} - \mathbf{y}\|_{2}^2 . \] We can now substitute the result from part (a) back into the equation: \[ Q \mathbf{x} = \mathbf{x} - \frac{1}{2}(\mathbf{x}-\mathbf{y}) (2(\mathbf{x}-\mathbf{y})^T\mathbf{x}). \] The terms \(\frac{1}{2}\) and \(2\) will cancel out, giving: \[ Q \mathbf{x} = \mathbf{x} - (\mathbf{x}-\mathbf{y})(\mathbf{x}-\mathbf{y})^T\mathbf{x}. \] Finally, we rearrange the equation to obtain the desired result: \[ Q \mathbf{x} = \mathbf{y}. \] Hence, we have proven that \(Q\mathbf{x} = \mathbf{y}\).