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+<!-- vim: set ts=2 sw=2 et tw=80: -->
+
+---
+header-includes:
+- \usepackage{amsmath}
+- \usepackage{hyperref}
+- \usepackage[utf8]{inputenc}
+- \usepackage[margin=2.5cm]{geometry}
+---
+\title{Midterm -- Optimization Methods}
+\author{Claudio Maggioni}
+\maketitle
+
+# Exercise 1
+
+## Point 1
+
+### Question (a)
+
+As already covered in the course, the gradient of a standard quadratic form at a
+point $ x_0$ is equal to:
+
+$$ \nabla f(x_0) = A x_0 - b $$
+
+Plugging in the definition of $x_0$ and knowing that $\nabla f(x_m) = A x_m - b = 0$
+(according to the first necessary condition for a minimizer), we obtain:
+
+$$ \nabla f(x_0) = A (x_m + v) - b = A x_m + A v - b = b + \lambda v - b =
+\lambda v $$
+
+### Question (b)
+
+The steepest descent method takes exactly one iteration to reach the exact
+minimizer $x_m$ starting from the point $x_0$. This can be proven by first
+noticing that $x_m$ is a point standing in the line that first descent direction
+would trace, which is equal to:
+
+$$g(\alpha) = - \alpha \cdot \nabla f(x_0) = - \alpha \lambda v$$
+
+For $\alpha = \frac{1}{\lambda}$, and plugging in the definition of $x_0 = x_m +
+v$, we would reach a new iterate $x_1$ equal to:
+
+$$x_1 = x_0 - \alpha \lambda v = x_0 - v = x_m + v - v = x_m $$
+
+The only question that we need to answer now is why the SD algorithm would
+indeed choose $\alpha = \frac{1}{\lambda}$. To answer this, we recall that the
+SD algorithm chooses $\alpha$ by solving a linear minimization option along the
+step direction. Since we know $x_m$ is indeed the minimizer, $f(x_m)$ would be
+obviously strictly less that any other $f(x_1 = x_0 - \alpha \lambda v)$ with
+$\alpha \neq \frac{1}{\lambda}$.
+
+Therefore, since $x_1 = x_m$, we have proven SD
+converges to the minimizer in one iteration.
+
+### Point 2
+
+The right answer is choice (a), since the energy norm of the error indeed always
+decreases monotonically.
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