--- 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. To prove that this is true, we first consider a way to express any iterate $x_k$ in function of the minimizer $x_s$ and of the missing iterations: $$x_k = x_s + \sum_{i=k}^{N} \alpha_i A^i p_0$$ This formula makes use of the fact that step directions in CG are all A-orthogonal with each other, so the k-th search direction $p_k$ is equal to $A^k p_0$, where $p_0 = -r_0$ and $r_0$ is the first residual. Given that definition of iterates, we're able to express the error after iteration $k$ $e_k$ in a similar fashion: $$e_k = x_k - x_s = \sum_{i=k}^{N} \alpha_i A^i p_0$$ We then recall the definition of energy norm $\|e_k\|_A$: $$\|e_k\|_A = \sqrt{\langle Ae_k, e_k \rangle}$$ We then want to show that $\|e_k\|_A = \|x_k - x_s\|_A > \|e_{k+1}\|_A$, which in turn is equivalent to claim that: $$\langle Ae_k, e_k \rangle > \langle Ae_{k+1}, e_{k+1} \rangle$$ Knowing that the dot product is linear w.r.t. either of its arguments, we pull out the sum term related to the k-th step (i.e. the first term in the sum that makes up $e_k$) from both sides of $\langle Ae_k, e_k \rangle$, obtaining the following: $$\langle Ae_{k+1}, e_{k+1} \rangle + \langle \alpha_k A^{k+1} p_0, e_k \rangle + \langle Ae_{k+1},\alpha_k A^k p_0 \rangle > \langle Ae_{k+1}, e_{k+1} \rangle$$ which in turn is equivalent to claim that: $$\langle \alpha_k A^{k+1} p_0, e_k \rangle + \langle Ae_{k+1},\alpha_k A^k p_0 \rangle > 0$$ From this expression we can collect term $\alpha_k$ thanks to linearity of the dot-product: $$\alpha_k (\langle A^{k+1} p_0, e_k \rangle + \langle Ae_{k+1}, A^k p_0 \rangle) > 0$$ and we can further "ignore" the $\alpha_k$ term since we know that all $\alpha_i$s are positive by definition: $$\langle A^{k+1} p_0, e_k \rangle + \langle Ae_{k+1}, A^k p_0 \rangle > 0$$ Then, we convert the dot-products in their equivalent vector to vector product form, and we plug in the definitions of $e_k$ and $e_{k+1}$: $$p_0^T (A^{k+1})^T (\sum_{i=k}^{N} \alpha_i A^i p_0) + p_0^T (A^{k})^T (\sum_{i=k+1}^{N} \alpha_i A^i p_0) > 0$$ We then pull out the sum to cover all terms thanks to associativity of vector products: $$\sum_{i=k}^N (p_0^T (A^{k+1})^T A^i p_0) \alpha_i+ \sum_{i=k+1}^N (p_0^T (A^{k})^T A^i p_0) \alpha_i > 0$$ We then, as before, can "ignore" all $\alpha_i$ terms since we know by definition that they are all strictly positive. We then recalled that we assumed that A is symmetric, so $A^T = A$. In the end we have to show that these two inequalities are true: $$p_0^T A^{k+1+i} p_0 > 0 \; \forall i \in [k,N]$$ $$p_0^T A^{k+i} p_0 > 0 \; \forall i \in [k+1,N]$$ To show these inequalities are indeed true, we recall that A is symmetric and positive definite. We then consider that if a matrix A is SPD, then $A^i$ for any positive $i$ is also SPD[^1]. Therefore, both inequalities are trivially true due to the definition of positive definite matrices. [^1]: source: [Wikipedia - Definite Matrix $\to$ Properties $\to$ Multiplication]( https://en.wikipedia.org/wiki/Definite_matrix#Multiplication) Thanks to this we have indeed proven that the delta $\|e_k\|_A - \|e_{k+1}\|_A$ is indeed positive and thus as $i$ increases the energy norm of the error monotonically decreases.