142 lines
5.0 KiB
Markdown
142 lines
5.0 KiB
Markdown
<!-- vim: set ts=2 sw=2 et tw=80: -->
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---
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header-includes:
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- \usepackage{amsmath}
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- \usepackage{hyperref}
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- \usepackage[utf8]{inputenc}
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- \usepackage[margin=2.5cm]{geometry}
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---
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\title{Midterm -- Optimization Methods}
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\author{Claudio Maggioni}
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\maketitle
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# Exercise 1
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## Point 1
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### Question (a)
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As already covered in the course, the gradient of a standard quadratic form at a
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point $x_0$ is equal to:
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$$ \nabla f(x_0) = A x_0 - b $$
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Plugging in the definition of $x_0$ and knowing that $\nabla f(x_m) = A x_m - b
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= 0$ (according to the first necessary condition for a minimizer), we obtain:
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$$ \nabla f(x_0) = A (x_m + v) - b = A x_m + A v - b = b + \lambda v - b =
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\lambda v $$
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### Question (b)
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The steepest descent method takes exactly one iteration to reach the exact
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minimizer $x_m$ starting from the point $x_0$. This can be proven by first
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noticing that $x_m$ is a point standing in the line that first descent direction
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would trace, which is equal to:
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$$g(\alpha) = - \alpha \cdot \nabla f(x_0) = - \alpha \lambda v$$
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For $\alpha = \frac{1}{\lambda}$, and plugging in the definition of $x_0 = x_m +
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v$, we would reach a new iterate $x_1$ equal to:
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$$x_1 = x_0 - \alpha \lambda v = x_0 - v = x_m + v - v = x_m $$
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The only question that we need to answer now is why the SD algorithm would
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indeed choose $\alpha = \frac{1}{\lambda}$. To answer this, we recall that the
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SD algorithm chooses $\alpha$ by solving a linear minimization option along the
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step direction. Since we know $x_m$ is indeed the minimizer, $f(x_m)$ would be
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obviously strictly less that any other $f(x_1 = x_0 - \alpha \lambda v)$ with
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$\alpha \neq \frac{1}{\lambda}$.
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Therefore, since $x_1 = x_m$, we have proven SD
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converges to the minimizer in one iteration.
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## Point 2
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The right answer is choice (a), since the energy norm of the error indeed always
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decreases monotonically.
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To prove that this is true, we first consider a way to express any iterate $x_k$
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in function of the minimizer $x_s$ and of the missing iterations:
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$$x_k = x_s + \sum_{i=k}^{N} \alpha_i A^i p_0$$
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This formula makes use of the fact that step directions in CG are all
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A-orthogonal with each other, so the k-th search direction $p_k$ is equal to
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$A^k p_0$, where $p_0 = -r_0$ and $r_0$ is the first residual.
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Given that definition of iterates, we're able to express the error after
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iteration $k$ $e_k$ in a similar fashion:
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$$e_k = x_k - x_s = \sum_{i=k}^{N} \alpha_i A^i p_0$$
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We then recall the definition of energy norm $\|e_k\|_A$:
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$$\|e_k\|_A = \sqrt{\langle Ae_k, e_k \rangle}$$
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We then want to show that $\|e_k\|_A = \|x_k - x_s\|_A > \|e_{k+1}\|_A$, which
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in turn is equivalent to claim that:
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$$\langle Ae_k, e_k \rangle > \langle Ae_{k+1}, e_{k+1} \rangle$$
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Knowing that the dot product is linear w.r.t. either of its arguments, we pull
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out the sum term related to the k-th step (i.e. the first term in the sum that
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makes up $e_k$) from both sides of $\langle Ae_k, e_k \rangle$,
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obtaining the following:
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$$\langle Ae_{k+1}, e_{k+1} \rangle + \langle \alpha_k A^{k+1} p_0, e_k \rangle
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+ \langle Ae_{k+1},\alpha_k A^k p_0 \rangle > \langle Ae_{k+1}, e_{k+1}
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\rangle$$
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which in turn is equivalent to claim that:
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$$\langle \alpha_k A^{k+1} p_0, e_k \rangle
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+ \langle Ae_{k+1},\alpha_k A^k p_0 \rangle > 0$$
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From this expression we can collect term $\alpha_k$ thanks to linearity of the
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dot-product:
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$$\alpha_k (\langle A^{k+1} p_0, e_k \rangle
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+ \langle Ae_{k+1}, A^k p_0 \rangle) > 0$$
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and we can further "ignore" the $\alpha_k$ term since we know that all
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$\alpha_i$s are positive by definition:
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$$\langle A^{k+1} p_0, e_k \rangle
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+ \langle Ae_{k+1}, A^k p_0 \rangle > 0$$
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Then, we convert the dot-products in their equivalent vector to vector product
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form, and we plug in the definitions of $e_k$ and $e_{k+1}$:
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$$p_0^T (A^{k+1})^T (\sum_{i=k}^{N} \alpha_i A^i p_0) +
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p_0^T (A^{k})^T (\sum_{i=k+1}^{N} \alpha_i A^i p_0) > 0$$
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We then pull out the sum to cover all terms thanks to associativity of vector
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products:
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$$\sum_{i=k}^N (p_0^T (A^{k+1})^T A^i p_0) \alpha_i+ \sum_{i=k+1}^N
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(p_0^T (A^{k})^T A^i p_0) \alpha_i > 0$$
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We then, as before, can "ignore" all $\alpha_i$ terms since we know by
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definition that
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they are all strictly positive. We then recalled that we assumed that A is
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symmetric, so $A^T = A$. In the end we have to show that these two
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inequalities are true:
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$$p_0^T A^{k+1+i} p_0 > 0 \; \forall i \in [k,N]$$
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$$p_0^T A^{k+i} p_0 > 0 \; \forall i \in [k+1,N]$$
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To show these inequalities are indeed true, we recall that A is symmetric and
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positive definite. We then consider that if a matrix A is SPD, then $A^i$ for
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any positive $i$ is also SPD[^1]. Therefore, both inequalities are trivially
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true due to the definition of positive definite matrices.
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[^1]: source: [Wikipedia - Definite Matrix $\to$ Properties $\to$
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Multiplication](
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https://en.wikipedia.org/wiki/Definite_matrix#Multiplication)
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Thanks to this we have indeed proven that the delta $\|e_k\|_A - \|e_{k+1}\|_A$
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is indeed positive and thus as $i$ increases the energy norm of the error
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monotonically decreases.
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