hw2: paraculation

Claudio_Maggioni_2/Claudio_Maggioni_2.tex

@@ -55,25 +55,24 @@ The matrix $A$ and the vector $b$ appear in the following form:
 	0 & 0 & 0 & 0 &   \ldots & 1 \\
 	\end{bmatrix}\;\;\;b = \begin{bmatrix}0\\h^2\\h^2\\\vdots\\0\end{bmatrix}\]
 
-For $N = 4$, we can attempt to build a minimizer to the solve the system $Ax = b$. In order to find an $x$ such that $Ax = b$, we could define a minimizer $\phi(x)$ such that:
+For $N = 4$, we have the following $A$ and $b$:
 
-\[\phi(x) = \|b - Ax\|^2\]
+\[A = \begin{bmatrix}
+	1 & 0 & 0 & 0 \\
+	-1 & 2 & -1 & 0 \\
+	0 & -1 & 2 & -1 \\
+	0 & 0 & 0 & 1 \\
+	\end{bmatrix}\;\;\;b = \begin{bmatrix}0\\\frac19\\\frac19\\0\end{bmatrix}\]
 
-Here $\phi(x) = 0$ would mean that $x$ is an exact solution of the system $Ax = b$. We can then attempt to write such minimizer for $N = 4$:
+In order to solve the minimization problem we need to minimize the energy function:
 
-\[\phi(x) = \|b - Ax\|^2 = \left|\begin{bmatrix}0 - x_1\\\frac19 -x_1 + 2x_2 -x_3\\
-\frac19 -x_2 +2x_3 -x_4\\0 - x_4\end{bmatrix}\right|^2 =\]\[= x_1^2 + \left(\frac19 - x_1 + 2x_2 - x_3\right)^2 + \left(\frac19 - x_2 + 2x_3 - x_4\right)^2 + x_4^2\]
+\[\phi(x) = \frac12 x^T A x - b^T x\]
 
-\[\Delta \phi(x) = \begin{bmatrix}4x_1 - 4x_2 + 2x_3 -\frac29\\
--4x_1 +10x_2 -8x_3 + 2x_4 +\frac29\\
-2x_1 -8x_2 +10x_3 + -4x_4 +\frac29\\
-	2x_2 - 4x_3 + 4x_4 -\frac29\end{bmatrix}\;\;\;\Delta^2 \phi(x) = \begin{bmatrix}4&-4&2&0\\-4&10&-8&2\\2&-8&10&-4\\0&2&-4&4\end{bmatrix}\]
+Computing the gradient of the minimizer, and considering that $A$ is clearly not symmetric as shown above, we find:
 
-		As it can be seen from the Hessian calculation, the Hessian is positive definite forall $x$s. This means, by the sufficient condition of minimizers, that we can find a minimizer by solving $\Delta \phi(x) = 0$ (i.e. finding stationary points in the hyperspace defined by $\phi(x)$. Solving that, we find:
+\[\Delta \phi(x) = \frac12A^T x + \frac12Ax - b\]
 
-\[x = \begin{bmatrix}0\\\frac19\\\frac19\\0\end{bmatrix}\;\;\]
-
-	which is indeed the minimizer and solution of $Ax = b$. Therefore, $\phi(x)$ is a valid energy function. Although the $\phi(x)$ given here is not strictly in a matrix vector product quadratic form, it is indeed a valid energy function and for different values of $N$ similar fashioned $\phi(x)$ could be derived. Therefore, we can say that the problem has an energy function.
+If $A$ would be symmetric, $\Delta \phi(x)$ is equal to $Ax - b$ and therefore the semantic equivalence between this energy function and the solution of $Ax = b$ is straightforward. However, if $A$ is not symmetric then this does not hold and an enegry function therefore does not exist.
 
 \subsection{Once the new matrix has been derived, write the energy function related to the new problem
 and the corresponding gradient and Hessian.}
@@ -138,26 +137,6 @@ To sum up our claim, we can say CG is indeed a Krylov subspace method because:
 
 These statements have been already proven in class and the proof can be found in Theorem 5.3 of Nocedal's book.
 
-% These statements can be proven by induction over $k$. The base case holds trivially for $k=0$, while we have by the induction hypothesis for any $k$ that:
-
-% \[r_k \in \text{span}\{r_0, A r_0, \ldots, A^k r_0\}\;\;\; p_k \in \text{span}\{r_0, A r_0, \ldots, A^k r_0\}\]
-
-% We'd like to prove that the two properties hold from $k+1$ starting from the hypothesis on $k$. We first multiply the first hypothesis by $A$ from the left:
-
-% \[A r_k \in \text{span}\{A r_0, A^2 r_0, \ldots, A^{k+1} r_0\}\]
-
-% By the alternative definition of the residual for the CG method (i.e. $r_{k+1} = r_k + \alpha_k A p_k$), we find that:
-
-% \[r_{k+1} \in \text{span}\{r_0, A r_0, A^2 r_0, \ldots, A^{k+1} r_0\}\]
-
-% We need to add $r_0$ in the span again since one of the components that defines $r_1$ is indeed $r_0$. We don't need to add other terms in the span since $r_1$ to $r_k$ are in the span already by the induction hypothesis.
-
-% Combining this expression with the induction hypothesis we prove induction of the first statement by having:
-
-% \[\text{span}\{r_0, r_1, \ldots, r_n, r_{n+1}\} \subseteq \text{span}\{r_0, A r_0, \ldots, A^k r_0, A^{k+1} r_0\}\]
-
-% To prove $\supseteq$ as well to achieve equality, we use the induction hypothesis for the second statement to find that $A^kr_0 \in \text{span}\{A$:
-
 \section{Exercise 2}
 
 Consider the linear system $Ax = b$, where the matrix $A$ is constructed in three different ways: