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@@ -22,11 +22,67 @@ The purpose of this assignment\footnote{This document is originally based on a S
 
 \subsection{Theory [20 points]}
 
+\subsubsection{Show that the order of convergence of the power method is linear,
+and state what the asymptotic error constant is.}
+
+First of all, we show the the sequence of vectors computed by power iteration
+indeed converges to $\lambda_1$ or the biggest eigenvector (we assume we name
+eigenvectors in decreasing order of magnitude, with $|\lambda_1| > |\lambda_i|$
+for $i \in 2..n$).
+
+We can express the seed for the eigenvector (i.e. the initial value of $v$ of
+the power iteration) as a linear combination of eigenvalues:
+
+\[v_0 = \sum_{i=1}^n a_i x_i\]
+
+We can then express the result of the n-th power method as
+
+\[v_n = \gamma A v_{n-1} = A^n v_0 = \sum_{i=1}^n \gamma a_i \lambda_i^n x_i =
+\lambda_1^n  \sum_{i=1}^n \gamma a_i \left( \frac{\lambda_i}{\lambda_1} \right)^n x_i  =
+\gamma a_1 \lambda_1^n x_1 + \lambda_1^n \sum_{i=2}^n \gamma a_i
+\left(\frac{\lambda_i}{\lambda_1}\right)^n x_i \]
+
+Here, $\gamma$ is just a normalization term to make $||v_n|| = 1$. $v_n$ clearly
+converges to $x_1$ since all the terms in the $\sum_{i=2}^n$ contain
+$\frac{\lambda_i}{\lambda_1}$, which is always less than 0 if $i > 1$ for the
+sorting of eigenvalues we did before. Therefore, these terms to the power of n
+converge to 0, and $\gamma$ will cancel out $a_1 \lambda_1^k$ due to the
+normalization, thus making the sequence converge to $\lambda_1$.
+
+To see if the sequence converges linearly we use the definitions of rate of
+convergence:
+
+\[\lim_{n \to \infty}\frac{|x_{n+1} - \lambda_1|}{|x_n - \lambda_1|^1} = \mu\]
+
+If this limit has a finite solution then the sequence converges linearly with
+rate $\mu$.
+
+\[\lim_{n \to \infty}\frac{\left| a_1 \lambda_1^{n+1} x_1 + \lambda_1^{n+1}
+\sum_{i=2}^n  a_i \left(\frac{\lambda_i}{\lambda_1}\right)^{n+1}
+x_i - \beta_{n+1} x_1\right|}
+{\left| a_1 \lambda_1^n x_1 + \lambda_1^n \sum_{i=2}^n  a_i
+\left(\frac{\lambda_i}{\lambda_1}\right)^n x_i - \beta_n x_1\right|^1} = \mu\]
+
+To simplify calculations, we consider the sequence without the normalization
+factor $\gamma$ that will converge to a denormalized version of $x_1$, named
+$\beta x_1$. We can then simplify the $a_1\lambda_1^{i}x_1$ terms in the
+sequences with $\beta_{i} x_1$ since $\beta_i$ can be set freely.
+
+Now we consider that if $|\lambda_2| > |\lambda_i| \forall i \in 3..n$ (since we
+sorted the eigenvalues), then
+$\left(\frac{\lambda_i}{\lambda_1}\right)^n$ for $i > 2$ will always converge faster to
+0 than $\left(\frac{\lambda_2}{\lambda_1}\right)^n$ thus all terms other than
+$i=2$ can be ignored in the limit computation. Therefore, the limit has finite
+solution and the convergence rate
+is
+
+\[\mu = \frac{\lambda_2}{\lambda_1}\]
+
 \subsubsection{What assumptions should be made to guarantee convergence of the power method?}
 The first assumption to make is that the biggest eigenvalue in terms of absolute values should (let's name it $\lambda_1$)
 be strictly greater than all other eigenvectors, so:
 
-$$|\lambda_1| < |\Lambda_i| \forall i \in \{2..n\}$$
+\[|\lambda_1| < |\lambda_i| \forall i \in \{2..n\}\]
 
 Also, the eigenvector \textit{guess} from which the power iteration starts must have a component in the direction of $x_i$, the eigenvector for the eigenvalue $\lambda_1$ from before.
 
@@ -38,11 +94,11 @@ The shift and invert approach is a variant of the power method that may signific
 
 where $\alpha$ is an arbitrary constant that must be chosen wisely in order to increase the rate of convergence. Since the eigenvalues $u_i$ of B can be derived from the eigenvalues $\lambda_i$ of A, namely:
 
-$$u_i = \frac{1}{\lambda_i - \alpha}$$
+\[u_i = \frac{1}{\lambda_i - \alpha}\]
 
 the rate of convergence of the power method on B is:
 
-$$\left|\frac{u_2}{u_1}\right| = \left|\frac{\frac1{\lambda_2 - \alpha}}{\frac1{\lambda_1 - \alpha}}\right| = \left|\frac{\lambda_1 - \alpha}{\lambda_2 - \alpha}\right|$$
+\[\left|\frac{u_2}{u_1}\right| = \left|\frac{\frac1{\lambda_2 - \alpha}}{\frac1{\lambda_1 - \alpha}}\right| = \left|\frac{\lambda_1 - \alpha}{\lambda_2 - \alpha}\right|\]
 
 By choosing $\alpha$ close to $\lambda_1$, the convergence is sped up. To further increase the rate of convergence (up to a cubic rate), a new $\alpha$, and thus a new $B$, may be chosen for every iteration.