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\author{ {\fontspec{Trebuchet MS}Marcin Chrz\k{a}szcz, Danny van Dyk} (UZH)}
\institute{UZH}
\title[Linear equation systems: exact methods]{Linear equation systems:exact methods}
\date{21 September 2016}
\begin{document}
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\flushright\fontspec{Trebuchet MS}\bfseries \Huge {Linear equation systems: exact methods}
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\quad
\vspace{3em}
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\flushright \vspace{-1.8em} {\fontspec{Trebuchet MS} \large Marcin Chrząszcz, Danny van Dyk\\\vspace{-0.1em}\small \href{mailto:mchrzasz@cern.ch}{mchrzasz@cern.ch}, \href{mailto:dany.van.dyk@gmail.com}{danny.van.dyk@gmail.com}}
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\vspace{1em}
% \footnotesize\textcolor{gray}{With N. Serra, B. Storaci\\Thanks to the theory support from M. Shaposhnikov, D. Gorbunov}\normalsize\\
\vspace{0.5em}
\textcolor{normal text.fg!50!Comment}{Numerical Methods, \\ 10 October, 2016}
\end{center}
\end{frame}
}
\begin{frame}\frametitle{Linear eq. system}
\ARROW This and the next lecture will focus on a well known problem. Solve the following equation system:
\begin{align*}
A \cdot x =b,
\end{align*}
\ARROWR $A = a_{ij} \in \mathbb{R}^{n\times n}$ and $\det(A) \neq 0$\\
\ARROWR $b=b_i \in \mathbb{R}^n$.\\
\ARROW The problem: Find the $x$ vector.
\end{frame}
\begin{frame}\frametitle{Error digression}
\begin{small}
\ARROW There is enormous amount of ways to solve the linear equation system.\\
\ARROW The choice of one over the other of them should be gathered by the {\it condition} of the matrix $A$ denoted at $cond(A)$.
\ARROW If the $cond(A)$ is small we say that the problem is well conditioned, otherwise we say it's ill conditioned.\\
\ARROW The {\it condition} relation is defined as:
\begin{align*}
cond(A) = \Vert A \Vert \cdot \Vert A^{-1} \Vert
\end{align*}
\ARROW Now there are many definitions of different norms... The most popular one (so-called ''column norm''):
\begin{align*}
\Vert A \Vert_1 = \max_{1 \leq j \leq n} \sum_{i=1}^n \vert a_{i,j} \vert,
\end{align*}
where $n$ -is the dimension of $A$, $i,j$ are columns and rows numbers.
\end{small}
\end{frame}
\begin{frame}\frametitle{More norms}
\begin{small}
\ARROW A different norm is a spectral norm:
\begin{align*}
\Vert A \Vert_2 &= \sqrt{\rho(A^T A)}\\
\rho(M) &= \max \lbrace \vert\lambda_i \vert: \det{M- \lambda I} =0,~i=1,...n \rbrace
\end{align*}
where $\rho(M)$ - spectral radius of $M$ matrix, $I$ unit matrix, $\lambda_i$ eigenvalues of $M$.\\
\ARROW Row norm:
\begin{align*}
\Vert A \Vert_{\infty} &= \max_{1 \leq i \leq n} \sum_{j=1}^n \vert a_{i,j} \vert,
\end{align*}
\begin{exampleblock}{Digression:}
\ARROWR Calculation of the matrix norms are not a simple process at all. There are certain class of matrices that make the calculations easier.\\
\ARROWR The spectral norm can be also defined:
\begin{align*}
cond_2(A) = \frac{\max_{1\leq i \leq n}\vert \lambda_i \vert }{\min_{1\leq i \leq n}\vert \lambda_i \vert },
\end{align*}
\end{exampleblock}
\end{small}
\end{frame}
\begin{frame}\frametitle{Example, ill-conditioned matrix}
\begin{small}
\ARROW The text-book example of wrongly conditioned matrix is the Hilbert matrix:
\begin{align*}
h_{i,j} = \frac{1}{i+j-1}
\end{align*}
\ARROWR Example:
\begin{align*}
h_{i,j}^{4 \times 4} = \begin{pmatrix}
1 & \frac{1}{2} & \frac{1}{3} & \frac{1}{4} \\
\frac{1}{2} & \frac{1}{3} & \frac{1}{4} & \frac{1}{5} \\
\frac{1}{3} & \frac{1}{4} & \frac{1}{5} & \frac{1}{6}\\
\frac{1}{4} & \frac{1}{5} & \frac{1}{6} & \frac{1}{7}
\end{pmatrix}
\end{align*}
\ARROW The condition of this matrix:
\begin{align*}
cond(A)=\mathcal{O}\left(\frac{e^{3.52N}}{\sqrt{N}}\right)
\end{align*}
\ARROW For $8 \times 8$ matrix we get:
\begin{align*}
cond_1(A)=3.387\cdot 10^{10},~~~cond_2(A)=1.526\cdot 10^{10},~~~~~cond_{\infty}(A)=3.387\cdot 10^{10}
\end{align*}
\ARROW Clearly large numbers ;)
\end{small}
\end{frame}
\begin{frame}\frametitle{Exact methods: Cramer method}
\begin{small}
\ARROW If $\det{A} \neq 0$ then the solutions are given by:
\begin{align*}
x_i =\frac{\det{A_i}}{\det{A}}
\end{align*}
\ARROW So calculate the solutions one needs to calculate $n+1$ determinants. To calculate each determinate one needs $(n-1)n!$ multiplications. \\
\ARROW Putting it all together one needs $(n+1)(n-1)n! = n^{n+2}$ \\
\ARROW Brute force but works ;)
\end{small}
\end{frame}
\begin{frame}\frametitle{Example, ill-conditioned matrix}
\begin{small}
\ARROW The text-book example of wrongly conditioned matrix is the Hilbert matrix:
\begin{align*}
h_{i,j} = \frac{1}{i+j-1}
\end{align*}
\ARROWR Example:
\begin{align*}
h_{i,j}^{4 \times 4} = \begin{pmatrix}
1 & \frac{1}{2} & \frac{1}{3} & \frac{1}{4} \\
\frac{1}{2} & \frac{1}{3} & \frac{1}{4} & \frac{1}{5} \\
\frac{1}{3} & \frac{1}{4} & \frac{1}{5} & \frac{1}{6}\\
\frac{1}{4} & \frac{1}{5} & \frac{1}{6} & \frac{1}{7}
\end{pmatrix}
\end{align*}
\ARROW The condition of this matrix:
\begin{align*}
cond(A)=\mathcal{O}\left(\frac{e^{3.52N}}{\sqrt{N}}\right)
\end{align*}
\ARROW For $8 \times 8$ matrix we get:
\begin{align*}
cond_1(A)=3.387\cdot 10^{10},~~~cond_2(A)=1.526\cdot 10^{10},~~~~~cond_{\infty}(A)=3.387\cdot 10^{10}
\end{align*}
\ARROW Clearly large numbers ;)
\end{small}
\end{frame}
\begin{frame}\frametitle{Exact methods: Cramer method}
\begin{small}
\ARROW If $\det{A} \neq 0$ then the solutions are given by:
\begin{align*}
x_i =\frac{\det{A_i}}{\det{A}}
\end{align*}
\ARROW So calculate the solutions one needs to calculate $n+1$ determinants. To calculate each determinate one needs $(n-1)n!$ multiplications. \\
\ARROW Putting it all together one needs $(n+1)(n-1)n! = n^{n+2}$ \\
\ARROW Brutal force but works ;)
\end{small}
\end{frame}
\begin{frame}\frametitle{Exact methods: Gauss method}
\begin{small}
\ARROW The idea besides the Gauss method is simple: transform the $A x =b$ to get the equvalent matrix $A^{\left[n\right]} x = b^{\left[n\right]}$ where $A^{\left[n\right]}$ is triangular matrix:
\begin{align*}
A^{\left[n\right]} = \begin{pmatrix}
a^{\left[n\right]}_{11} & a^{\left[n\right]_{12}} & ... & a^{\left[n\right]}_{1n}\\
0 & a^{\left[n\right]}_{22} & ... & a^{\left[n\right]}_{2n} \\
...\\
0 & 0 & ... & a^{\left[n\right]}_{nn} \\
\end{pmatrix}
\end{align*}
\ARROWR The algorithm:
\ARROW To do so we calculate the: $d^{\left[1\right]}_{i,1}=\frac{a^{\left[1\right]}_{i1}}{a^{\left[1\right]}_{11}}$ \\
\ARROW The first row multiplied by the $d^{\left[1\right]}_{i,1}$ we subtract from the $i^{th}$ row.
\ARROW After this we get:
\begin{align*}
\begin{pmatrix}
a^{\left[ 1 \right]}_{11} & a^{\left[ 1 \right]_{12}} & ... & a^{\left[ 1 \right]}_{1n}\\
0 & a^{\left[ 1 \right]}_{22} & ... & a^{\left[ 1 \right]}_{2n} \\
...\\
0 & a^{\left[ 1 \right]}_{n2} & ... & a^{\left[ 1 \right]}_{nn} \\
\end{pmatrix} \overrightarrow{x}=
\begin{pmatrix}
b^{\left[ 1 \right]}_{1} \\
b^{\left[ 1 \right]}_{1} \\
... \\
b^{\left[ 1 \right]}_{1} \\
\end{pmatrix}
\end{align*}
\end{small}
\end{frame}
\begin{frame}\frametitle{Exact methods: Gauss method 2}
\begin{small}
\ARROW Now one needs to repeat the above n times moving each time row down.
\begin{alertblock}{Cons:}
\ARROW The algoright can be stooped if you divide by zero.\\
\ARROW The method is very efficient to accumulate numerical errors.
\end{alertblock}
\begin{exampleblock}{Pros:}
\ARROWR The number of needed floating point operations is less then Cramer.\\
\ARROWR Example for 15 equations: $1345$ vs $5 \cdot 10^{12}$.
\end{exampleblock}
\end{small}
\end{frame}
\begin{frame}\frametitle{Exact methods: modified Gauss method}
\begin{small}
\ARROW The biggest disadvantage of Gauss method is the fact that we can have zero elements :( \\
\ARROW The modified method fixes this problem :)\\
\ARROW The modification is as follows:
\begin{itemize}
\item In each step before we do the elimination we look for main element:
\begin{align*}
\vert a_{mk}^{\vert k \vert } = \max \lbrace \vert a_{jk}^{\vert k \vert} : j=k, k+1,..., n \rbrace
\end{align*}
\item We exchange the rows $m$ and $k$.
\item We do the standard elimination.
\end{itemize}
\end{small}
\end{frame}
\begin{frame}\frametitle{Exact methods: Jordan elimination method}
\begin{small}
\ARROW The Jordan elimination method is similar to Guass method but the idea is to transform the matrix from $Ax =b$ to $I x = b^{\lfloor n+1 \rfloor}$.
\begin{enumerate}
\item We start from eliminating the as in Gauss method in the first row.
\item When we move to the second row we eliminate the $x_2$ element also from the first raw.
\item The third raw we eliminate the first and second raw.
\item We repeat this $n$ times.
\end{enumerate}
\ARROW After this we will get new system $I x = b^{\lfloor n+1 \rfloor}$. \\
\ARROW The $b^{\lfloor n+1 \rfloor}$ is already the solution! No need to do more.
\end{small}
\end{frame}
\begin{frame}\frametitle{Exact methods: LU method}
\begin{small}
\ARROW The most popular of the exact methods is so-called LU method.\\
\ARROW The idea is very simple; we represent the matrix in a form:
\begin{align*}
A = & \begin{pmatrix}
a_{11} & a_{12} & a_{13} & ... & a_{1n} \\
a_{21} & a_{22} & a_{23} & ... & a_{2n} \\
a_{31} & a_{32} & a_{33} & ... & a_{2n} \\
& & ... & & \\
_{n1} & a_{n2} & a_{n3} & ... & a_{nn} \\
\end{pmatrix} = \\ & \begin{pmatrix}
1 & 0 & 0 & ... & 0 \\
l_{21} & 1 & 0 & ... & 0 \\
l_{31} & l_{32} & 1 & ... &0 & \\
& & ... & & \\
l_{n1} & l_{n2} & l_{n3} & ... & l_{nn}\\
\end{pmatrix} \cdot
\begin{pmatrix}
u_{11} & u{12} & u_{13} & ... & u_{1n} \\
0 & u{22} & u_{23} & ... & u_{2n} \\
0 & 0 & u_{33} & ... & u_{3n} \\
& & ... & & \\
0 & 0 & 0 & ... & u_{nn}
\end{pmatrix} = L \cdot U
\end{align*}
\ARROW After this decomposition we need to solve:
\begin{align*}
\begin{cases}
L y & =b\\
U x & =y
\end{cases}
\end{align*}
\end{small}
\end{frame}
\begin{frame}\frametitle{Exact methods: LU method, algorithms}
\begin{small}
\ARROW We solve the following matrix: $A^{\lfloor 1 \rfloor} x = b^{\lfloor1 \rfloor}$\\
\ARROW We start by preparing the matrix $L^{\lfloor 1 \rfloor}$:
\begin{align*}
L^{\lfloor 1 \rfloor }=\begin{pmatrix}
1 & 0 & 0 & ... & 0 \\
-l_{21} & 1 & 0 & ... & 0\\
-l_{31} & 0 & 1 & ... & 0\\
& & ... & & \\
-l_{n1} & 0 & 0 & ... & 1
\end{pmatrix}
\end{align*}
where the $l_{i1}$ are defined:
\begin{align*}
l_{i1} = \frac{a_{i1}^{\lfloor 1 \rfloor}}{a_{11}^{\lfloor 1 \rfloor}}
\end{align*}
\ARROW Now we take our base equation by $L^{\lfloor 1 \rfloor }$:
\begin{align*}
L^{\lfloor 1 \rfloor } A^{\lfloor 1 \rfloor } x = L^{\lfloor 1 \rfloor } b^{\lfloor1 \rfloor}
\end{align*}
\ARROW We get the a new system:
\begin{align*}
A^{\lfloor 2 \rfloor} x = b^{\lfloor2 \rfloor}~~~~~~~~\Leftrightarrow~~~~~~~~~~L^{\lfloor 1 \rfloor } A^{\lfloor 1 \rfloor } x = L^{\lfloor 1 \rfloor } b^{\lfloor1 \rfloor}
\end{align*}
\end{small}
\end{frame}
\begin{frame}\frametitle{Exact methods: LU method, algorithms}
\begin{small}
\ARROW In the second step we construct the $L^{\lfloor 2 \rfloor}$ in the form:
\begin{align*}
L^{\lfloor 2 \rfloor}=\begin{pmatrix}
1 & 0 & 0 & ... & 0 \\
0 & 1 & 0 & ... & 0\\
0 & -l_{32} & 1 & ... & 0\\
& & ... & & \\
0 & -l_{n2} & 0 & ... & 1
\end{pmatrix}
\end{align*}
where :
\begin{align*}
l_{i2} = \frac{a_{i2}^{\lfloor 2 \rfloor}}{a_{22}^{\lfloor 2 \rfloor}}
\end{align*}
\ARROW Now we take the $A^{\lfloor 2 \rfloor} x = b^{\lfloor2 \rfloor}$ and multiply it by $L^{\lfloor 2 \rfloor}$:
\begin{align*}
L^{\lfloor 2 \rfloor } A^{\lfloor 2 \rfloor } x = L^{\lfloor 2 \rfloor } b^{\lfloor2 \rfloor}
\end{align*}
\ARROW We get the a new system:
\begin{align*}
A^{\lfloor 3 \rfloor} x = b^{\lfloor3 \rfloor}~~~~~~~~\Leftrightarrow~~~~~~~~~L^{\lfloor 2 \rfloor } L^{\lfloor 1 \rfloor } A^{\lfloor 1 \rfloor } x = L^{\lfloor 2 \rfloor } L^{\lfloor 1 \rfloor } b^{\lfloor1 \rfloor}
\end{align*}
\end{small}
\end{frame}
\begin{frame}\frametitle{Exact methods: LU method, algorithms}
\begin{small}
\ARROW Now the we repeat the above steps $n-1$ times after which we get:
\begin{align*}
& L^{\lfloor n-1 \rfloor } L^{\lfloor n-2 \rfloor } ... L^{\lfloor 2 \rfloor } L^{\lfloor 1 \rfloor } A^{\lfloor 1 \rfloor } = A^{\lfloor n \rfloor } = U\\
& L^{\lfloor n-1 \rfloor } L^{\lfloor n-2 \rfloor } ... L^{\lfloor 2 \rfloor } L^{\lfloor 1 \rfloor } b^{\lfloor1 \rfloor} = b^{\lfloor n \rfloor}
\end{align*}
\ARROW From the above we can calculate:
\begin{align*}
A^{\lfloor 1 \rfloor } = \left( L^{\lfloor 1 \rfloor } \right)^{-1} \left( L^{\lfloor 2 \rfloor } \right)^{-1} ... \left( L^{\lfloor n-1 \rfloor } \right)^{-1} A^{\lfloor n \rfloor }
\end{align*}
\ARROW So the matrix $L$ we search for is:
\begin{align*}
L= \left( L^{\lfloor 1 \rfloor } \right)^{-1} \left( L^{\lfloor 2 \rfloor } \right)^{-1} ... \left( L^{\lfloor n-1 \rfloor } \right)^{-1}
\end{align*}
\ARROW The $\left( L^{\lfloor k \rfloor } \right)^{-1}$ can be easy calculated:
\begin{footnotesize}
\begin{align*}
\left( L^{\lfloor k \rfloor } \right)^{-1} = \begin{pmatrix}
1 & 0 & ... & 0 & ... & 0 \\
0 & 1 & ... & 0 & ... & 0 \\
& & & ... & & \\
0 & 0 & ... & 1 & ... & 0 \\
0 & 0 & ... & l_{k+1 k} & ... & 0 \\
0 & 0 & ... & l_{k+2 k} & ... & 0 \\
& & & ... & & \\
0 & 0 & ... & l_{n k} & ... & 1 \\
\end{pmatrix}
\end{align*}
\end{footnotesize}
\end{small}
\end{frame}
\begin{frame}\frametitle{Exact methods: LU method, algorithms}
\begin{small}
\ARROW Now the only thing left is to solve the simple linear system:
\begin{align*}
\begin{cases}
L y & =b\\
U x & =y
\end{cases}
\end{align*}
\ARROW Because of the triangular matrix the solution is straightforward:
\begin{align*}
y_1=\frac{b_1}{L_11},~~~~~~~~~y_i=\frac{b_1 - \sum_{j=1^{i-1}} L_{ij} y_j }{L_{ii}}~~{ i\geq 2}
\end{align*}
\end{small}
\end{frame}
\begin{frame}\frametitle{Summary}
\begin{small}
\ARROW This lecture we learn the exact methods of solving linear equation system.\\
\ARROW The three most popular one are: Gauss, Jordan, LU.\\
\ARROW By default the LU method should be used.\\
\ARROW And remember: be sure the system is well conditioned!
\end{small}
\end{frame}
\backupbegin
\begin{frame}\frametitle{Backup}
\end{frame}
\backupend
\end{document}
mchrzasz