ICS/hw3/hw3.tex

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\title{Howework 3 -- Introduction to Computational Science}

\author{Claudio Maggioni}

\begin{document} \maketitle 
\section*{Question 1}
\[i=1 \hspace{1cm} l_1 = \begin{bmatrix}1\\1\\5\\-3\\\end{bmatrix} \hspace{1cm} u_1 = \begin{bmatrix}
2 & 1 & 1 & -2 \\\end{bmatrix}\]
\[A_2 = \begin{bmatrix}
2 & 1 & 1 &-2 \\ 2 & 2 & -2 & -1 \\ 10 & 4 & 23 & -8 \\
-6 & -2 & 4 & 6 \\\end{bmatrix} - \begin{bmatrix}
2 & 1 & 1 & -2 \\ 2 &1& 1 &-2 \\ 10 & 5 & 5 & -10 \\ -6 & -3 & -3 & 6 \\\end{bmatrix} = \begin{bmatrix}\\& 1 & -3 & 1 \\ & -1 & 18 & 2 \\ & 1 & 7 & 0 \end{bmatrix}\] 

\[i = 2 \hspace{1cm} l_2 = \begin{bmatrix}\\1 \\-1 \\1\\\end{bmatrix} \hspace{1cm} u_2 = \begin{bmatrix} & 1& -3& 1\\\end{bmatrix}\]
\[A_3 = \begin{bmatrix}\\& 1 & -3 & 1 \\ & -1 & 18 & 2 \\ & 1 & 7 & 0 \end{bmatrix} - \begin{bmatrix}\\&1 & -3 & 1 \\ & -1 & 3 & -1 \\ & 1 & -3 & 1 \\\end{bmatrix} = \begin{bmatrix}
\\
\\
&&15&3\\&&10&-1\\\end{bmatrix}\]

\[i = 3 \hspace{1cm} l_3 = \begin{bmatrix}\\\\1\\2/3\\\end{bmatrix} \hspace{1cm} u_3 = \begin{bmatrix}&& 15 & 3 \\ \end{bmatrix}\]
\[A_4 = \begin{bmatrix}
\\
\\
&&15&3\\&&10&-1\\\end{bmatrix} - \begin{bmatrix}\\\\&&15 & 3 \\&&10 & 2 \\\end{bmatrix} = \begin{bmatrix}\\\\\\&&&-3\\\end{bmatrix}\]
\[i = 4 \hspace{1cm} l_4 = \begin{bmatrix}\\\\\\1\\\end{bmatrix} \hspace{1cm} u_4 = \begin{bmatrix}&&& -3 \\ \end{bmatrix}\]
\[L = \begin{bmatrix}1\\1&1\\5&-1&1\\-3 &1 &2/3&1\\\end{bmatrix} \hspace{1cm} U = \begin{bmatrix}2 &1 &1&-2\\&1 &-3&1\\&&15&3\\&&&-3\\\end{bmatrix}\]
\[Ly = B \Rightarrow \begin{bmatrix}1\\1&1\\5&-1&1\\-3 &1 &2/3&1\\\end{bmatrix} \begin{bmatrix}y_1\\y_2\\y_3\\y_4\\\end{bmatrix} = \begin{bmatrix}-1\\-3\\36\\18\\\end{bmatrix} 
\]
\[y_1 = -1\]
\[y_2 = -3 - (-1) \cdot 1 = -2\]
\[y_3 = 36 - 1 \cdot 2 - (-1) \cdot 5 = 39\]
\[y_4 = 18 - \frac{2}{3} \cdot 39 - (-2) - (-3) \cdot (-1) = -9\]
\[Ux = Y \Rightarrow \begin{bmatrix}2 &1 &1&-2\\&1 &-3&1\\&&15&3\\&&&-3\\\end{bmatrix} \begin{bmatrix}x_1\\x_2\\x_3\\x_4\\\end{bmatrix} = \begin{bmatrix}-1\\-2\\39\\-9\\\end{bmatrix} 
\]
\[x_4 = 3\]
\[x_3 = \frac{39 - 3 - 3}{15} = 2\]
\[x_2 = \frac{-2 -3 - (-3) \cdot 2}{1} = 1\]
\[x_1 = \frac{ -1 - (-2) \cdot 3 - 1 \cdot 2 - 1 \cdot 1}{2} = 1\]
\[x = \begin{bmatrix}1\\1\\2\\3\\\end{bmatrix}\]

\section*{Question 2}

\[i = 1 \hspace{1cm} k = 4 \hspace{1cm} \begin{bmatrix}4&2&3&1\\\end{bmatrix}\]
\[
l_1 = 
\begin{bmatrix}
       1/8 \\
       1/4 \\
       1/2 \\
       1 \\
\end{bmatrix}
\hspace{1cm}
u_1 = \begin{bmatrix} 32 & 24 & 10 & 11\end{bmatrix}
\hspace{1cm}
\]\[
A_2 = \begin{bmatrix}4 &3 &2& 1\\ 8& 8& 5& 2\\ 16& 12& 10& 5\\ 32& 24& 20 &11 \\\end{bmatrix} - \begin{bmatrix}
       4         &     3 &             5/2 &          11/8  \\   
       8          &    6  &            5    &         11/4    \\ 
      16           &  12   &          10     &        11/2     \\
      32            & 24    &         20      &       11       \\
\end{bmatrix} =
\begin{bmatrix}
       0     &         0  &           -1/2    &       -3/8 \\    
       0      &        2   &           0       &      -3/4   \\  
       0       &       0    &          0        &     -1/2     \\
       0        &      0      &        0         &     0       \\
\end{bmatrix}
\]

\[i = 2 \hspace{1cm} k = 2 \hspace{1cm} p = \begin{bmatrix}4&2&3&1\\\end{bmatrix}\]
\[
l_2 =\begin{bmatrix}
0\\1\\0\\0\\
\end{bmatrix}
\hspace{1cm}
u_2 =\begin{bmatrix} 0  &            2   &           0     &        -3/4 \end{bmatrix}\]
\[    
A_3 =  \begin{bmatrix}
       0     &         0  &           -1/2    &       -3/8 \\    
       0      &        2   &           0       &      -3/4   \\  
       0       &       0    &          0        &     -1/2     \\
       0        &      0      &        0         &     0       \\
\end{bmatrix} - 
\begin{bmatrix}
       0       &       0  &            0        &      0       \\
       0        &      2   &           0         &    -3/4     \\
       0         &     0    &          0          &    0       \\
       0          &    0     &         0           &   0       \\
\end{bmatrix}
 =
\begin{bmatrix}
       0       &       0      &       -1/2     &      -3/8  \\   
       0        &      0       &       0        &      0      \\ 
       0         &     0        &      0         &    -1/2     \\
       0          &    0         &     0          &    0       \\
\end{bmatrix}
\]

\[ i = 3 \hspace{1cm} k = 4 \hspace{1cm} p = \begin{bmatrix}
4&2&1&3\\
\end{bmatrix}
\]
\[
l_3 = \begin{bmatrix}
1 \\0\\0\\0\\
\end{bmatrix}
\hspace{1cm}
u_3 = \begin{bmatrix} 0 & 0& -1/2 & -3/8\\\end{bmatrix}     
\]\[
A_4 = \begin{bmatrix}
       0       &       0      &       -1/2     &      -3/8  \\   
       0        &      0       &       0        &      0      \\ 
       0         &     0        &      0         &    -1/2     \\
       0          &    0         &     0          &    0       \\
\end{bmatrix} -
\begin{bmatrix}
       0       &       0 &            -1/2&           -3/8\\     
       0        &      0  &            0   &           0    \\   
       0         &     0   &           0    &          0      \\ 
       0          &    0    &          0     &         0       \\
\end{bmatrix} =
\begin{bmatrix}
       0       &       0  &            0 &             0 \\      
       0        &      0   &           0  &            0   \\    
       0         &     0    &          0   &          -1/2   \\  
       0          &    0     &         0    &          0       \\
       \end{bmatrix}
       \]

\[ i =4 \hspace{1cm} k = 4 \hspace{1cm} p = \begin{bmatrix}
4&2&1&3\\
\end{bmatrix}\]
\[l_4 = \begin{bmatrix} 0 \\0\\1\\0\\\end{bmatrix}
u_4 = \begin{bmatrix}0&0&0&-1/2\end{bmatrix}
\]

\[
P =
\begin{bmatrix}
       0       &       0  &            0    &          1   \\   
       0        &      1   &           0      &        0    \\  
       1         &     0    &          0     &         0     \\  
       0          &    0     &         1       &       0 \\
\end{bmatrix}
\]
\[
L = P *       
\begin{bmatrix}
 1/8       &     0         &     1       &       0       \\
       1/4  &          1    &          0  &            0   \\    
       1/2   &         0     &         0   &           1     \\  
       1       &       0      &        0    &          0       \\
\end{bmatrix}
=
\begin{bmatrix}
       1       &       0    &          0     &         0   \\    
       1/4      &      1     &         0      &        0     \\  
       1/8       &     0      &        1       &       0       \\
       1/2        &    0       &       0        &      1       \\
       \end{bmatrix}
       \]
       
\[
U =
\begin{bmatrix}
      32      &      24     &        20    &         11      \\ 
       0        &      2     &         0    &         -3/4     \\
       0         &     0      &       -1/2   &        -3/8     \\
       0          &    0       &       0      &       -1/2     \\
\end{bmatrix}
\]

\section*{Question 4}
\[
A_1 = \begin{bmatrix}

       1    &          4   &           8    &          3  \\     
       4     &        20     &        40     &        28  \\
       8      &       40      &       89      &       71  \\
       3       &      28       &      71       &     114  \\
\end{bmatrix}
\]

\[ i = 1 \hspace{1cm} l_1 = \begin{bmatrix}1\\4\\8\\3\\\end{bmatrix}\]

\[
A_2 = \begin{bmatrix}

       1    &          4   &           8    &          3  \\     
       4     &        20     &        40     &        28  \\
       8      &       40      &       89      &       71  \\
       3       &      28       &      71       &     114  \\
\end{bmatrix}
-
\begin{bmatrix}
       1      &        4  &            8   &           3\\
       4       &      16   &          32    &         12\\
       8        &     32    &         64     &        24\\
       3         &    12     &        24      &        9\\
\end{bmatrix}
=
\begin{bmatrix}
\\
& 4 & 8 & 16\\       
& 8 & 25 & 47 \\      
& 16 & 47 & 105 \\
\end{bmatrix}
\]      

\[i=2 \hspace{1cm} l_2 = \begin{bmatrix}\\2\\4\\8\\\end{bmatrix}\]

\[A_3 = 
\begin{bmatrix}
\\
& 4 & 8 & 16\\       
& 8 & 25 & 47 \\      
& 16 & 47 & 105 \\
\end{bmatrix}
-
\begin{bmatrix}
\\       
& 4 & 8 & 16 \\       
& 8 & 16 & 32  \\     
& 16 & 32 & 64   \\    
\end{bmatrix}
=
\begin{bmatrix}\\\\&& 9 & 15 \\&&15 & 41\\\end{bmatrix}
\]

\[i = 3 \hspace{1cm} l_3 = \begin{bmatrix}\\\\3\\5\\\end{bmatrix}\]
\[A_4 = 
\begin{bmatrix}\\\\&& 9 & 15 \\&&15 & 41\\\end{bmatrix}
-
\begin{bmatrix}
\\
\\
&&9&15\\
&& 15 & 25\\       
\end{bmatrix}
=
\begin{bmatrix}
\\\\\\&&&16\\
\end{bmatrix}       
\]

\[i = 4 \hspace{1cm} l_4 = \begin{bmatrix}\\\\\\4\\\end{bmatrix}\]

\[L = \begin{bmatrix}1\\4&2\\8&4&3\\3&8&5&4\\\end{bmatrix}\]

\section*{Question 5}

\subsection*{Point a)}
First of all, to show that $A_{1,1}$ is symmetric, we say:

\[\begin{bmatrix}A_{1,1} & A_{1,2}\\A_{2,1} & A_{2,2}\\\end{bmatrix} = A = A^T = \begin{bmatrix}A_{1,1}^T & A_{2,1}\\A_{1,2} & A_{2,2}^T\\\end{bmatrix}\]

Therefore we can say that

$$A_{1,1} = A_{1,1}^T$$

and thus $A_{1,1}$ is shown to be symmetric.

Then, since $A$ is positive definite, $x^TAx > 0$ for any vector $x \in R^n$. We choose a vector $x$ such that $x_k = 0$ for any $p < k \leq n$.

Then:

$$ x^TAx = \sum_{j=1}^n x_j \cdot \sum_{i=1}^n x_i a_{i,j} = \sum_{j=1}^p x_j \cdot \sum_{i=1}^n x_i a_{i,j} + 
\sum_{j=p+1}^n 0 \cdot \sum_{i=1}^n 0 a_{i,j} = 
\sum_{j=1}^p x_j \cdot (\sum_{i=i}^p x_i a_{i,j} + 
\sum_{i=p+1}^n 0 a_{i,j}) =$$
$$=\sum_{j=1}^p x_j \cdot \sum_{i=i}^p x_i a_{i,j} =
\begin{bmatrix}x_1 & x_2 & \ldots & x_n\end{bmatrix}
A_{1,1}\begin{bmatrix}x_1 & x_2 & \ldots & x_n\end{bmatrix}^T$$

Then:

$$\begin{bmatrix}x_1 & x_2 & \ldots & x_n\end{bmatrix}A_{1,1}\begin{bmatrix}x_1 & x_2 & \ldots & x_n\end{bmatrix}^T>0$$ 

which is the definition of positive definiteness for $A_{1,1}$.

\end{document}