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\noindent
\textsc{\Large CS 4220 / MATH 4260: Homework 2}\\
Instructor: Anil Damle\\
Due: February 20, 2019\\
\subsection*{Policies}
You may discuss the homework problems freely with other students, but please
refrain from looking at their code or writeups (or sharing your own).
Ultimately, you must implement your own code and write up your own solution to
be turned in. Your solution, including plots and requested output from your code
should be typeset and submitted via the CMS as a pdf file. Additionally, please
submit any code written for the assignment via the CMS as well. This can be done
by either including it in your solution as an appendix, or uploading it as a zip
file via the CMS.\\
\hrule
\subsection*{Question 1:}
Let $A$ and $B$ be $n\times n$ matrices with real entries (you may assume the entries are explicitly representable in floating point if you like\textemdash though it does not change the answer in a meaningful way). For any $i$ and $j$ with $1 \leq i,j \leq n$ bound the floating point error of computing $C_{i,j}$ where $C=AB.$ Your bound should be in terms of $n,$ $\mu$ (machine precision), and entries of $A$ and $B.$ Given your solution, when might you expect large error in a relative sense?
\subsection*{Question 2:}
When discussing Gaussian elimination and then $LU$ decompositions in class, we
considered matrices $G^{(k)}$ (so-called Gauss Transforms) that when applied to a matrix $A$ introduced zeros
in the $k^{th}$ column below the diagonal. Assuming all of the matrices involved in
this problem are $n \times n,$ we may write these matrices compactly as
\[
G^{(k)} = I - \ell^{(k)}e_k^T,
\]
where $\ell^{(k)} = \left[0,\ldots,0, \ell_{k+1,k},\ldots \ell_{n,k} \right]^T$
is zero in the first $k$ entries. We will now prove two properties of these
matrices outlined in class.
\begin{enumerate}[label=(\alph*)]
\item Prove that $\left(G^{(k)}\right)^{-1} = I + \ell^{(k)}e_k^T$
\item Prove that $\displaystyle \left(G^{(1)}\right)^{-1}\left(G^{(2)}\right)^{-1}
\cdots\left(G^{(n-1)}\right)^{-1} = I + \sum_{k=1}^{n-1} \ell^{(k)}e_k^T$
\end{enumerate}
\subsection*{Question 3:}
Assume that you are given an $n \times n$ non-singular matrix of the form
\[
A = D + ue_{n-1}^T + e_{n-1}v^T
\]
where $u$ and $v$ are length $n$ vectors, and $D$ is a diagonal matrix whose
diagonal entries we will encode in the vector $d,$ \emph{i.e.} $D_{i,i} = d_i.$
(In the course of thinking about this problem, you may stumble across the so-called
Sherman-Morrison-Woodbury formula and find it provides a potential solution.
You are welcome to try it out and see what happens, however it will not yield an
accurate solution for the given instance of this problem in part (c).)
\begin{enumerate}[label=(\alph*)]
\item Under the assumption that we may compute the $LU$ decomposition of $A$ without
requiring any pivoting, devise a means to solve a linear system of the form $Ax=b$
using $\mathcal{O}(n)$ time. For the purposes of this problem we will also consider
the cost of memory allocation and therefore you cannot form $A$ explicitly as that
would take $\mathcal{O}(n^2)$ time.
\item Implement your method and demonstrate that it achieves the desired scaling
\item Download the files on the course website containing vectors $d,v,b$ and the
true solution $x_{\text{true}}$. Solve the associated linear system $Ax=b$ for $x.$ Using a logarithmic scale for the error plot both
the absolute error of the difference between your computed solution and
$x_{\text{true}}$ and the absolute value of the residual vector $r=b-Ax$
generated by your solution. What do you observe? (If you are using Julia rather
than Matlab, there is a package you can use to read the input file available
at: \mbox{\url{https://github.com/JuliaIO/MAT.jl}}. Similarly, both NumPy and SciPy provide means to read the input file.)
\item By avoiding pivoting, we were able to get a fast, $\mathcal{O}(n)$ algorithm.
However, what do you observe about the accuracy of the solution and how might
you explain it?
\item If we introduce partial pivoting into our algorithm can we guarantee that
the algorithm will still have $\mathcal{O}(n)$ complexity? If not, what about
$\mathcal{O}(n^2)$ complexity?
\item \textbf{Ungraded, but interesting to try:} Sometimes, though not always,
we may be able to pair the faster, albeit problematic, algorithm with an iterative
method (a class of algorithms we will talk more about later in this course) to
clean up the solution. One simple methodology is called iterative refinement.
Specifically, given an approximate solution $\tilde{x}$ such that
$A\tilde{x}\approx b,$ we can refine the solution via the following procedure:
\begin{enumerate}
\item Compute $r = b - A\tilde{x}$
\item Solve the linear system $Az = r$
\item Let $x = \tilde{x}+z$
\end{enumerate}
If we solve $Az = r$ exactly, then $x$ will be a solution. However, in general
since that is also a problem involving $A$ the solution for $z$ is only approximate
and it requires a more careful analysis to determine how much we may improve the
solution $\tilde{x}.$ In fact, it could even be repeated multiple times. If you
are so inclined, try this out for this problem and see if it helps.
\end{enumerate}
\subsection*{Question 4:}
Consider the following $n \times n$ matrix
\[
A = \begin{bmatrix}
1 & 0 & 0 & \cdots & 0 & 1 \\
-1 & 1 & 0 & \cdots & 0 & 1 \\
\vdots & \ddots & \ddots & \ddots & \vdots & \vdots \\
-1 & \cdots & -1 & 1 & 0 & 1 \\
-1 & \cdots & -1 & -1 & 1 & 1 \\
-1 & \cdots & -1 & -1 & -1 & 1 \\
\end{bmatrix}
\]
that has all ones one the diagonal and in the last column, and all the entries
below the diagonal are $-1.$
\begin{enumerate}[label=(\alph*)]
\item Consider computing an $LU$ decomposition of $A$ (with or without partial
pivoting, the answer will be the same either way). Work out an expression for the largest
entry of $U$ in terms of $n$.
\item If we computed an $LU$ factorization in practice and encountered an entry
of this size do you think it would be problematic? why or why not?
\end{enumerate}
Note that this matrix demonstrates the worst case behavior for the size of elements
in $U$ when computing an $LU$ factorization with partial pivoting. Nevertheless,
such cases are considered rare in practice and $LU$ with partial pivoting is
widely used.
\end{document}