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\noindent
\parbox{2.25in}{\bf Math 5230}
\hfill {\Large \bf Homework \#1} \hfill
\parbox{2.25in}{\bf \hfill Due: Wed., Aug. 28${}^\mathbf{th}$\!, 2024}

\vspace{0.2in}

%\noindent {\bf\large Name:} \quad \underline{\large\qquad\sc\color{red}MY NAME GOES HERE\qquad}
%
%\vspace{0.05in}

RREF Handout: \href{https://billcookmath.com/courses/math5230-fall2016/math5230-RREF.pdf}{https://billcookmath.com/courses/math5230-fall2016/math5230-RREF.pdf}

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\begin{description}
\item[\#1 RREF] \quad First, (1) use Gaussian elimination (as defined on the handout) to compute the RREF of $A$. Then, (2) give a vector version of the general solution of $A\B x = \B 0$. Finally, (3) suppose $A=[C\; : \; \B b]$ ($C$ is all but the last column and $\B b$ is the last column of $A$). Give a vector form of the general solution of $C\B x = \B b$. 

$$\mbox{(a)} \ \  A=\begin{bmatrix} 1 & -2 & 0 & 2 & 3 \\ 2 & -4 & 1 & 5 & 4 \\ -1 & 2 & 1 & -1 & -5 \end{bmatrix} \qquad \qquad \qquad
    \mbox{(b)} \ \  A=\begin{bmatrix} 0 & 1 & 3 & 1 \\ 2 & 0 & 4 & 2 \\ 1 & -1 & -1 & 3 \end{bmatrix}$$

%\begin{enumerate}[(a)]
%\item $A=\begin{bmatrix} 1 & -2 & 0 & 2 & 3 \\ 2 & -4 & 1 & 5 & 4 \\ -1 & 2 & 1 & -1 & -5 \end{bmatrix}$
%\item $A=\begin{bmatrix} 0 & 1 & 3 & 1 \\ 2 & 0 & 4 & 2 \\ 1 & -1 & -1 & 3 \end{bmatrix}$
%\end{enumerate}

\item[\#2 PLU and Partial Pivoting] \quad First, (1) use partial pivoting to row reduce $A$ (to REF not RREF), and compute a PLU decomposition for $A$. Then (2) use this decomposition to solve $A\B x = \B b$.

$$\mbox{(a)} \ \  A= \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \mbox{\   and \   } \B b=\begin{bmatrix} 5 \\ 6 \end{bmatrix} \qquad \qquad \qquad
    \mbox{(b)} \ \  A=\begin{bmatrix} 1 & 4 & 1 & 0 \\ 2 & 3 & 0 & 1 \\ 3 & 2 & 1 & 1 \\ 4 & 1 & 0 & 1 \end{bmatrix} \mbox{\   and \   } \B b= \begin{bmatrix} 2 \\ 1 \\ 1 \\ 3 \end{bmatrix}$$

%\begin{enumerate}[(a)]
%\item $A= \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$ and $\B b=\begin{bmatrix} 5 \\ 6 \end{bmatrix}$
%\item $A=\begin{bmatrix} 1 & 4 & 1 & 0 \\ 2 & 3 & 0 & 1 \\ 3 & 2 & 1 & 1 \\ 4 & 1 & 0 & 1 \end{bmatrix}$ and $\B b= \begin{bmatrix} 2 \\ 1 \\ 1 \\ 3 \end{bmatrix}$
%\end{enumerate}


\item[\#3 Matrix Mechanics] \quad First, (1) give a formula for a matrix which performs the given list of row operations. For example: Swapping rows 1 and 2 is $E=I-E_{11}-E_{22}+E_{12}+E_{21}$.  Then, (2) give a formula for its inverse (for my example $E^{-1}=E=I-E_{11}-E_{22}+E_{12}+E_{21}$). Next, (3) what happens if you multiply a matrix by your $E$ on the right? (my example swaps columns 1 and 2). Finally, (4) give a concrete example to show off this behavior (for my example I could take some $2 \times 2$ matrix and show how $E$ swaps its rows and columns when multiplying on the left or right). 

\begin{enumerate}[(a)]
   \item Find a matrix $E$ which swaps rows 3 and 4 and then scales row 2 by 5.
   \item Find a matrix $E$ which swaps rows 1 and 4 and then adds -2 times row 2 to row 3.
\end{enumerate}

\item[\#4 More Matrix Mechanics] \quad Let $B=I-E_{11}+3E_{22}-E_{33}-E_{99}+E_{39}+E_{93}$. What does $BA$ do to $A$? What does $AB$ do to $A$? (Assume that $A$ and $B$ are appropriately sized.


%\item[\#5 Matrix Mechanics 3] \quad Let $\mathbb{R}^{n \times n}$ be the collection of all square $n \times n$ matrices with real entries. We say that $A \in \mathbb{R}^{n \times n}$ is in the {\bf center} of $\mathbb{R}^{n \times n}$ if $AB=BA$ for all $B \in \mathbb{R}^{n \times n}$. In other words, central matrices commute with all matrices of the same size. 
%
%Show that the $A$ is in the center if and only if $A=cI$ for some $c \in \mathbb{R}$ (i.e. the central matrices are exactly the scalar multiples of the identity). Please do this using $E_{ij}$'s.

\item[\#5 \sout{Poof!} Proof!] First, let's review some notation and terminology\dots

Let $A$ be an $m \times n$ matrix with $(i,j)$-entry denoted $a_{ij}$. Likewise, let $B$ be a $n \times \ell$ matrix with $(i,j)$-entry denoted $b_{ij}$. Recall that $AB$ is an $m \times \ell$ matrix whose $(i,j)$-entry is given by $$\displaystyle\sum\limits_{k=1}^n a_{ik}b_{kj}=\begin{bmatrix} a_{i1} & a_{i2} & \cdots & a_{in} \end{bmatrix} \begin{bmatrix} b_{1j} \\ b_{2j} \\ \vdots \\ b_{nj} \end{bmatrix} = a_{i1}b_{1j}+a_{i2}b_{2j}+\cdots+a_{in}b_{nj}$$ (i.e., the $i^\text{th}$ row of $A$ dotted with the $j^\text{th}$ column of $B$). 

We say that a matrix is {\it lower triangular} if all of its entries below the main diagonal are zero. In other words, $A$ is lower triangular if $a_{ij}=0$ whenever $i<j$ (down more rows than over in terms columns). Thus the matrix $A$ is lower triangular if $$A = \begin{bmatrix} a_{11} & 0 & 0 & \cdots & 0 & \cdots & 0 \\ a_{21} & a_{22} & 0 & \cdots & 0 & \cdots & 0 \\ \vdots & \vdots & \ddots & \ddots & \vdots & & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mm} & 0 & \cdots & 0 \end{bmatrix}$$
It is {\it strictly lower triangular} if $a_{ij}=0$ whenever $i \leq j$ (i.e., lower triangular with zeros on the diagonal as well).

{\bf Problem:} Let $A$ and $B$ be $n \times n$ (square) lower triangular matrices. Prove that $AB$ is lower triangular. 

\end{description}

% Additional IDEAS: 
% Solve the same Ax=b using regular, partial, full. Also, find PLU decomp doing regular, partial, full.

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