\section{Matrices} \definition{} A \textit{matrix} is a two-dimensional array of numbers: \\ $$ A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} $$ The above matrix has two rows and three columns. It is thus a $2 \times 3$ matrix. \problem{} Draw a $3 \times 2$ matrix. \vfill \definition{} We can define the product of a matrix $A$ and a vector $v$: $$ Av = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} \begin{bmatrix} a \\ b \\ c \end{bmatrix} = \begin{bmatrix} 1a + 2b + 3c \\ 4a + 5b + 6c \end{bmatrix} $$ Each element of the resulting $2 \times 1$ matrix is the dot product of a row of $A$ with $v$: $$ Av = \begin{bmatrix} \text{---} a_1 \text{---} \\ \text{---} a_2 \text{---} \end{bmatrix} \begin{bmatrix} | \\ v \\ | \\ \end{bmatrix} = \begin{bmatrix} r_1v \\ r_2v \end{bmatrix} $$ Naturally, a vector can only be multiplied by a matrix if the number of rows in the vector equals the number of columns in the matrix. \problem{} Compute the following: $$ \begin{bmatrix} 1 & 2 \\ 3 & 4 \\ 5 & 6 \end{bmatrix} \begin{bmatrix} 5 \\ 3 \end{bmatrix} $$ \vfill \pagebreak \generic{Remark:} It is a bit more interesting to think of matrix-vector multiplication in the following way: \\ \begin{minipage}[t]{0.48\textwidth}\vspace{0pt} \begin{center} The problem: \vspace{2mm} $$ \begin{bmatrix} 1 & 2 \\ 3 & 4 \\ 5 & 6 \end{bmatrix} \begin{bmatrix} 5 \\ 3 \end{bmatrix} = \begin{bmatrix} 11 \\ 27 \\ 43 \end{bmatrix} $$ \end{center} \end{minipage}% \hfill \begin{minipage}[t]{0.48\textwidth}\vspace{0pt} \begin{center} Top-input, right-output: \vspace{2mm} \begin{tikzpicture}[>=stealth,thick,baseline] \matrix [ matrix of math nodes, left delimiter={[}, right delimiter={]} ] (A) { 1 & 2 \\ 3 & 4 \\ 5 & 6 \\ }; \node[ fit=(A-1-1)(A-1-1), inner xsep=0mm,inner ysep=3mm, label=above:5 ] (L) {}; \draw[->, gray] (L.north) -- ([yshift=0mm]A-1-1.north); \node[ fit=(A-1-2)(A-1-2), inner xsep=0mm,inner ysep=3mm, label=above:3 ] (R) {}; \draw[->, gray] (R.north) -- ([yshift=0mm]A-1-2.north); \node[ fit=(A-1-2)(A-1-2), inner xsep=8mm,inner ysep=0mm, label=right:{$5 + 6 = 11$} ](Y) {}; \draw[->, gray] ([xshift=3mm]A-1-2.east) -- (Y); \node[ fit=(A-2-2)(A-2-2), inner xsep=8mm,inner ysep=0mm, label=right:{$15 + 12 = 27$} ](H) {}; \draw[->, gray] ([xshift=3mm]A-2-2.east) -- (H); \node[ fit=(A-3-2)(A-3-2), inner xsep=8mm,inner ysep=0mm, label=right:{$25 + 18 = 43$} ](N) {}; \draw[->, gray] ([xshift=3mm]A-3-2.east) -- (N); \end{tikzpicture} \end{center} \end{minipage}% \vspace{2mm} Be aware that this is only a model for intuition. \\ Make sure you understand the dot product definition on the previous page. \vspace{5mm} \theorem{} Any linear map $T: \mathbb{R}^n \to \mathbb{R}^m$ can be written as an $n \times m$ matrix. \\ Conversely, every $n \times m$ matrix represents a liner map $T: \mathbb{R}^n \to \mathbb{R}^m$ \\ \vspace{2mm} In other words, \textbf{matrices are linear transformations}. \\ If you only learn only one thing today, this should be it. \vfill \problem{} Show that the transformation $T: \mathbb{R}^n \to \mathbb{R}^m$ defined by $T(v) = Av$ is linear. \\ Before you start, answer the following questions: \begin{itemize} \item What is $A$? \item What is $v$? \item What are their sizes? \end{itemize} \vfill \problem{} Show that any linear transformation can be written as a matrix. \vfill \problem{} Consider the transformation $D: \mathbb{P}^3 \to \mathbb{P}^2$ defined by $D(p) = \frac{d}{dx}p$. \\ Find a matrix that corresponds to $D$. \\ \hint{$\mathbb{P}^3$ and $\mathbb{R}^4$ are isomorphic. How solutions?} \vfill \pagebreak \problem{} Does \ref{thebigtheorem} hold in arbitrary vector spaces? \\ Repeat \ref{prooffwd} and \ref{proofback} using only axioms, without assuming that we're working in $\mathbb{R}^n$. \vfill \pagebreak