Finished LA 101 handout
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@ -115,7 +115,7 @@ AB =
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$$
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\begin{center}
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\begin{tikzpicture}[>=stealth,thick,baseline]
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\begin{tikzpicture}
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\begin{scope}[layer = nodes]
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\matrix[
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@ -148,16 +148,16 @@ $$
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};
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\end{scope}
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\draw[rounded corners,fill=black!30!white,draw=none] ([xshift=-2mm,yshift=3mm]A-1-1) rectangle ([xshift=2mm,yshift=-3mm]A-2-1) {};
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\draw[rounded corners,fill=black!30!white,draw=none] ([xshift=-2mm,yshift=2mm]A-1-1) rectangle ([xshift=2mm,yshift=-2mm]A-1-2) {};
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\draw[rounded corners,fill=black!30!white,draw=none] ([xshift=-3mm,yshift=2mm]B-1-1) rectangle ([xshift=3mm,yshift=-2mm]B-1-2) {};
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\draw[rounded corners,fill=black!30!white,draw=none] ([xshift=-3mm,yshift=2mm]B-1-1) rectangle ([xshift=3mm,yshift=-2mm]B-2-1) {};
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\draw[rounded corners,fill=black!30!white,draw=none] ([xshift=-4mm,yshift=2mm]C-1-1) rectangle ([xshift=4mm,yshift=-2mm]C-1-1) {};
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\draw[rounded corners] ([xshift=-2mm,yshift=3mm]A-1-2) rectangle ([xshift=2mm,yshift=-3mm]A-2-2) {};
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\draw[rounded corners] ([xshift=-2mm,yshift=2mm]A-2-1) rectangle ([xshift=2mm,yshift=-2mm]A-2-2) {};
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\draw[rounded corners] ([xshift=-3mm,yshift=2mm]B-2-1) rectangle ([xshift=3mm,yshift=-2mm]B-2-2) {};
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\draw[rounded corners] ([xshift=-3mm,yshift=2mm]B-1-2) rectangle ([xshift=3mm,yshift=-2mm]B-2-2) {};
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\draw[rounded corners] ([xshift=-4mm,yshift=2mm]C-2-2) rectangle ([xshift=4mm,yshift=-2mm]C-2-2) {};
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\end{tikzpicture}
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@ -183,8 +183,7 @@ $$
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\problem{}
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Consider the following matrix product. \\
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Compute it or explain why you can't.
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Compute the following matrix product or explain why you can't.
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$$
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\begin{bmatrix}
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@ -207,13 +206,76 @@ If $A$ is an $m \times n$ matrix and $B$ is a $p \times q$ matrix, when does the
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\vfill
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\pagebreak
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\problem{}
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Is matrix multiplication commutative? \\
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\note{Does $AB = BA$ for all $A, B$? \\ You only need one counterexample to show this is false.}
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\vfill
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\definition{}
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Say we have a matrix $A$. The matrix $A^T$, pronounced \say{A-transpose}, is created by turning rows of $A$ into columns, and columns into rows:
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$$
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\begin{bmatrix}
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1 & 2 & 3 \\
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4 & 5 & 6
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\end{bmatrix} ^ T
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=
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\begin{bmatrix}
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1 & 4 \\
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2 & 5 \\
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3 & 6
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\end{bmatrix}
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$$
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\problem{}
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Look back to \ref{matvec}. \\
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Convince yourself that vectors are matrices. \\
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Compute the following:
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\hfill
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$
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\begin{bmatrix}
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a & b \\
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c & d
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\end{bmatrix} ^ T
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$\hfill
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$
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\begin{bmatrix}
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1 \\
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3 \\
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3 \\
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7 \\
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\end{bmatrix} ^ T
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$\hfill
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$
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\begin{bmatrix}
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1 & 2 & 4 & 8 \\
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\end{bmatrix} ^ T
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$
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\hfill~
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\vfill
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\pagebreak
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The \say{transpose} operator is often used to write column vectors compactly. \\
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Vertical arrays don't look good in horizontal text.
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\problem{}
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Consider the vectors $a = [1, 2, 3]^T$ and $b = [40, 50, 60]^T$ \\
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\begin{itemize}
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\item Compute the dot product $ab$.
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\item Can you redefine the dot product using matrix multiplication?
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\end{itemize}
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\note{As you may have noticed, a vector is a special case of a matrix.}
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\vfill
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\problem{}
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A \textit{column vector} is an $m \times 1$ matrix. \\
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A \textit{row vector} is a $1 \times m$ matrix. \\
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We usually use column vectors. Why? \\
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\hint{How does vector-matrix multiplication work?}
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Can you multiply a matrix by a vector, as in $vA$? \\
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How does the dot prouduct relate to matrix multiplication? (transpose)
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\vfill
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\pagebreak
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