MAT223 Lecture 14

• 1

  • For any two vectors in ${\mathbb{R}}^3$
    • $|\overrightarrow{v}\times \overrightarrow{w}|=\left|\overrightarrow{v}\right|\left|\overrightarrow{w}\right|\sin \theta$
  • Draw two non-parallel vectors with a common tail.
  • $\sin \theta$ will give us $\left|A^{\prime }B\right|$
  • $\sin \theta =\frac{\left|A^{\prime }B\right|}{\left|W\right|}$
  • $|v\times w|=\left|v\right|\left|w\right|\left(\frac{\left|A^{\prime }B\right|}{\left|w\right|}\right)=\left|v\right|\left|A^{\prime }B\right|$
  • Since $\left|A^{\prime }B\right|=\left|w\right|\sin \theta$
  • We get $|v\times w|=\left|v\right|\left|w\right|\sin \theta$
  • The area of the parallelogram formed by $v$ and $w$ is $\left|v\right|\left|A^{\prime }B\right|$

• 3

  • linear transformation
  • Zero transformation is allowed.

• 4

  • $T(\overrightarrow{x})=\left[\begin{array}{c}2x-3z\\ x+2y-z\end{array}\right]$
  • $T(\overrightarrow{x})=A\overrightarrow{x}$
  • We can infer:
    • $A\overrightarrow{x}=\left[\begin{array}{ccc}2& -3& 0\\ 1& 2& -1\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$

• 5

  • $T(\overrightarrow{x})=\left[\begin{array}{c}4x+y\\ x+4y\end{array}\right]$
  • Compute $T({\overrightarrow{e}}_1)$ and $T({\overrightarrow{e}}_2)$
  • $T\left[\begin{array}{c}1\\ 0\end{array}\right]=\left[\begin{array}{c}4\\ 1\end{array}\right]$
  • $T\left[\begin{array}{c}0\\ 1\end{array}\right]=\left[\begin{array}{c}1\\ 4\end{array}\right]$
  • Showcase the transformation

  \begin{document}
    \begin{tikzpicture}[domain=0:4]
      \draw[very thin,color=gray] (-0.1,-1.1) grid (3.9,3.9);
      \draw[->] (-0.2,0) -- (4.2,0) node[right] {$x$};
      \draw[->] (0,-1.2) -- (0,4.2) node[above] {$y$};
      \draw[color=red] plot (\x,\x) node[right] {$y=x$};
      \draw[color=blue] plot (\x,{4*\x}) node[right] {$y=4x$};
    \end{tikzpicture}
  \end{document}
  

  • Take points in the co-domain, note they will lie on the parallelogram formed by $T({\overrightarrow{e}}_1)$ and $T({\overrightarrow{e}}_2)$
    • $T\left[\begin{array}{c}0\\ \frac{1}{4}\end{array}\right]$
      • $=\left[\begin{array}{c}\frac{1}{4}\\ 1\end{array}\right]$

• 5

  • $a=T\left[\begin{array}{c}1\\ 1\end{array}\right]=\left[\begin{array}{c}2\\ 3\end{array}\right]$
  • $b=T\left[\begin{array}{c}0\\ 1\end{array}\right]=\left[\begin{array}{c}1\\ 5\end{array}\right]$
  • $T\left[\begin{array}{c}2\\ 0\end{array}\right]=?$
  • $a+a=\left[\begin{array}{c}2\\ 2\end{array}\right]$
  • $2a-2b=\left[\begin{array}{c}2\\ 0\end{array}\right]$
  • $T\left[\begin{array}{c}2\\ 0\end{array}\right]=2a-2b=2\left[\begin{array}{c}2\\ 3\end{array}\right]-2\left[\begin{array}{c}1\\ 5\end{array}\right]=\left[\begin{array}{c}2\\ -4\end{array}\right]$

• 6

  • $T(\overrightarrow{0})=\overrightarrow{0}$