MAT223 Lecture 16

• 1

  • Let $A$ be an $n\times n$
  • Which means it's diagonalizable?
  • $A=PD{P}^{-1}$

• 2

  • $A$ has rank $1$
  • $A$ is $3\times 3$
  • $c_A(x)=x^2(x-1)$
  • So it's diagonalizable
  • This is because $\mathrm{rank}\left(A\right)=1$ meaning we have $2$ parameters. Meaning we have $A\overrightarrow{x}=\overrightarrow{0}$ has 2 solutions.
  • $(A-0I_n)\overrightarrow{x}=\overrightarrow{0}$ is the same as $A\overrightarrow{x}=\overrightarrow{0}$ so we have 2 eigenvectors for the eigenvalue $0$.
  • So that's why $\lambda =0$ has $2$ eigenvectors.

• 3

  • $T_A:{\mathbb{R}}^2\to {\mathbb{R}}^2$
  • $A=\left[\begin{array}{cc}1& -2\\ 1& 4\end{array}\right]$
  • $A$ has $\lambda =2$ and $\mu =3$
  • $\lambda =2$ the basic eigenvector is:
    • $A\overrightarrow{v}=\lambda \overrightarrow{v}$
    • $\left[\begin{array}{cc}1& -2\\ 1& 4\end{array}\right]\left[\begin{array}{c}?\\ 1\end{array}\right]=2\left[\begin{array}{c}?\\ 1\end{array}\right]$
    • $\left[\begin{array}{c}?-2\\ ?+4\end{array}\right]=2\left[\begin{array}{c}?\\ 1\end{array}\right]$
    • $a-2=2a$
    • $a=-2$
    • $a+4=2a$
    • $a=4$
  • $T(\overrightarrow{x})=A_T\overrightarrow{x}$
  • ???
  • Solve $(A-2I_n)\overrightarrow{x}=\overrightarrow{0}$
    • $\left[\begin{array}{ccc}-1& -2& 0\\ 1& 2& 0\end{array}\right]$
    • $x=-2y$
    • $y=t$
    • $\dots$

• 4

  • $T_A:{\mathbb{R}}^2\to {\mathbb{R}}^2$
  • $A=\left[\begin{array}{cc}1& -2\\ 1& 4\end{array}\right]$
  • $\lambda =2$
    • $\overrightarrow{v}=\left[\begin{array}{c}-2\\ 1\end{array}\right]$
    • $T(\overrightarrow{v})=A_T\overrightarrow{v}$
    • $T(\overrightarrow{v})=\left[\begin{array}{cc}1& -2\\ 1& 4\end{array}\right]\left[\begin{array}{c}-2\\ 1\end{array}\right]$
    • $T(\overrightarrow{v})=⟨-4,2⟩$
  • $\mu =3$
    • $\overrightarrow{w}=\left[\begin{array}{c}-1\\ 1\end{array}\right]$
    • $T(\overrightarrow{w})=A_T\overrightarrow{w}$
    • $T(\overrightarrow{w})=\left[\begin{array}{cc}1& -2\\ 1& 4\end{array}\right]\left[\begin{array}{c}-1\\ 1\end{array}\right]$
    • $T(\overrightarrow{w})=⟨-3,3⟩$

• 5

  • $T_A:{\mathbb{R}}^2\to {\mathbb{R}}^2$

  • $A=\left[\begin{array}{cc}1& -2\\ 1& 4\end{array}\right]$

    left=-5; right=5;
    top=5; bottom=-5;
    ---
    (-2,1)
    (-1,1)
    (-3,3)
    (-4,2)
    

• Activity

  • $A$ is diagonalizable if $A=PD{P}^{-1}$

  • Reflection across $y=0$

  • $T:{\mathbb{R}}^2\to {\mathbb{R}}^2$

  • This is reflection across x axis.

  • $T({\overrightarrow{e}}_1)$ and $T({\overrightarrow{e}}_2)$ are?

    • $T({\overrightarrow{e}}_1)={\overrightarrow{e}}_1$
      • $x$ is still $x$
    • $T({\overrightarrow{e}}_2)=-{\overrightarrow{e}}_2$
      • $y$ is now $-y$

  • $A_T=\left[\begin{array}{cc}1& 0\\ 0& -1\end{array}\right]$

  • Plot everything

    left=-5; right=5;
    top=5; bottom=-5;
    ---
    (1,0)
    (0,1)
    (1,0)
    (0,-1)
    

  • $A_T=\left[\begin{array}{cc}1& 0\\ 0& -1\end{array}\right]$

  • it's already diagonal. So $P=I_n$ and $D=A_T$.

  • We know $\lambda =1,-1$

  • We know the basic eigenvector for $\lambda =1$ is ${\overrightarrow{e}}_1$ and the basic eigenvector for $\lambda =-1$ is ${\overrightarrow{e}}_2$.

• Reflection across $y=x$

  • $T:{\mathbb{R}}^2\to {\mathbb{R}}^2$
  • $T\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}y\\ x\end{array}\right]$
  • $T({\overrightarrow{e}}_1)={\overrightarrow{e}}_2$
  • $T({\overrightarrow{e}}_2)={\overrightarrow{e}}_1$
  • $T(\overrightarrow{x})=A_T\overrightarrow{x}$
  • $T(\overrightarrow{x})=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$
  • Notice $A_T=I_n^T$
  • Here ${\overrightarrow{e}}_1$ and ${\overrightarrow{e}}_2$ are not eigenvectors.
  • Vectors on $y=x$ and $y=-x$ are the same. So they're eigenvectors.
  • So we can do ${\overrightarrow{e}}_1+{\overrightarrow{e}}_2$ and ${\overrightarrow{e}}_1-{\overrightarrow{e}}_2$ as the eigenvectors.
  • Our Eigenvectors are $⟨1,1⟩$ and $⟨1,-1⟩=-⟨-1,1⟩$
  • $PD{P}^{-1}=A_T$
  • $P=\left[\begin{array}{cc}1& 1\\ 1& -1\end{array}\right]$
  • $D=\left[\begin{array}{cc}1& 0\\ 0& -1\end{array}\right]$
  • $P^{-1}=\left[\begin{array}{cccc}1& 1& 1& 0\\ 1& -1& 0& 1\end{array}\right]$
  • $R_2=R_2-R_1$
  • $P^{-1}=\left[\begin{array}{cccc}1& 1& 1& 0\\ 0& -2& -1& 1\end{array}\right]$
  • $R_2=\frac{R_2}{-1}$
  • $P^{-1}=\left[\begin{array}{cccc}1& 1& 1& 0\\ 0& 2& 1& -1\end{array}\right]$
  • $R_2=\frac{R_2}{2}$
  • $P^{-1}=\left[\begin{array}{cccc}1& 1& 1& 0\\ 0& 1& \frac{1}{2}& -\frac{1}{2}\end{array}\right]$
  • $R_1=R_1-R_2$
  • $P^{-1}=\left[\begin{array}{cccc}1& 0& \frac{1}{2}& \frac{1}{2}\\ 0& 1& \frac{1}{2}& -\frac{1}{2}\end{array}\right]$
  • $P^{-1}=\left[\begin{array}{cc}\frac{1}{2}& \frac{1}{2}\\ \frac{1}{2}& -\frac{1}{2}\end{array}\right]$
  • $\left[\begin{array}{cc}1& 1\\ 1& -1\end{array}\right]\left[\begin{array}{cc}1& 0\\ 0& -1\end{array}\right]\left[\begin{array}{cc}\frac{1}{2}& \frac{1}{2}\\ \frac{1}{2}& -\frac{1}{2}\end{array}\right]$
  • $\left[\begin{array}{cc}1& -1\\ 1& 1\end{array}\right]\left[\begin{array}{cc}\frac{1}{2}& \frac{1}{2}\\ \frac{1}{2}& -\frac{1}{2}\end{array}\right]$
  • $\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$

• Activity

  • General reflections

  • $T:{\mathbb{R}}^2\to {\mathbb{R}}^2$

  • $y=mx$

  • $y=mx$

  • $y=-\frac{1}{m}x$

  • ${\overrightarrow{v}}_1=\left[\begin{array}{c}1\\ m\end{array}\right]$

  • ${\overrightarrow{v}}_2=\left[\begin{array}{c}-1\\ \frac{1}{m}\end{array}\right]$

  • $T({\overrightarrow{v}}_1)$

  • $T({\overrightarrow{v}}_2)$

    left=-5; right=5;
    top=5; bottom=-5;
    ---
    y=0.5x
    y=-(1/0.5)x
    (1,0.5)
    (-1, 1/0.5)
    (1, -1/0.5)
    

  • Based on this, we know $⟨1,m⟩$ is an eigenvector, so is $⟨-1,\frac{1}{m}⟩$

  • When we reflect across $y=mx$, the two eigenvectors don't move. The $⟨-1,\frac{1}{m}⟩$ just reflects to $-⟨-1,\frac{1}{m}⟩\to ⟨1,-\frac{1}{m}⟩$

  • $P=\left[\begin{array}{cc}1& -1\\ m& \frac{1}{m}\end{array}\right]$

  • $D=\left[\begin{array}{cc}1& 0\\ 0& -1\end{array}\right]$

  • So then $A=\frac{1}{m^2+1}\left[\begin{array}{cc}1-m^2& 2m\\ 2m& m^2-1\end{array}\right]$