MAT223 Lecture 11

• 1
  • Let $D=\left[\begin{array}{ccc}2& 0& 0\\ 0& -1& 0\\ 0& 0& 3\end{array}\right]$
  • $D^3=\left[\begin{array}{ccc}8& 0& 0\\ 0& -1& 0\\ 0& 0& 27\end{array}\right]$
• 2
  • $A$ is $n\times n$
  • What systems have the same solutions as:
  • $A\overrightarrow{x}=2\overrightarrow{x}$
  • $\lambda =2$
  • $(A-2I_n)\overrightarrow{x}=\overrightarrow{0}$
  • $A\overrightarrow{x}=2\overrightarrow{x}$
    • $A\overrightarrow{x}-2\overrightarrow{x}=\overrightarrow{0}$
    • $A\overrightarrow{x}-2I_n\overrightarrow{x}=\overrightarrow{0}$
    • $(A-2I_n)\overrightarrow{x}=\overrightarrow{0}$
• Activity 3
  • If every $\lambda$ of $A$ is positive, then $A$ is invertible.
    • $A\overrightarrow{x}=\lambda \overrightarrow{x}$
    • $A\overrightarrow{x}-\lambda \overrightarrow{x}=\overrightarrow{0}$
    • $(A-\lambda I_n)\overrightarrow{x}=\overrightarrow{0}$
    • True.
    • Incorrect counter example:
      • $\left[\begin{array}{cc}1& 1\\ 1& 1\end{array}\right]$, not invertible.
      • But we have $\lambda =0$. So it's not positive.
      • You're invertible if you don't have $\lambda =0$
      • You're non-invertible if you have $\lambda =0$
    • If $\lambda >0$ for all $\lambda$ of $A$, then $A$ doesn't have $\lambda =0$. So $A$ is invertible.
  • If $A$ is diagonalizable, then $A$ has $n$ distinct $\lambda$s
    • $A=PD{P}^{-1}$
    • $A$ has $n$ distinct $\lambda$s
    • True.
    • Because $A=PD{P}^{-1}$ if we have $n$ eigenvectors.
    • $n$ $\lambda$s means we have minimum $n$ eigenvectors.
    • Above proves the other direction.
    • But it's false:
      • $I_2=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$, $\lambda =1$
      • It is diagonalizable. But we only have $1$ $\lambda$
  • If $A$ is diagonalizable, then $A^{2026}$ is diagonalizable.
    • $A=PD{P}^{-1}$
    • $A^k=P{D}^k{P}^{-1}$
    • True
  • If $B$ is a $2\times 2$, with one $\lambda$. For all eigenvectors $\overrightarrow{u}$ and $\overrightarrow{v}$ of $B$, $\overrightarrow{u}=k\overrightarrow{v}$ for some $k\in \mathbb{R}$. Then $B$ is not diagonalizable.
    • $B=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$, $\lambda =1$
    • True
    • If $B$ is diagonalizable, $B=PD{P}^{-1}$
    • Then we have $P=\left[\begin{array}{cc}{\overrightarrow{v}}_1& {\overrightarrow{v}}_2\end{array}\right]$
    • $D=\left[\begin{array}{cc}\lambda & 0\\ 0& \lambda \end{array}\right]$
    • $P^{-1}={\left[\begin{array}{cc}{\overrightarrow{v}}_1& {\overrightarrow{v}}_2\end{array}\right]}^{-1}$
    • Then $v_1,v_2$ are basic solutions to $B\overrightarrow{x}=\lambda \overrightarrow{x}$.
    • But if ${\overrightarrow{v}}_1=k{\overrightarrow{v}}_2$, they're the same basic solution.
    • So then we get that $P$ isn't invertible, so $B$ isn't diagonalizable.
• Polls 1:
  • $P(1,3,-1)$
  • $Q(1,-2,3)$
  • $\overrightarrow{QP}=P-Q$
  • $\overrightarrow{QP}=\left[\begin{array}{c}0\\ 5\\ -4\end{array}\right]$
• 2
  • If $\overrightarrow{x}=\left[\begin{array}{c}1\\ 3\\ 2\end{array}\right]$
  • $\overrightarrow{y}=\left[\begin{array}{c}2\\ 1\\ 1\end{array}\right]$
  • $|\overrightarrow{x}+2\overrightarrow{y}|$
  • $\overrightarrow{x}+2\overrightarrow{y}$
  • $\left[\begin{array}{c}5\\ 5\\ 4\end{array}\right]$
  • $\sqrt{5^2+5^2+4^2}=\sqrt{66}$