6.2 Diagonalization

1. Show that $A=PD{P}^{-1}$ for the given matrices, in other words, in other words, show that $A$ is diagonalizable.

$$A=\left[\begin{array}{ll}3& 1\\ 1& 3\end{array}\right],D=\left[\begin{array}{ll}4& 0\\ 0& 2\end{array}\right],P=\left[\begin{array}{cc}1& 1\\ 1& -1\end{array}\right]$$ $$PD{P}^{-1}=\left[\begin{array}{cc}4& 2\\ 4& -2\end{array}\right]P^{-1}=\left[\begin{array}{cc}3& 1\\ 1& 3\end{array}\right]$$

2. Determine whether or not $A$ is diagonalizable. If so, find $D$ and $P$ such that $A=PD{P}^{-1}$.
   a. $A=\left[\begin{array}{cc}1& -2\\ 0& 1\end{array}\right]$
   $m\pm \sqrt{m^2-p}$
   $1\pm \sqrt{1-1}$
   Only one eigenvalue

$$Av=1v$$

A is not diagonolizable

b. $A=\left[\begin{array}{cc}-4& -1\\ 5& 2\end{array}\right]$
$m\pm \sqrt{m^2-p}$
$-1\pm \sqrt{1+3}$

$$\lambda =-3,1$$ $$(A-\lambda I)v=0$$ $$A_1=\left[\begin{array}{cc}-5& -1\\ 5& 1\end{array}\right]$$ $$v_1=[5,-1]$$ $$A_{-3}=\left[\begin{array}{cc}-1& -1\\ 5& 5\end{array}\right]$$ $$v_{-3}=[1,-5]$$

c. $A=\left[\begin{array}{ccc}2& 0& 0\\ 9& -1& 3\\ 6& -2& 4\end{array}\right]$

$$2(-4{\lambda }^2+6)=0$$ $$\lambda =\pm \sqrt{\frac{3}{2}}$$ $$A_{+}=\left[\begin{array}{ccc}2\sqrt{3/2}& 0& 0\\ 9& -1\sqrt{3/2}& 3\\ 6& -2& 4\sqrt{3/2}\end{array}\right]$$ $$V=[-2\sqrt{3/2},\sqrt{3/2}-4\sqrt{3/2}]$$

d. $A=\left[\begin{array}{ccc}17& -3& -21\\ -3& 1& 5\\ 12& -2& -15\end{array}\right]$
3. Compute $A^5$ where $A=\left[\begin{array}{cc}-4& -1\\ 5& 2\end{array}\right]$.
4. Suppose that a $5\times 5$ matrix $A$ has eigenvalues $\lambda =-3,\frac{1}{2},1,2,4$. Compute $\det (A)$.
5. Find a $3\times 3$ matrix whose eigenvalues are $-3,4,4$ and associated eigenvectors are $\left[\begin{array}{l}1\\ 1\\ 0\end{array}\right],\left[\begin{array}{c}1\\ 2\\ -1\end{array}\right],\left[\begin{array}{c}1\\ 3\\ -1\end{array}\right]$.
6. Suppose that ${\lambda }_1\ne {\lambda }_2$ are eigenvalues of a $2\times 2$ matrix $A$ with associated eigenvectors ${\mathbf{u}}_1$ and ${\mathbf{u}}_2$. Prove that $\det (U)\ne 0$ for $U=\left[\begin{array}{ll}{\mathbf{u}}_1& {\mathbf{u}}_2\end{array}\right]$.