Consider the multiplication operator $L_A:{\mathbb R}^2\to {\mathbb R}^2$ defined by $L_A(x)=Ax$ where $A=\left[\begin{array}{cc} -2 &1\cr -2 &2 \end{array}\right]$. Find an ordered basis $B=(b_1,b_2)$ for ${\mathbb R}^2$ such that $[L_A]_B^B=\left[\begin{array}{cc} -10 &-14\cr 7 &10 \end{array}\right]$. Hint: $[L_A]_E^E=A$ where $E$ is the standard basis.
$b_1 =$ $\left[\Rule{0pt}{2.4em}{0pt}\right.$$\left]\Rule{0pt}{2.4em}{0pt}\right.$, $b_2 =$ $\left[\Rule{0pt}{2.4em}{0pt}\right.$$\left]\Rule{0pt}{2.4em}{0pt}\right.$