Let $\alpha=\{(1,1,1), (1,1,0), (1,0,0)\}$ be a basis for $R^3$. Let $T$ satisfy $T((1,1,1))=(2,2,2), T((1,1,0))=(3,3,0)$ and $T((1,0,0))=(-1,0,0)$. How to find $[T]_\beta ^\beta$ for $\beta$ the standard basis for $R^3$?
So I think, by change of basis, $[T]_\beta ^\beta$ can be written as $[T]_\beta ^\beta$ = $[I]_\alpha^\beta$ $[T]_\alpha ^\alpha$ $[I]_\beta^\alpha$ =$[I]_\alpha^\beta$ $[T]_\alpha ^\alpha$ $([I]_\alpha^\beta )^{-1}$
The problem here I don't know how to locate $[I]_\alpha^\beta$ and $[T]_\alpha ^\alpha$.
Could someone please help
Let \begin{align*} v_1 &= \langle1,1,1\rangle & v_2 &= \langle 1,1,0\rangle & v_3 &=\langle 1,0,0\rangle \end{align*} so $\alpha=\{v_1,v_2,v_3\}$. Then note that \begin{array}{rcrcrcrcrcrccrcrcrc} T(v_1) & = & \langle 2,2,2\rangle & = & \color{purple}{2}\, v_1 &+& \color{blue}{0}\,v_2 &+& \color{red}{0}\,v_3 \\ T(v_2) & = & \langle 3,3,0\rangle & = & \color{purple}{0}\, v_1 &+& \color{blue}{3}\,v_2 &+& \color{red}{0}\,v_3 \\ T(v_3) & = & \langle -1,0,0\rangle & = & \color{purple}{0}\, v_1 &+& \color{blue}{0}\,v_2 &+& (\color{red}{-1})\,v_3 \end{array} This implies that $$ [T]_\alpha^\alpha= \left[\begin{array}{rrr} \color{purple}{2} & \color{purple}{0} & \color{purple}{0} \\ \color{blue}{0} & \color{blue}{3} & \color{blue}{0} \\ \color{red}{0} & \color{red}{0} & \color{red}{-1} \end{array}\right] $$ Now, the matrix $[I]_\alpha^\beta$ is obtained by inserting the vectors in $\alpha$ into columns $$ [I]_\alpha^\beta= \left[\begin{array}{rrr} 1 & 1 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 0 \end{array}\right] $$ Finally, your formula becomes \begin{align*} [T]_\beta^\beta &= [I]_\alpha^\beta[T]_\alpha^\alpha[I]_\beta^\alpha \\ &= \left[\begin{array}{rrr} 1 & 1 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 0 \end{array}\right] \left[\begin{array}{rrr} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & -1 \end{array}\right] \left[\begin{array}{rrr} 0 & 0 & 1 \\ 0 & 1 & -1 \\ 1 & -1 & 0 \end{array}\right] \\ &= \left[\begin{array}{rrr} -1 & 4 & -1 \\ 0 & 3 & -1 \\ 0 & 0 & 2 \end{array}\right] \end{align*}