Finding Basis Transformation Matrix Let $A=
\begin{pmatrix} 
1 & 1 & 0 \\
0 & 2 & 0 \\
0 & 0 & 3 
\end{pmatrix}$ be a transformation $T:\mathbb{R}^3\to \mathbb{R}^3$ by the elementary basis $B$ and let $V=\{(1 ,0 ,0),(1 ,1 ,0),(0 ,0 ,1)\}$ basis such that the $A$ is diagonal.
I am trying to show that $[I]^{B}_{V}[T]_{B}=[T]_V$
So we have to solve 
$(1,0,0)=\alpha(1 ,0 ,0)+\beta(1 ,1 ,0)+\gamma(0 ,0 ,1)$
$(0,1,0)=\alpha(1 ,0 ,0)+\beta(1 ,1 ,0)+\gamma(0 ,0 ,1)$
$(0,01,1)=\alpha(1 ,0 ,0)+\beta(1 ,1 ,0)+\gamma(0 ,0 ,1)$
which is $\begin{pmatrix} 
1 & -1 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 
\end{pmatrix}$ but $\begin{pmatrix} 
1 & 1 & 0 \\
0 & 2 & 0 \\
0 & 0 & 3 
\end{pmatrix}\cdot \begin{pmatrix} 
1 & 1 & 0 \\
0 & 2 & 0 \\
0 & 0 & 3 
\end{pmatrix}\neq \text{diagonal}$

Let
$$S=
\begin{pmatrix} 
1 & 1 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 
\end{pmatrix}$$
then by the change of basis $y=Sv$ and $x=Su$, we have that
$$y=Ax \implies Sv=ASu \implies v=S^{-1}ASu$$
