# 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 $\begin{pmatrix} 1 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix}\cdot \begin{pmatrix} 1 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix}\neq \text{diagonal}$

• For a base change you have to compute $SAS^{-1}$ and not $A^2$. – Dietrich Burde May 28 '18 at 15:14
• So $S=[I]^B_V$ and $S^{-1}=[I]_{B}^V$? – newhere May 28 '18 at 15:17
• See for example here, how it works. – Dietrich Burde May 28 '18 at 15:19
• Yes, exactly. $\,$ – Berci May 28 '18 at 15:20

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