<s>[INST] transformation changing basis Let $T:R^3->R^3 $ be defined by $T(a_1,a_2,a_3)=(3a_1+a_2,a_1+a_3,a_1-a_3)
$
so here first i do transformation to the bases $T \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}=\begin{pmatrix}3\\1\\1\end{pmatrix}$, $T \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}=\begin{pmatrix}1\\0\\0\end{pmatrix}$ , $T \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}=\begin{pmatrix}0\\1\\-1\end{pmatrix}$
then express it as linear combination of standard basis and find coordinate respect to it such as $T \begin{pmatrix} 3 \\ 1 \\ 1 \end{pmatrix}=a_1\begin{pmatrix}1\\0\\0\end{pmatrix}+a_2  \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}+a_3 \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$ and do it for three of them then got the coefficient for $a_1,a_2,a_2$
this matrix with respect to standard basis is $A=\begin{bmatrix}3&1&0\\1&0&1\\1&0&-1\end{bmatrix}$
is this the right idea behind with respect to standard basis?
suppose we choose as a basis $V=R^3$ set  ${\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix},  \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}}$
the matrix representation of T with respect to this basis is? so if I'm doing the same thing as above such as transform, $T \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}=\begin{pmatrix}3\\1\\1\end{pmatrix}$, $T \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}=\begin{pmatrix}4\\1\\1\end{pmatrix}$, $T \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}=\begin{pmatrix}4\\2\\0\end{pmatrix}$ and it turns out this is the matrix

A=\begin{bmatrix}3&4&4\\1&1&2\\1&1&0\end{bmatrix}
but why can t i represent the transformation as linear combination of basis b?
$\begin{pmatrix}3\\1\\1\end{pmatrix}=a_1{\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}+a_2\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}+a_3  \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}}$ for matrix column? i got different answer
but i know also the detour way $V^{-1}AV$
 [/INST] For $A$ in the standard basis you are correct.
The matrix $\bar A$ in the new basis is given by $V^{-1}AV$ where $V$ is the matrix which has for columns the vectors of the new basis. Can you see why and how it works?
Simply note that matrix $V$ changes the coordinates from the new basis to the standard and $V^{-1}$ changes the coordinates from standard to the new basis. 
Then we have


*

*$w=Av$

*$v=V\bar v \qquad w=V\bar w$


and then 
$$w=V\bar w=Av=AV\bar v\iff \bar w=V^{-1}AV\bar v$$
that is
$$\begin{bmatrix}1&-1&0\\0&1&-1\\0&0&1\end{bmatrix}\begin{bmatrix}3&1&0\\1&0&1\\1&0&-1\end{bmatrix}\begin{bmatrix}1&1&1\\0&1&1\\0&0&1\end{bmatrix}=\begin{bmatrix}2&3&2\\0&0&2\\1&1&0\end{bmatrix}$$
To obtain the result with a different method note that if we indicate with 
$$u={\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, v=\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix},  w=\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}}$$ 
the vectors of the new basis, we need first to find the expression of the vectors of the standard basis in term of the new basis vectors that is 


*

*$e_1=u$

*$e_2=-u+v$

*$e_3=-v+w$


Then we need to find the expression for $T(e_1)$, $T(e_2)$, $T(e_3)$ with respect to the new basis that is


*

*$a_1u+a_2v+a_3w=T(e_1)=(3,1,1)\implies (a_1,a_2,a_3)=(2,0,1)$

*$b_1u+b_2v+b_3w=T(e_2)=(1,0,0)\implies (b_1,b_2,b_3)=(1,0,0)$

*$c_1u+c_2v+c_3w=T(e_3)=(0,1,-1)\implies (c_1,c_2,c_3)=(-1,2,-1)$


Then for the transformation in the new basis we know that


*

*$T(e_1)=T(u)=2u+w$

*$T(e_2)=T(-u+v)=-T(u)+T(v)=u$

*$T(e_3)=T(-v+w)=-T(v)+T(w)=-u+2v-w$


from wich we obtain


*

*$T(u)=2u+w$

*$T(v)=3u+w$

*$T(w)=2u+2v$


which leads to the same result.
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