Find matrix of linear transformation in a different basis and write it in terms of the original matrix

Question

Problem asks how does matrix $A$ of some linear transformation in basis $B=\{v_1, v_2, v_3\}$ look like in basis $B'=\{v_1+v_2,v_2+v_3,v_3\}$ so I found transition matrices that are lower triangular and the first one from $B' \to B$ is: $$T= \begin{bmatrix} 1 & 0 & 0 \\ 1 & 1& 0 \\ 0 &1 &1 \end{bmatrix} $$ and its inverse ie. $B\to B'$ : $$T^{-1}= \begin{bmatrix} 1 & 0 & 0 \\ -1 & 1& 0 \\ 1 &-1 &1 \end{bmatrix} $$ So the problem says only about $A$ like they want the result in $A$. However I could not solving without writing down its entries: So I say $$A=\sum_{i=0,j=0}^{i=3,j=3}a_{ij}$$. And I multiplied it out and I got something really ugly so : $$T^{-1}AT= \begin{bmatrix} a_{11}+a_{12} & a_{12}+a_{13} & a_{13} \\ a_{21}+a_{22}-(a_{11}+a_{12})& a_{22}+a_{23}-(a_{12}+a_{13})& a_{23}-a_{13} \\ a_{11}+a_{12}+a_{31}+a_{32}-(a_{22}+a_{21}) &a_{12}+a_{13}+a_{32}+a_{33}-(a_{22}+a_{23}) &a_{13}+a_{33}-a_{23} \end{bmatrix} $$ I cannot really see how could I write it in terms of $A$ so I assume there must be some mistake, but as I am saying the exercise does not explicitly say "write it in terms of $A$", but I would expect it from how it is phrased.

Answer 1

It should be $TAT^{-1}$, so that first you get back to the standard basis, apply $A$, and then go back to the basis you want.

In the end, you want a map $B^{\prime} \to B^{\prime}$. $T:B \to B^{\prime}$ is the matrix you describe, so doing $T$ firsts does not make sense

Answer 2

It is known that the new components of a vector $v$ under a change of base is $T^{-1}v$, then we have $$Av=ATT^{-1}v,$$ and $$T^{-1}(Av)=(T^{-1}AT)(T^{-1}v)$$ where we can see how the new components $T^{-1}(Av)$ are related by $T^{-1}AT$ with the new components $T^{-1}v$