transformation changing basis

Question

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 $V^{-1}AV$

Answer 1

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.