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 $V^{-1}AV$

• Sorry for the confusion today, I was completely uncorrect with some hint given, finally I've revised all my work and now I hope it can be useful. Do not hesitate to ask for any further clarification. – user Mar 11 '18 at 21:39
• @gimusi thankyou so much for really clear explanation and notation!!!! But if you have time can you see this picture a little bit? This is the problem that i asked, i just wondering why doing Transformation directly to the basis can produced the matrix transformation(?) such as T(1,0,0)=(3,1,1) T(1,1,0)=(4,1,1) And what is the matrix transformation here exactly they asked(?) Maybe i was wrong in the interpretation of the problem? ibb.co/hgHK07 ibb.co/cC0uYS Thankyou so much!!! Sorry for asking many things！ Anyway i see this one too. ibb.co/hTw1V7 – fiksx Mar 12 '18 at 11:19
• Yes of course, I will take a look! – user Mar 12 '18 at 11:21
• Now it is clear what the example was talking about, matrix $\begin{bmatrix}3&4&4\\1&1&2\\1&1&0\end{bmatrix}$ represent the transformation from the new basis (as input) to the standard basis (as ouput) indeed ot is equal to AV. Thus it is not completely correct when it is said that the matrix represent T in the new basis. – user Mar 12 '18 at 14:05
• @gemusi got it, thankyou so much all is clear now!!!! and I understood the idea behind all this! really thankyou for clarifying all my confusion too!!!!!! learn a lot from your explanation!!!!! thankyou ^^ – fiksx Mar 12 '18 at 15:38

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.

• yeah i know that way , changing basis from b to standart basis, $[X]_E=V_{b->e}[X]_b$ so $V$ is matrix with basis b as column, its easier to find from some bases to standard bases, just using that bases as matrix column, so suppose $[T(x)]_b$ is the matrix transformation in basis b, $V^{-1}AV$ right? but in my question above why only apply transformation to basis $b$ and got the matrix $[T(x)]_b$? – fiksx Mar 11 '18 at 15:38
• because in the new basis the vector (1,1,1) becomes (0,0,1) and the vector (1,1,0) becomes (0,1,0) – user Mar 11 '18 at 15:42
• but isn't new basis $u,v,w$ is the same as basis b?${\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}}$ I'm confuse what is exactly new basis here? and how did you get $(1,1,1)$ become $0,0,1$ ? – fiksx Mar 11 '18 at 15:57
• but i count it and got $a_1=3, a_2=0,a_3=1$ and it suppose to be in the column of matrix transformation? and how come vector $(1,1,1)$ become $(0,0,1)$ in same basis(?) – fiksx Mar 11 '18 at 16:06
• As an example consider the vector (3,2,1) in the standard basis, what is its representation in the b basis? – user Mar 11 '18 at 16:07