<s>[INST] Question about basis change/transformation Let
$B=\left[ \begin{pmatrix} 1 \\ 0 \\ 0  \end{pmatrix},   \begin{pmatrix} 0 \\ 1 \\ 0\end{pmatrix},  \begin{pmatrix} 0 \\ 0\\ 1   \end{pmatrix}\right]$
$B_1=\left[ \begin{pmatrix} 1 \\ -2 \\ 0  \end{pmatrix},   \begin{pmatrix} 1\\ 0 \\ -2\end{pmatrix},  \begin{pmatrix} 2 \\ 5\\ 6   \end{pmatrix}\right]$
and the following map
$f:\mathbb{R}^3 \rightarrow \mathbb{R}^3$, $f\left[\begin{pmatrix} \alpha \\ \beta \\ \gamma  \end{pmatrix}\right]=\begin{pmatrix} \alpha+\gamma   \\ 2\alpha+2\beta+\gamma  \\ 4\alpha+2\beta+\gamma  \end{pmatrix}$
I need to determine $\left[f\right]_{B,B}$ and $T=[id]_{B,B_1}$
So I have an idea how to start with  $\left[f\right]_{B,B}$
I need to take the first column vector and
$f\begin{pmatrix} 1 \\ 0 \\ 0  \end{pmatrix}=\begin{pmatrix} 1 \\ 2 \\ 4  \end{pmatrix}$
and then I need to disassemble the resulting vector  $\begin{pmatrix} 1 \\ 2 \\ 4  \end{pmatrix}$ into = $1 \cdot\begin{pmatrix} 1 \\ 0 \\ 0  \end{pmatrix}+2\cdot \begin{pmatrix} 0 \\ 1 \\ 0\end{pmatrix}+4 \cdot \begin{pmatrix} 0 \\ 0\\ 1   \end{pmatrix}$.
and the same also with the second and third column vector. Is this correct?
In this case the breakdown into the basis vectors was very easy, but what if it's not the standard basis? How can I see how to disassemble in such cases? Can anyone recommend me any good .pdf with examples provided where I can read about this topic?
my second question how to determine $T=[id]_{B,B_1}$?
Do I need to invert something here?
 [/INST] Yes, you are correct. In general, given a linear transformation from $V$ to $W$ vector spaces $f:V \to W$, $\alpha=\{v_1, \cdots, v_n\}$ basis of $V$ and $\beta=\{w_1, \cdots, w_n\}$ basis of $W$ to construct $[f]_{\alpha,\beta}$ you proceed like this: The $i$-th column of the matrix $[f]_{\alpha,\beta}$ is given by the coordinates of $f(v_i)$ with respect to the basis $\beta$. That is the $i$-th column of $[f]_{\alpha,\beta}$
$$
\begin{bmatrix}
    c_1 \\
    c_2\\
    \cdot\\
 \cdot\\
 \cdot\\
    c_m
\end{bmatrix}
$$
is such that $f(v_i)=c_1 w_1+ c_2 w_2 + \cdots + c_m w_m$. As you noticed this is not always so simple as it was in this case with the canonical basis of $\mathbb{R}^3$. But in your case with $V=W=\mathbb{R}^3$ and $\alpha=\beta=\{u_,u_2,u_3\}$ we must find $[f]_{\alpha,\alpha}$. Then we need to decompose $f(u_i)$ with respect to the basis $\{u_1,u_2,u_3\}$. Hence we must find $a_{1,i}$, $a_{2,i}$ and $a_{3,i}$ such that $f(u_i)=a_{1,i} u_1 + a_{2,i} u_2 + a_{3,i} u_3$. Now denoting the vectors as
$$
f(u_i)=\begin{pmatrix}
    y_{1,i}\\
    y_{2,i}\\
    y_{3,i}
\end{pmatrix}
, u_i=\begin{pmatrix}
    x_{1,i}\\
    x_{2,i}\\
   x_{3,i}
\end{pmatrix}
$$
we have that
$$
\begin{pmatrix}
   y_{1,i}\\
    y_{2,i}\\
    y_{3,i}
\end{pmatrix}=\begin{pmatrix}
    a_{1,i}x_{1,1}\\
    a_{1,i}x_{2,1}\\
   a_{1,i}x_{3,1}
\end{pmatrix}+\begin{pmatrix}
    a_{2,i}x_{1,2}\\
    a_{2,i}x_{2,2}\\
   a_{2,i}x_{3,2}
\end{pmatrix}+\begin{pmatrix}
    a_{3,i}x_{1,3}\\
    a_{3,i}x_{2,3}\\
   a_{3,i}x_{3,3}
\end{pmatrix}
$$
$$
\begin{pmatrix}
   y_{1,i}\\
    y_{2,i}\\
    y_{3,i}
\end{pmatrix}=\begin{pmatrix}
    a_{1,i}x_{1,1}+a_{2,i}x_{1,2}+a_{3,i}x_{1,3}\\
    a_{1,i}x_{2,1}+a_{2,i}x_{2,2}+a_{3,i}x_{2,3}\\
   a_{1,i}x_{3,1}+a_{2,i}x_{3,2}+a_{3,i}x_{3,3}
\end{pmatrix}
$$
$$
\begin{pmatrix}
   y_{1,i}\\
    y_{2,i}\\
    y_{3,i}
\end{pmatrix}=\begin{pmatrix}
    x_{1,1} && x_{1,2} && x_{1,3}\\
    x_{2,1}  && x_{2,2} && x_{2,3}\\
   x_{3,1} && x_{3,2} && x_{3,3}
\end{pmatrix} \begin{pmatrix}
    a_{1,i}\\
    a_{2,i}\\
   a_{3,i}
\end{pmatrix}.
$$
And in order to find $a_{1,i}$, $a_{2,i}$ and $a_{3,i}$ you solve the $i$-th linear system. Now $[id]_{B,B_1}$ is the representation of the identity transformation $id:\mathbb{R}^3 \to \mathbb{R}^3$ relative with the basis $B$ and $B_1$. The first column of the matrix is the coordinates of the identity applied to $(1, 0, 0)$ with respect to the basis $B_1$. That is we must find $a, b$ and $c$ such that
$$
\begin{pmatrix}
    1\\
    0\\
   0
\end{pmatrix}=id \begin{pmatrix}
    1\\
    0\\
   0
\end{pmatrix}=a\begin{pmatrix}
    1\\
    -2\\
   0
\end{pmatrix}+b\begin{pmatrix}
    1\\
    0\\
   -2
\end{pmatrix}+c\begin{pmatrix}
    2\\
    5\\
   6
\end{pmatrix}
$$
$$
\begin{pmatrix}
  1\\
    0\\
    0
\end{pmatrix}=\begin{pmatrix}
    1 && 1 && 2\\
    -2  && 0 && 5\\
   0 && -2 && 6
\end{pmatrix} \begin{pmatrix}
    a\\
    b\\
   c
\end{pmatrix}.
$$
Then you can apply the same procedure to $(0,1,0)$ and $(0,0,1)$ in order to find the second and third columns of $[id]_{B,B_1}$. Another way to calculate $[id]_{B,B_1}$ is to find $[id]_{B_1,B}$ and since $B$ is the canonical basis
$$
[id]_{B_1,B}=\begin{pmatrix}
    1 && 1 && 2\\
    -2  && 0 && 5\\
   0 && -2 && 6
\end{pmatrix}.
$$
And then you can use the fact that $[id]_{B,B_1}=([id]_{B_1,B})^{-1}$ and calculate the inverse matrix.
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