<s>[INST] linear algebra- how to find the representation matrix of a linear transformation when given the base of the origin and its transformations as the title states i want to find the matrix representing some linear transformations $T:V\rightarrow W$ but all i know is the base for $V$is $ (b_1,b_2,b_3)$ and $T(b_1),T(b_2),T(b_3)$
i have done this for the vectors
but in a convoluted way and not in the short and beautiful way my teacher showed me can anyone help me understand if i am taking any redundant steps or if i am missing the point all together
thank you for your time
here is my solution:
$\left[\begin{matrix}x\\
y\\
z
\end{matrix}\right]=\lambda_{1}\left[\begin{matrix}1\\
3\\
1
\end{matrix}\right]+\lambda_{2}\left[\begin{matrix}1\\
2\\
-1
\end{matrix}\right]+\lambda_{3}\left[\begin{matrix}1\\
0\\
5
\end{matrix}\right]\\ T\left(\left[\begin{matrix}x\\
y\\
z
\end{matrix}\right]\right)=T\left(\lambda_{1}
\left[\begin{matrix}1\\
3\\
1
\end{matrix}\right]+\lambda_{2}\left[\begin{matrix}1\\
2\\
-1
\end{matrix}\right]+\lambda_{3}\left[\begin{matrix}1\\
0\\
5
\end{matrix}\right]\right)\\ T\left(\left[\begin{matrix}x\\
y\\
z
\end{matrix}\right]\right)=T\left(\lambda_{1}\left[\begin{matrix}1\\
3\\
1
\end{matrix}\right]\right)+T\left(\lambda_{2}\left[\begin{matrix}1\\
2\\
-1
\end{matrix}\right]\right)+T\left(\lambda_{3}\left[\begin{matrix}1\\
0\\
5
\end{matrix}\right]\right)\\ T\left(\left[\begin{matrix}x\\
y\\
z
\end{matrix}\right]\right)=\lambda_{1}T\left(\left[\begin{matrix}1\\
3\\
1
\end{matrix}\right]\right)+\lambda_{2}T\left(\left[\begin{matrix}1\\
2\\
-1
\end{matrix}\right]\right)+\lambda_{3}T\left(\left[\begin{matrix}1\\
0\\
5
\end{matrix}\right]\right)\\ T\left(\left[\begin{matrix}x\\
y\\
z
\end{matrix}\right]\right)=\lambda_{1}\left[\begin{matrix}1\\
-1\\
0
\end{matrix}\right]+\lambda_{2}\left[\begin{matrix}0\\
1\\
1
\end{matrix}\right]+\lambda_{3}\left[\begin{matrix}2\\
-5\\
-3
\end{matrix}\right]\\ \left[\begin{matrix}1 & 0 & 0 & | & -x+\frac{3y}{5}+\frac{z}{5}\\
0 & 1 & 0 & | & \frac{3x}{2}-\frac{2y}{5}-\frac{3z}{10}\\
0 & 0 & 1 & | & \frac{x}{2}-\frac{y}{5}+\frac{z}{10}
\end{matrix}\right]\\ T\left(\left[\begin{matrix}x\\
y\\
z
\end{matrix}\right]\right)=\left(-x+\frac{3y}{5}+\frac{z}{5}\right)\left[\begin{matrix}1\\
-1\\
0
\end{matrix}\right]+\left(\frac{3x}{2}-\frac{2y}{5}-\frac{3z}{10}\right)\left[\begin{matrix}0\\
1\\
1
\end{matrix}\right]+\left(\frac{x}{2}-\frac{y}{5}+\frac{z}{10}\right)\left[\begin{matrix}2\\
-5\\
-3
\end{matrix}\right]\\ \begin{cases}
\left(-x+\frac{3y}{5}+\frac{z}{5}\right)+2\left(\frac{x}{2}-\frac{y}{5}+\frac{z}{10}\right)=x'\\
-\left(-x+\frac{3y}{5}+\frac{z}{5}\right)+\left(\frac{3x}{2}-\frac{2y}{5}-\frac{3z}{10}\right)-5\left(\frac{x}{2}-\frac{y}{5}+\frac{z}{10}\right)=y'\\
\left(\frac{3x}{2}-\frac{2y}{5}-\frac{3z}{10}\right)-3\left(\frac{x}{2}-\frac{y}{5}+\frac{z}{10}\right)=z'
\end{cases}\\ \begin{cases}
\frac{y}{5}+\frac{2z}{5}=x'\\
-z=y'\\
\frac{y}{5}-\frac{6z}{10}=z'
\end{cases}\\ \left[\begin{matrix}0 & \frac{1}{5} & \frac{2}{5}\\
0 & 0 & -1\\
0 & \frac{1}{5} & \frac{-6}{10}
\end{matrix}\right]$
 [/INST] Your method is correct and the preferable method in this case.
Usually you first try inspection, so you try to find out what happens to the standard basis vectors $e_1,e_2,e_3$, so you try write $e_1$ as a linear combination of $b_1,b_2$ and $b_3$, and then you can directly calculate $T(e_1)$, which is the first column of the matrix, $T(e_2)$ the second and $T(e_3)$ the third.
If inspection doesn't work (because it's too difficult, which is the case here: at least I cannot see directly how to write $e_1$ as a linear combination of $b_1,b_2,b_3$), then you would have to solve three $Ax=b$ equations, so in that case do what you have done: solving these three equations at the same time by taking them together in a matrix and finding the inverse, etc.
</s>