Find the matrix in non-standard basis.

Question

Consider the two bases for $\mathbb{R}^3$
$$\begin{array}{l}
B:=\{(1,0,0),(0,1,0),(0,0,1)\} \\
B^{\prime}:=\{(1,0,0),(1,1,0),(1,1,1)\}
\end{array}$$
And the map $T\in End(\mathbb{R}^3)$ given by:
$$T\left(x_{1}, x_{2}, x_{3}\right)=\left(2 x_{1}-x_{2}, x_{2}+x_{3}, 4 x_{3}\right)$$
I want to find the matrix in both bases.
Now in the standard basis, we can easily see that:
$$T(1,0,0)=(2,0,0),\quad T(0,1,0)=(-1,1,0)\quad T(0,0,1)=(0,1,4)$$
$$\Longrightarrow [T]_B=\left[\begin{array}{ccc}
2 & -1 & 0 \\
0 & 1 & 1 \\
0 & 0 & 4
\end{array}\right]$$
To determine $[T]_{B'}$, can I simply look at how the standard basis transforms, e.g.:
$$T(1,1,0)=T(1,0,0)+T(0,1,0)=(1,1,0)$$
Which then gives the matrix:
$$[T]_{B'}=\left[\begin{array}{lll}
2 & 1 & 1 \\
0 & 1 & 2 \\
0 & 0 & 4
\end{array}\right]$$
Is this a valid procedure to find the matrix in basis $B'$?

Answer 1

Given two bases $B,B'\in \mathbb{R}^3$ and a linear transformation $$\varphi:\mathbb{R}^3_B \to \mathbb{R}^3_{B'},$$ what you want is to somehow express the images of your basis vectors $B$ with respect to $B'$.

Let $e_1,e_2,e_3 \in B$ denote the basis vectors of $B$ and $b_1,b_2,b_3\in B'$ the basis vectors of $B'$. Your goal is to find the coordinates of $\varphi(e_1)$ in $\mathbb{R}^3_{B'}$ with respect to the basis $B'$, that is

$$\varphi(e_i) = \lambda_1b_1+ \lambda_2b_2+ \lambda_3b_3, \ \text{for}\ i=1,2,3$$

That leads to solving three linear equations $$\varphi(e_i) = \begin{pmatrix} 1 & 1 & 1\\ 0 & 1 & 1\\ 0 & 0 & 1 \end{pmatrix}\cdot\begin{pmatrix} \lambda_1 \\ \lambda_2 \\ \lambda_3\end{pmatrix}$$

Note that the columns of the matrix $A_\varphi$ are precisely the basis vectors $b_1,b_2,b_2$ of $B'$ and $\varphi(e_i)$ are simply the images of the basis vectors of $B$.