Linear transformation matrix using the standard bases in 3D

Question

Let T : $\mathbb{R^3}$$\to$$\mathbb{R^3}$ be the linear transformation $$T(x_1,x_2,x_3) = (2x_1-x_2,2x_2+3x_3,3x_1+4x_3)^T$$ for all $(x_1,x_2,x_3) \in \mathbb{R^3}$

I need to calculate $[T]_\varepsilon$ of the operator $T$ corresponding to the standard basis $\varepsilon$ of $\mathbb{R^3}$, where $\varepsilon={\{e_1,e_2,e_3\}}$, $e_1=(1,0,0)^T , e_2=(0,1,0)^T , e_3=(0,0,1)^T$.

I also need to show whether $T$ is one-to-one or not, and explain why.

What I did so far (not sure if I am correct):

$[T(e)]_\varepsilon = A[e]_\varepsilon$ $= Ae_1 + Ae_2 + Ae_3$ $$=\left[{\begin{array}{ccc}2&-1&0\\0&2&3\\3&0&4\\\end{array}}\right]\left[{\begin{array}{c}1\\0\\0\\\end{array}}\right]+\left[{\begin{array}{ccc}2&-1&0\\0&2&3\\3&0&4\\\end{array}}\right]\left[{\begin{array}{c}0\\1\\0\\\end{array}}\right]+\left[{\begin{array}{ccc}2&-1&0\\0&2&3\\3&0&4\\\end{array}}\right]\left[{\begin{array}{c}0\\0\\1\\\end{array}}\right]= A$$

I have just gone in a circle and come back out with A.

Answer 1

What is standard is that the matrix of a linear operator in a finite dimensional vector space, relative to a given basis, has column vectors equal to the coordinates of the images of the basis vectors, expressed in that basis. So it is not surprising that, doing the concatenation of the three matrices $Ae_1, Ae_2$ and $Ae_3$ (and not their sum), you obtain $A$ again.