Linear transformation matrix using the standard bases in 3D

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.

• What do you denote $e$? Apr 7, 2019 at 9:58
• Try to see that $T(e_1),T(e_2),T(e_3)$ are linearly independent. Hence $\text {Im} (T) = \Bbb R^3.$ So $\ker (T) = \{(0,0,0) \}.$ Therefore $T$ is invertible. Apr 7, 2019 at 10:21
• What is $e$? Is $e = (1,1,1)$? Also how do you get $A$ back? $Ae_1+Ae_2 +Ae_3$ is a $3 \times 1$ matrix while $A$ is a $3 \times 3$ matrix. Apr 7, 2019 at 10:23
• @Bernard our teacher uses e to denote the vectors within the basis $\varepsilon$. Apr 7, 2019 at 10:47
• @Dbchatto67 is that for finding out if it is one-to-one? Also, how is $Ae_1+Ae_2+Ae_3$ a 3x1 matrix?I get three lots of 3x1 matrices but i then put them together and they make the same matrix A with dimensions 3x3. Apr 7, 2019 at 10:47

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.
• You just have to explain (in plain language) that the vectors $Ae1, \dots$ are the column vectors of the matrix of $T$ in the basis $\varepsilon$. It's a standard result. Apr 8, 2019 at 8:13