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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.

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    $\begingroup$ What do you denote $e$? $\endgroup$
    – Bernard
    Apr 7, 2019 at 9:58
  • $\begingroup$ 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. $\endgroup$
    – little o
    Apr 7, 2019 at 10:21
  • $\begingroup$ 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. $\endgroup$
    – little o
    Apr 7, 2019 at 10:23
  • $\begingroup$ @Bernard our teacher uses e to denote the vectors within the basis $\varepsilon$. $\endgroup$
    – User1997
    Apr 7, 2019 at 10:47
  • $\begingroup$ @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. $\endgroup$
    – User1997
    Apr 7, 2019 at 10:47

1 Answer 1

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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.

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  • $\begingroup$ So does that mean what I have done already is all that needs to be done to answer this question? If not, could you please elaborate? $\endgroup$
    – User1997
    Apr 7, 2019 at 11:51
  • $\begingroup$ As far as I understand the question, yes, except you do not make a sum of three matrices. $\endgroup$
    – Bernard
    Apr 7, 2019 at 12:14
  • $\begingroup$ What do I make then? I used the three resultant vectors as the columns of the new matrix for the transformation.. thats how i got A back. Can you please explain what else I am meant to do? I dont understand where to go from obtaining these three vectors. $\endgroup$
    – User1997
    Apr 8, 2019 at 1:52
  • $\begingroup$ 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. $\endgroup$
    – Bernard
    Apr 8, 2019 at 8:13

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