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.

  • 1
    $\begingroup$ What do you denote $e$? $\endgroup$ – Bernard Apr 7 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. $\endgroup$ – Dbchatto67 Apr 7 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. $\endgroup$ – Dbchatto67 Apr 7 at 10:23
  • $\begingroup$ @Bernard our teacher uses e to denote the vectors within the basis $\varepsilon$. $\endgroup$ – Kayla Martin Apr 7 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. $\endgroup$ – Kayla Martin Apr 7 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.

  • $\begingroup$ 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? $\endgroup$ – Kayla Martin Apr 7 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 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. $\endgroup$ – Kayla Martin Apr 8 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 at 8:13

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