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Let T: $\mathbb R^3 \to \mathbb R^3$ be defined by:

$$T\left( \begin{bmatrix} x \\ y \\ z \end{bmatrix} \right) = \begin{bmatrix} x - y + 2z \\ 2x + 3y - z \\ -x +2y -2z \end{bmatrix} $$

a) Show that T is a linear transformation:

I think i can rewrite this as

$T[x,y,z] = T[x - y + 2z, 2x + 3y - z, -x + 2y - 2z]$

Let $u = [u_1, u_2, u_3]$ and $v = [v_1, v_2, v_3]$ where

$u,v, \in \mathbb R^3$, let $r \in \mathbb R$

i) T(u + v) = $T([u_1+v_1, u_2 + v_2, u_3 + v_3])$

$= [(u_1 - u_2 + 2u_3 + v_1 - v_2 + 2v_3, 2u_1 + 3u_2 - u_3 + 2v_1 + 3v_2 - v_3, -u_1 + 2u_2 - 2u_3 -v_1 + 2v_2 - 2v_3]$

$= T([u_1,u_2,u_3]) + T([v_1,v_2,v_3]) = T(u) + T(v)$

ii) T(ru) = $T[ru_1, ru_2, ru_3]$

$= [ru_2 - ru_2 + 2ru_3, 2ru_1 + 3ru_2 - ru_3, -ru_1 + 2ru_2 - 2ru_3]$

$ r[u_1 - u_2 + 2u_3, 2u_1 + 3u_2 - u_3, -u_1 + 2u_2 - 2u_3]$

= $rT(u)$

b) Find the matrix representation of T:

$$ \begin{bmatrix} 1 & -1 & 2 \\ 2 & 3 & -1 \\ -1 & 2 & -2 \end{bmatrix} $$

c) Determine if T is invertible and, if it is, give a formula for $T^{−1}$

$$T^{-1} \left( \begin{bmatrix} x \\ y \\ z \end{bmatrix} \right) = \begin{bmatrix} -\frac{4}{5}x + \frac{2}{5} y -1 z \\ x + z \\ \frac{7}{5}x - \frac{1}{5}y + z \end{bmatrix} $$

. d) Determine whether or not T is one-to-one and onto.

Idk

I'm just following the definition I'm not sure if im right. Could someone check?

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i) T(u + v) = $T([u_1+v_1, u_2 + v_2, u_3 + v_3])$

$= [(u_1 - u_2 + 2u_3 + v_1 - v_2 + 2v_3, 2u_1 + 3u_2 - u_3 + 2v_1 + 3v_2 - v_3, -u_1 + 2u_2 - 2u_3 -v_1 + 2v_2 - 2v_3]$

$= \color{red}{T(T[u_1,u_2,u_3])} = T(u) + T(v)$

You may mean well, but this is written in a strange way (definitely the part in red).

You want to show that $T(u+v)=T(u)+T(v)$, I would write it like this: $$\begin{align} T(u+v) & = \begin{bmatrix} u_1+v_1 - (u_2+v_2) + 2(u_3+v_3) \\ 2(u_1+v_1) + 3(u_2+v_2) - (u_3+v_3) \\ -(u_1+v_1) +2(u_2+v_2) -2(u_3+v_3) \end{bmatrix} \\[10pt] & = \begin{bmatrix} u_1- u_2 + 2u_3 \\ 2u_1 + 3u_2 - u_3+ \\ -u_1 +2u_2 -2u_3 \end{bmatrix}+ \begin{bmatrix} v_1- v_2 + 2v_3 \\ 2v_1 + 3v_2 - v_3+ \\ -v_1 +2v_2 -2v_3 \end{bmatrix}\\[10pt] & = T(u)+T(v) \end{align}$$

ii) T(ru) = $T[ru_1, ru_2, ru_3]$

$= [r\color{red}{u_2} - ru_2 + 2ru_3, 2ru_1 + 3ru_2 - ru_3, -ru_1 + 2ru_2 - 2ru_3]$

$ r[\color{red}{u_2} - u_2 + 2u_3, 2u_1 + 3u_2 - u_3, -u_1 + 2u_2 - 2u_3]$

Good, apart from a small typo in red (should be $\color{green}{u_1}$).

b) Find the matrix representation of T:

$$ \begin{bmatrix} 1 & -1 & 2 \\ 2 & 3 & -1 \\ -1 & 2 & -2 \end{bmatrix}$$

Correct!

c) Determine if T is invertible and, if it is, give a formula for $T^{−1}$

$$T^{-1} \left( \begin{bmatrix} x \\ y \\ z \end{bmatrix} \right) = \begin{bmatrix} -\frac{4}{5}x + \frac{2}{5} y -1 z \\ \color{red}{x - z} \\ \frac{7}{5}x - \frac{1}{5}y + z \end{bmatrix} $$

Almost, the part in red should be $\color{green}{x + z}$.

d) Determine whether or not T is one-to-one and onto.

Since $T$ is invertible (see part c), it is bijective and thus surely one-to-one and onto.

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  • $\begingroup$ Thank you. Could you look at d). I like ur way much better for a) $\endgroup$ – user349557 Apr 4 '17 at 8:18
  • $\begingroup$ @user349557 I was adding after your edit; answer updated. $\endgroup$ – StackTD Apr 4 '17 at 8:20
  • $\begingroup$ What does it mean to be one-to-one and onto, and what do u mean bijective? $\endgroup$ – user349557 Apr 4 '17 at 8:21
  • $\begingroup$ @user349557 If you don't know what these terms mean, you should first check your course notes and/or look up the definitions. See here or for the specific context of linear transformations, this pdf-document is nice as well. $\endgroup$ – StackTD Apr 4 '17 at 8:23

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