Validating if there is a linear transformation $S:\mathbb{R}^{3}\mapsto\mathbb{R}^{3}$ such that:

Question

Let $T:\mathbb{R}^{3}\mapsto\mathbb{R}^{3}$ be a linear transformation such that $T\begin{pmatrix}x_1\\ x_2\\x_3\end{pmatrix} = \begin{pmatrix}x_1-x_2\\ x_2-x_3\\ x_3-x_1\end{pmatrix}$

Is there a linear transformation $S:\mathbb{R}^{3}\mapsto\mathbb{R}^{3}$ such that:

$(T\circ S)\begin{pmatrix}0\\ 1\\2\end{pmatrix} = \begin{pmatrix}-2\\ 0\\ 2\end{pmatrix}$ and $(T\circ S)\begin{pmatrix}-2\\ 0\\2\end{pmatrix} = \begin{pmatrix}0\\ 1\\ 2\end{pmatrix}$

Hey guys!, how should one approach such question? I am trying to reach a contradiction, however I can't manage to put my hand on one.

Answer 1

The entire $S$ thing is a distraction.

As pointed out in the comments, you can start by proving that $\begin{pmatrix}0\\ 1\\ 2\end{pmatrix} \notin T(\mathbb{R^3})$, where $T(\mathbb{R^3})$ just means the image of $T$. In other words, the equation system $T\begin{pmatrix}x\\ y\\ z\end{pmatrix}= \begin{pmatrix}0\\ 1\\ 2\end{pmatrix}$ has no solutions.

OK, so now we know that $T$ cannot transform anything into $\begin{pmatrix}0\\ 1\\ 2\end{pmatrix}$, whether that "anything" is the image of a vector by $S$ or not

Answer 2

Let $S\begin{pmatrix}-2\\ 0\\2\end{pmatrix} = \begin{pmatrix}x_1\\ x_2\\x_3\end{pmatrix}$.

Given \begin{equation} \begin{split} (T\circ S)\begin{pmatrix}-2\\ 0\\2\end{pmatrix} &= \begin{pmatrix}0\\ 1\\ 2\end{pmatrix} \\ T\begin{pmatrix}x_1\\ x_2\\x_3\end{pmatrix} &= \begin{pmatrix}0\\ 1\\2\end{pmatrix} \\ \begin{pmatrix}x_1-x_2\\ x_2-x_3\\x_3-x_1\end{pmatrix} &= \begin{pmatrix}0\\ 1\\2\end{pmatrix} \end{split} \label{1} \tag{1} \end{equation}

$\ref{1} \implies $

$x_1 = x_2 \label{1a} \tag{1a}$

$x_2-x_3=1 \label{1b} \tag{1b}$

$x_3-x_1=2 \label{1c} \tag{1c}$

Combining $\ref{1a}$, $\ref{1b} \implies x_1-x_3=1 \implies x_3-x_1=-1$ which contradicts with $\ref{1c}$. Hence such a transformation $S$ doesn't exist.