Could use some help with this.
The linear transformation $T$: $\mathbb{R}^5 \longrightarrow \mathbb{R}^4$ is given by
$$ T \left[\begin{matrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \end{matrix}\right] = \left[\begin{matrix} 2x_1 - 4x_2 - x_3 - 3x_4 + 2x_5 \\ -x_1 + 2x_2 + x_3 + x_5 \\ x_1 - 2x_2 -x_3 - 3x_4 - x_5 \\ -x_1 + 4x_2 -x_3 + x_5 \\ \end{matrix}\right] , x = \left[\begin{matrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \end{matrix}\right] \in \mathbb{R}^5 $$
A linear transformation $S$: $\mathbb{R}^4 \longrightarrow \mathbb{R}^5$ which fullfill $(T \circ S)(y) = y$ for all $y \in \mathbb{R}^4$ is called a rightinverse to $T$.
Why is any rightinverse to T injective?
I've found two right inverses:
$$ S = \left(\begin{matrix} 3 & 1 & -3 & 1 \\ 1 & \frac{1}{2} & -1 & \frac{1}{2} \\ 1 & 1 & -1 & 0 \\ 0 & -\frac{1}{3} & -\frac{1}{3} & 0 \end{matrix}\right)^{(-1)} , x_5 = 0 $$
$$ S' = \left(\begin{matrix} 2 & -4 & -1 & -3 & 2 \\ -1 & 2 & 1 & 0 & 1 \\ 1 & -2 & -1 & -3 & -1 \\ -1 & 4 & -1 & 0 & 1 \\ 0 & 0 & 0 & 0 & \frac{1}{y_1+y_2} \end{matrix}\right) , x_5 = y_1 + y_2 \neq 0 $$
But I'm not sure how I argue and realize as to why any rightinverse to T is injective.
Let $S(x)=S(y)$.
By hypothesis, $T(S(x)) = x$ and $T(S(y))=y$.
Since $T$ is a mapping, we obtain $x=T(S(x)) = T(S(y)) = y$.