# Why is any rightinverse to T injective? The linear transformation $T$: $\mathbb{R}^5 \longrightarrow \mathbb{R}^4$

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.

• A rightinverse being injective is true for any functions, not just linear transformations. Let's say $T\circ S=id$. Suppose $S(x)=S(y)$. Can you conclude that $x=y$ from this? – Mark May 17 at 12:27
• First, how can you just suppose $S(x)=S(y)$? But if we suppose that, I guess that could be concluded, yes. Seems logical atleast. – user10829235 May 17 at 12:40
• Think about the definition of injective linear transformation, the comment by @Mark is made for any kind of functions but you can use the linear transformation characterization of injectivity S(x)=0 if and only if x = 0 and the argument is the same. – Manuel DaGeo May 17 at 12:46
• I don't understand, frankly. I'm sorry. But thank you for the help – user10829235 May 17 at 13:10

Let $$S(x)=S(y)$$.
By hypothesis, $$T(S(x)) = x$$ and $$T(S(y))=y$$.
Since $$T$$ is a mapping, we obtain $$x=T(S(x)) = T(S(y)) = y$$.