1
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ 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? $\endgroup$ – Mark May 17 at 12:27
  • $\begingroup$ 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. $\endgroup$ – user10829235 May 17 at 12:40
  • $\begingroup$ 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. $\endgroup$ – Manuel DaGeo May 17 at 12:46
  • $\begingroup$ I don't understand, frankly. I'm sorry. But thank you for the help $\endgroup$ – user10829235 May 17 at 13:10
1
$\begingroup$

Let $S(x)=S(y)$.

By hypothesis, $T(S(x)) = x$ and $T(S(y))=y$.

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

$\endgroup$
  • 1
    $\begingroup$ Thank you, Wuestenfux! $\endgroup$ – user10829235 May 18 at 6:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.