Prove that translation is linear transformation iff orthogonal projection on subspace $U^\bot$, is injective

Question

Let $u\in$ E some arbitrary vector in Euclidean space E. Prove that translation space E for some vector u is linear transformation iff operator orthogonal projection on subspace $\{u\}^\bot$ is injective

$=>$

First $E=U⊕U^\bot$ where U is some subspace, if we choose some vector $u\in$ E and we translation the space, since we know that translation is linear transformation $f:E\to E$ so it must hold that $f(0)=0$, so how we translation for some vector $u$, $f(0)=0+u=0$ so $u=0$, it will hold if we choose some vector from $U$ then $U=\{0\}$, $U^\bot=E$ so $kerP=\{0\}$

$<=$

Since $P$ is orthogonal projection on $U^\bot$ and it is injection $ker(P)={0}$, $ker(P)=U$ and $im(P)=U^\bot$, if we choose a some $u$ from U then if you translation every vector for that vector, then $f$ will be linear transformation, because $u=0$

But I think that it not hold what if someone choose some vector from $U^\bot$ then f will not be linear transformation, do you know some better option?

Answer 1

Apparently, you are not sure about your direction "<=". I'll try to restate your proof to highlight the relevant points. We have $u\in E$, a euclidean Space. We want to show that

The translation $T_u$ of $E$ by $u$ is linear if the orthogonal projection $P$ onto $\langle u\rangle ^\perp$ is injective.

Suppose that $P$ is injective and denote $U=\langle u \rangle$. We know that $U\subset \ker(P)$, as it is a projection. Since $P$ is injective, $\ker(P) = \{0\}$, so $u\in \{0\}$. Therefore $u=0$ and $T_u$ is linear. $\square$

Now, if $v\notin U$, the translation $T_v$ won't be linear. But that's not needed, we only have to show that $T_u$ is linear. In particular, $T_w$ is linear if and only if $w\in U$ (i.e. $w=0$).