Question

Let U and V be vector spaces over a field F. Let T:U→V is one-one if and only if ...

(A) rank (T) = 0
(B) rank (T) = 1
(C) Ker (T) = 0
(D) Ker(T) = 1

Answer 1

Question #22435 Let $U$ and $V$ be vector spaces over a field $F$. Let $T\colon U\to V$ is one-one if and only if ...

(A) $\operatorname{rank}(T) = 0$

(B) $\operatorname{rank}(T) = 1$

(C) $\ker (T) = 0$

(D) $\ker (T) = 1$

Please explain

Solution. Let us prove that the wright answer is C. Really, assume that $\ker T = 0$, then if one has $T(u_{1}) = T(u_{2})$, when $u_{1} \neq u_{2}$, then $T(u_{1} - u_{2}) = 0$, since $T$ is linear, so $u_{1} - u_{2} \in \ker T$, which contradicts the assumption that $\ker T = 0$. Now assume that $T$ is 1-1 mapping. Since $T$ is linear that $T(0) = 0$, and since $T$ is 1-1, then $T(u) = 0$, $u \neq 0$, thus $\ker T = 0$. Hence

Answer C.