if two linearly independent vectors are mapped to a same non zero vector by a linear transformation then it's nullspace is non trivial? [closed]
Solution 1:
$T\in {\scr{L}}{(V, W) }$
$v_1, v_2 \in V$ Linearly independent. (Here, $v_1 \neq v_2 $ is sufficient to prove the claim)
$T(v_1) =T(v_2) $
$T(v_1 -v_2) =0$
$v_1-v_2 \in Null(T) $
And ,$v_1 -v_2\neq 0$
$T$ is not injective linear map.