Proofs from Kernel, image and rank

Linearity gives $T(0)=0$, so $0\in\ker(T)$. If $u_1,u_2\in\ker(T)$, then

so $u_1+u_2\in\ker(T)$. If $u\in\ker(T)$ and $\lambda$ is a scalar, then

so $\lambda u\in\ker(T)$. The subspace test now applies.

The image contains $0$ because $T(0)=0$. If $w_1,w_2\in\operatorname{im}(T)$, write $w_1=T(v_1)$ and $w_2=T(v_2)$. Then

so $w_1+w_2\in\operatorname{im}(T)$. If $w=T(v)$, then

so $\lambda w\in\operatorname{im}(T)$. The subspace test now applies.