find a basis of $\ker T$ and find a basis of $\mathrm{Im} T$. You may assume that $T$ is linear.

Question

let $T:\Bbb R^3 → \Bbb R^3$; $T (x, y, z)= (x+y, x+y, 0)$ for $(x, y, z) \in\Bbb R^3$

Answer 1

$T(x,y,z)=0$, whenever $x+y=0$, i.e., when $(x,y,z)\cdot\pmatrix{1\\-1\\0}=0$. Thus $\ker T$ is the orthogonal complement of the span of $(1,-1,0)^T$.

Dimension of $\ker T$ is thus $3-1=2$. By rank nullity theorem we have that the dimension of $\mathrm{Im} T$ is $3-2=1$. Now let $x+y=k$, then $T(x,y,z)=\pmatrix{k\\k\\0}$. You can check that $\mathrm{Im} T=\mathrm{span}((1,1,0)^T)$.