A basis for $\{(x-y,y-z,z-x) : (x,y,z)\in \mathbb R^3\}$ The original question : Prove that $F: \mathbb R^3 \to \mathbb R^3$ such that $F(x,y,z)=(x-y,y-z,z-x)$ is a linear-map and find a basis for $imF$ and $kerF$. My try : I proved that $F$ holds the two conditions of being linear. Also, It was easy to find a basis for $kerF=\\{(x,y,z) \in \mathbb R^3 : x=y=z\\}$ ( The basis is $(1,1,1)$. ) The part i'm stuck on : I know that $dim(\mathbb R^3)=dim(imF)+dim(kerF)$ . So, a basis for $imF$ should have $2$ members. The problem is that i don't know what it is. I mean, i don't know which members of $\mathbb R^3$ are spanned by $imF$. Thanks in advance.

\begin{align} im(F)&=\\{(x-y,y-z,z-x) \mid x,y,z\in\mathbb{R}\\}\\\ &=\\{x(1,0,-1)+y(-1,1,0)+z(0,-1,1) \mid x,y,z\in\mathbb{R}\\}\\\ &=span\\{(1,0,-1),(-1,1,0),(0,-1,1)\\}\\\ &=span\\{(1,0,-1),(-1,1,0)\\}\\\ \end{align}

Show that $\\{(1,0,-1),(-1,1,0)\\}$ is linearly independent and you've got yourself a basis.

There's general way to figure out a basis for a spanning set which involves building a matrix whose rows are the vectors spanning the set, and then subsequently row reducing the matrix.