Given are the following transformations $$f_1:\mathbb{R}^n\to \mathbb{R},\; (x_1,\dots,x_n)\mapsto x_1+\dots +x_n$$
and
$$f_2:\mathbb{R}^2\to \mathbb{R},\; (x,y)\mapsto x\cdot y$$
First, I need to determine whether they're linear or not.
$f_1$ should be linear, because $\forall u,v \in \mathbb{R}^n:f(u+v)=f(u)+f(v)$ and $\forall \lambda \in \mathbb{R},\; \forall u \in \mathbb{R}^n:f(\lambda u) = \lambda f(u) $
$f_2$ is not linear, because $\forall u,v \in \mathbb{R}^2:f(u+v)\neq f(u)+f(v)$ and that's why we ignore $f_2$ for the rest of the task.
Now I should determine $\dim\operatorname{im}f$ and $\dim\operatorname{ker}f$ and one basis of $\operatorname{ker} f$, but I'm not sure how to do it in this case. If I had a transformation matrix from $f_1$, it would be simple.
Best Answer
Observe that
$$x_1+\ldots +x_n=0\iff x_n=-(x_1+\ldots+x_{n-1})\implies \dim\ker f=n-1$$
as we have $\;n-1\;$ free variables in the above equation. This is called sometimes hyperplane = maximal subspace of a linear space, in this case $\;\Bbb R^n\;$ . Any hyperplane is of this form: the kernel of a non-zero linear functional.
For a basis now is easy: for example, $\;(1,-1,0,...,0) , (1,0,-1,0,...,0),\ldots\;$ ...finish the argument and prove you get a l.i. set which is a then a basis of $\;\ker f\;$