# proving a natural projection is linear and finding its kernel

Let $V_i = 1,...,N$ be a collection of vector spaces over a field $F$. Consider the Cartesian product $V=V_1 \times V_2 \times ... \times V_N$ with the natural projections $\pi=V \rightarrow V_i$. Prove that $\pi$ is linear and compute its kernel.

I am pretty confused as to what I need to do for this question. So far I've chosen an arbitrary basis for $V$ but I don't know what to do after that.

Elements of $V$ can be written as $(v_1,\dots,v_n)$ for $v_i\in V_i$. If you want to show that $\pi$ is linear, you need to show that for any $v=(v_1,\dots,v_n),w=(w_1,\dots,w_n)\in V$ and any scalar $\alpha\in F$, $$\pi(\alpha v+w)=\alpha \pi(v)+\pi(w).$$ This should follow from a quick computation, using the fact that $$\alpha(v_1,\dots,v_n)+(w_1,\dots,w_n)=(\alpha v_1+w_1,\dots,\alpha v_n+w_n).$$ As for finding the kernel, by definition $$\ker{\pi}=\{(v_1,\dots,v_n)\in V\mid \pi(v_1,\dots,v_n)=0\}.$$ But this is just any vector with $v_i=0$.
So the question wants you to show that the map $\phi_i : (v_1, v_2, ..., v_n) \rightarrow v_i$ is linear and to find the kernel of each one. Now, we can certainly take a basis for the product by taking, in each coordinate, a basis for the factor spaces. Then, just show that $a\phi_i(v) = \phi_i(av)$. This amounts to checking that scaling commutes into the coordinates i.e. $a(v_1, v_2, \ldots, vn) = (av_1, av_2, \ldots, av_n)$. Similarly, just check that adding two vectors will commute with your projection maps. Then, as for the kernel, when is $(v_1, v_2, \ldots, v_n) \rightarrow v_i = 0$? That shouldn't be too difficult to see.