I would like to show this, here is my starting point
Consider the mapping $f : \mathbb{R}^d\to\mathbb{R}^{k}\times\mathbb{R}^{d-k}$
defined as $f(v)=((v_1,...,v_k),(v_{k+1},...,v_d))$
Then we have $f(\alpha v+\beta w)=((\alpha v_1+\beta w_1,..., \alpha v_k +\beta w_k), (\alpha v_{k+1}\beta w_{k+1},..., \alpha v_d +\beta w_d)) = (\alpha(v_1,...,v_k) + \beta(w_1,..., w_k), \alpha(v_{k+1},...,v_d) +\beta(w_{k+1},..., w_d))=\alpha f(v) +\beta f(w)$
where the two last equalities follows from the fact that $\mathbb{R}^{k}\times\mathbb{R}^{d-k}$ is a vector space. First, am I correct until now ?
Now I would like to know what kind of argument can I use to show that this linear mapping is bijective, if it is of course ?
Thank you a lot !
Since you have already proven, that $f$ is linear. Let us assume that $f(v)=0$, then $(v_1,\dots,v_k)=(v_{k+1},\dots, v_n)=0$. But $v=(v_1,\dots,v_k,v_{k+1},\dots, v_n)$, so it follows that $v=0$. Hence $f$ is injective. Now we know that the dimensions of both vector spaces are equal to $d$. Therefore $f$ is already a isomorphisms between the two spaces. Of course it is also easy to see that $f$ is surjective, for any $((v_1,\dots,v_k),(w_1,\dots, w_{d-k}))$ we choose to map the element $(v_1,\dots,v_k,w_1,\dots, w_{d-k})$ and get $f((v_1,\dots,v_k,w_1,\dots, w_{d-k})) = ((v_1,\dots,v_k),(w_1,\dots, w_{d-k}))$.