Let $\{v_1,\cdots,v_k\}$ be linearly independent. Then we have $f$ st. $f(v_i)=1$ for all $i$ Given a linearly independent set $\{v_1,\cdots,v_k\}$, do we always have an element $f$ st. $f(v_i)=1$ for all $i$? My textbook states this is the case, but I don't really see how we can construct such element.
 A: This is one of the key ideas of linear algebra, so it's worth being confused about.
For convenience, let's assume that $\{v_1,\cdots,v_k\}$ is a basis for the space.  (If it isn't, we can just add vectors as needed to make any linearly independent set of vectors into a basis.  This gets messier if the space is not finite-dimensional, but I won't go there.)
Then, for any vector $u$ in our space, there is exactly one $k$-tuple of real numbers $(c_{u_1},...,c_{u_k})$ such that $$u=\sum_{i=1}^kc_{u_i}v_i$$  Let's assume that this appropriate choice of "coordinates" (relative to the basis) has been made for each vector $u$ in the space.  Notice that the coordinates of $v_1$ are $(1,0,...0)$, the coordinates of $v_2$ are $(0,1,0,...,0)$, and so on.  Also notice that the coordinates of $v_1+v_2$ are $(1,1,0,...,0)$ and the coordinates of $3v_1$ are $(3,0,...,0)$, so this correspondence between our vector space and $\mathbb R^k$ turns out to be a linear isomorphism.  (Some people use this identification to argue that $\mathbb R^k$ are the only finite dimensional vector spaces worth studying.)
With all this out of the way, the function you are looking for is just $$f(u)=\sum_{i=1}^kc_{u_i}$$  It's just a bunch of one-line proofs to show that $f$ is a linear form with $f(v_i)=1$ for each member of the basis.
