Last question ever, I promise.
If T is a linear operator on the space V and the eigenvector of T is a non-zero vector v in V such that Tv=av for some scalar a. Suppose that V is finite-dimensional and that B is an ordered basis of V such that the first vector v is an eigenvector and a is the corresponding eigenvalue. Show that the first column of [T]B has a as its first entry and the rest of those entries are zero.
I have absolutely no idea what to do here.
Any help would be appreciated.