I suppose
The vectors that span form a basis for
if they are (1) linearly independent and (2) any vector in
can be expressed as a linear combination of those vectors (i.e. they span
).
- Independence:
Compute the Wronskian determinant:
The determinant is non-zero, so the vectors are linearly independent. For this reason, we also know the dimension of is 3.
- Span:
Write an arbitrary vector in as
. Then the given vectors span
if there is always a choice of scalars
such that
which is equivalent to the system
The coefficient matrix is non-singular, so it has an inverse. Multiplying both sides by that inverse gives
so the vectors do span .
The vectors comprising form a basis for it because they are linearly independent.