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The vectors in form a basis of
if they are mutually linearly independent and span
.
To check for independence, we can compute the Wronskian determinant:
The determinant is non-zero, so the vectors are indeed independent.
To check if they span , you need to show that any vector in
can be expressed as a linear combination of the vectors in
. We can write an arbitrary vector in
as
Then we need to show that there is always some choice of scalars such that
This is equivalent to solving
or the system (in matrix form)
This has a solution if the coefficient matrix on the left is invertible. It is, because
(that is, the coefficient matrix is not singular, so an inverse exists)
Compute the inverse any way you like; you should get
Then
A solution exists for any choice of , so the vectors in
indeed span
.
The vectors in are independent and span
, so
forms a basis of
.