7.2. SUBSPACES AND BASES 193

Definition 7.2.8 Let f1, · · · , fn be smooth functions defined on an interval [a, b] . TheWronskian of these functions is defined as follows.

W (f1, · · · , fn) (x) ≡

∣∣∣∣∣∣∣∣∣∣f1 (x) f2 (x) · · · fn (x)

f ′1 (x) f ′2 (x) · · · f ′n (x)...

......

f(n−1)1 (x) f

(n−1)2 (x) · · · f

(n−1)n (x)

∣∣∣∣∣∣∣∣∣∣Note that to get from one row to the next, you just differentiate everything in that row. Thenotation f (k) (x) denotes the kth derivative.

With this definition, the following is the theorem. The interesting theorem involving theWronskian has to do with the situation where the functions are solutions of a differentialequation. Then much more can be said and it is much more interesting than the followingtheorem.

Theorem 7.2.9 Let {f1, · · · , fn} be smooth functions defined on [a, b] . Then they are lin-early independent if there exists some point t ∈ [a, b] where W (f1, · · · , fn) (t) ̸= 0.

Proof: Form the linear combination of these vectors (functions) and suppose it equals0. Thus

a1f1 + a2f2 + · · ·+ anfn = 0

The question you must answer is whether this requires each aj to equal zero. If they allmust equal 0, then this means these vectors (functions) are independent. This is what itmeans to be linearly independent.

Differentiate the above equation n− 1 times yielding the equationsa1f1 + a2f2 + · · ·+ anfn = 0

a1f′1 + a2f

′2 + · · ·+ anf

′n = 0

...

a1f(n−1)1 + a2f

(n−1)2 + · · ·+ anf

(n−1)n = 0

Now plug in t. Then the above yields

f1 (t) f2 (t) · · · fn (t)

f ′1 (t) f ′2 (t) · · · f ′n (t)...

......

f(n−1)1 (t) f

(n−1)2 (t) · · · f

(n−1)n (t)



a1

a2...

an

 =

0

0...

0

Since the determinant of the matrix on the left is assumed to be nonzero, it follows thismatrix has an inverse and so the only solution to the above system of equations is to haveeach ak = 0. ■

Here is a useful lemma.

Lemma 7.2.10 Suppose v /∈ span (u1, · · · ,uk) and {u1, · · · ,uk} is linearly independent.Then {u1, · · · ,uk,v} is also linearly independent.

Proof: Suppose∑k

i=1 ciui + dv = 0. It is required to verify that each ci = 0 and thatd = 0. But if d ̸= 0, then you can solve for v as a linear combination of the vectors,{u1, · · · ,uk},

v = −k∑

i=1

(cid

)ui