11.2. THE GRAM SCHMIDT PROCESS 275

It turns out that the validity of this identity is equivalent to the existence of an innerproduct which determines the norm as described above. These sorts of considerations aretopics for more advanced courses on functional analysis.

Definition 11.1.10 A basis for an inner product space, {u1, · · · , un} is an orthonormalbasis if

(uk, uj) = δkj ≡

{1 if k = j

0 if k ̸= j.

Note that if a list of vectors satisfies the above condition for being an orthonormal set,then the list of vectors is automatically linearly independent. To see this, suppose

n∑j=1

cjuj = 0

Then taking the inner product of both sides with uk,

0 =

n∑j=1

cj (uj , uk) =

n∑j=1

cjδjk = ck.

11.2 The Gram Schmidt Process

Lemma 11.2.1 Let X be an inner product space and let {x1, · · · , xn} be linearly indepen-dent. Then there exists an orthonormal basis for X, {u1, · · · , un} which has the propertythat for each k ≤ n, span(x1, · · · , xk) = span (u1, · · · , uk) .

Proof: Let u1 ≡ x1/ |x1| . Thus for k = 1, span (u1) = span (x1) and {u1} is anorthonormal set. Now suppose for some k < n, u1, · · · , uk have been chosen such that(uj , ul) = δjl and span (x1, · · · , xk) = span (u1, · · · , uk). Then define

uk+1 ≡xk+1 −

∑kj=1 (xk+1, uj)uj∣∣∣xk+1 −

∑kj=1 (xk+1, uj)uj

∣∣∣ , (11.1)

where the denominator is not equal to zero because the xj form a basis and so

xk+1 /∈ span (x1, · · · , xk) = span (u1, · · · , uk)

Thus by induction,

uk+1 ∈ span (u1, · · · , uk, xk+1) = span (x1, · · · , xk, xk+1) .

Also, xk+1 ∈ span (u1, · · · , uk, uk+1) which is seen easily by solving 11.1 for xk+1 and itfollows

span (x1, · · · , xk, xk+1) = span (u1, · · · , uk, uk+1) .

If l ≤ k,

(uk+1, ul) = C

(xk+1, ul)−k∑

j=1

(xk+1, uj) (uj , ul)

= C

(xk+1, ul)−k∑

j=1

(xk+1, uj) δlj

= C ((xk+1, ul)− (xk+1, ul)) = 0.