5.3. Soluciones en el l´ımite Λ → 0
5.5.2. Propuesta al problema de jerarqu´ıas para Λ > 0
Example 9.4.5. The list ((√1
2, 1 √ 2),( 1 √ 2,− 1 √
2)) is an orthonormal basis for R 2.
The next theorem allows us to use inner products to find the coefficients of a vectorv ∈V
in terms of an orthonormal basis. This result highlights how much easier it is to compute with an orthonormal basis.
Theorem 9.4.6. Let (e1, . . . , en) be an orthonormal basis for V. Then, for all v ∈ V, we
have
v =hv, e1ie1+· · ·+hv, enien
and kvk2 =Pn
k=1|hv, eki|2.
Proof. Letv ∈V. Since (e1, . . . , en) is a basis forV, there exist unique scalarsa1, . . . , an∈F
such that
v =a1e1+· · ·+anen.
Taking the inner product of both sides with respect to ek then yields hv, eki=ak.
9.5
The Gram-Schmidt orthogonalization procedure
We now come to a fundamentally important algorithm, which is called theGram-Schmidt orthogonalization procedure. This algorithm makes it possible to construct, for each list of linearly independent vectors (resp. basis) in an inner product space, a corresponding orthonormal list (resp. orthonormal basis).Theorem 9.5.1. If (v1, . . . , vm) is a list of linearly independent vectors in an inner product
space V, then there exists an orthonormal list (e1, . . . , em) such that
span(v1, . . . , vk) = span(e1, . . . , ek), for all k= 1, . . . , m. (9.4) Proof. The proof is constructive, that is, we will actually construct vectorse1, . . . , em having
the desired properties. Since (v1, . . . , vm) is linearly independent, vk 6= 0 for each k =
1,2, . . . , m. Set e1 = kvv1
1k. Then e1 is a vector of norm 1 and satisfies Equation (9.4) for
k = 1. Next, set
e2 =
v2− hv2, e1ie1
kv2− hv2, e1ie1k
130 CHAPTER 9. INNER PRODUCT SPACES
This is, in fact, the normalized version of the orthogonal decomposition Equation (9.3). I.e.,
w=v2− hv2, e1ie1,
where w⊥e1. Note that ke2k= 1 and span(e1, e2) = span(v1, v2).
Now, suppose that e1, . . . , ek−1 have been constructed such that (e1, . . . , ek−1) is an or-
thonormal list and span(v1, . . . , vk−1) = span(e1, . . . , ek−1). Then define
ek =
vk− hvk, e1ie1− hvk, e2ie2− · · · − hvk, ek−1iek−1
kvk− hvk, e1ie1− hvk, e2ie2− · · · − hvk, ek−1iek−1k
.
Since (v1, . . . , vk) is linearly independent, we know that vk 6∈span(v1, . . . , vk−1). Hence, we
also know that vk 6∈span(e1, . . . , ek−1). It follows that the norm in the definition ofek is not
zero, and so ek is well-defined (i.e., we are not dividing by zero). Note that a vector divided
by its norm has norm 1 so that kekk= 1. Furthermore,
hek, eii= vk− hvk, e1ie1− hvk, e2ie2− · · · − hvk, ek−1iek−1 kvk− hvk, e1ie1− hvk, e2ie2− · · · − hvk, ek−1iek−1k , ei = hvk, eii − hvk, eii kvk− hvk, e1ie1− hvk, e2ie2− · · · − hvk, ek−1iek−1k = 0,
for each 1≤i < k. Hence, (e1, . . . , ek) is orthonormal.
From the definition of ek, we see that vk ∈ span(e1, . . . , ek) so that span(v1, . . . , vk) ⊂
span(e1, . . . , ek). Since both lists (e1, . . . , ek) and (v1, . . . , vk) are linearly independent, they
must span subspaces of the same dimension and therefore are the same subspace. Hence Equation (9.4) holds.
Example 9.5.2. Take v1 = (1,1,0) and v2 = (2,1,1) in R3. The list (v1, v2) is linearly
independent (as you should verify!). To illustrate the Gram-Schmidt procedure, we begin by setting e1 = v1 kv1k = √1 2(1,1,0). Next, set e2 = v2− hv2, e1ie1 kv2− hv2, e1ie1k .
9.5. THE GRAM-SCHMIDT ORTHOGONALIZATION PROCEDURE 131 The inner product hv2, e1i= √12h(1,1,0),(2,1,1)i= √32, so
u2 =v2− hv2, e1ie1 = (2,1,1)−
3
2(1,1,0) = 1
2(1,−1,2). Calculating the norm ofu2, we obtainku2k=
q
1
4(1 + 1 + 4) =
√
6
2 . Hence, normalizing this
vector, we obtain e2 = u2 ku2k = √1 6(1,−1,2).
The list (e1, e2) is therefore orthonormal and has the same span as (v1, v2).
Corollary 9.5.3. Every finite-dimensional inner product space has an orthonormal basis.
Proof. Let (v1, . . . , vn) be any basis for V. This list is linearly independent and spans V.
Apply the Gram-Schmidt procedure to this list to obtain an orthonormal list (e1, . . . , en),
which still spans V by construction. By Proposition 9.4.2, this list is linearly independent and hence a basis of V.
Corollary 9.5.4. Every orthonormal list of vectors in V can be extended to an orthonormal basis of V.
Proof. Let (e1, . . . , em) be an orthonormal list of vectors in V. By Proposition 9.4.2, this
list is linearly independent and hence can be extended to a basis (e1, . . . , em, v1, . . . , vk) of V
by the Basis Extension Theorem. Now apply the Gram-Schmidt procedure to obtain a new orthonormal basis (e1, . . . , em, f1, . . . , fk). The first m vectors do not change since they are
already orthonormal. The list still spansV and is linearly independent by Proposition 9.4.2 and therefore forms a basis.
Recall Theorem 7.5.3: given an operator T ∈ L(V, V) on a complex vector spaceV, there exists a basisB forV such that the matrixM(T) ofT with respect toB is upper triangular. We would like to extend this result to require the additional property of orthonormality. Corollary 9.5.5. Let V be an inner product space over F and T ∈ L(V, V). If T is up- per triangular with respect to some basis, then T is upper triangular with respect to some orthonormal basis.
132 CHAPTER 9. INNER PRODUCT SPACES
Proof. Let (v1, . . . , vn) be a basis of V with respect to which T is upper triangular. Apply
the Gram-Schmidt procedure to obtain an orthonormal basis (e1, . . . , en), and note that
span(e1, . . . , ek) = span(v1, . . . , vk), for all 1≤k≤n.
We proved before thatT is upper triangular with respect to a basis (v1, . . . , vn) if and only if
span(v1, . . . , vk) is invariant underT for each 1≤k ≤n. Since these spans are unchanged by
the Gram-Schmidt procedure, T is still upper triangular for the corresponding orthonormal basis.