If a basis
for a subspace of 
is also an
set it is called an
.
An orthogonal basis can always be made into
an orthonormal basis by normalizing each of the vectors.
Examples
, 
is an orthogonal basis since
and is orthonormal since 
.
V is orthogonal, since
but not
orthonormal since the length
of each vector is 3.
V spans 
, by Prop. 
,
since
Thus V is an orthogonal basis for .
It is possible to transform any given basis
for a subspace S of
into an orthogonal basis for S
by repeated application of Equation .
A procedure which systematically accomplishes this is the
celebrated Gram-Schmidt Orthogonalization Process ,
which actually produces an orthonormal basis
for S from V.
In part a of Steps 1-m the next orthogonal vector in produced.
In part b of Steps 1-m the orthogonal vector just produced
is normalized.
If normalized vectors are not required,
part b of Steps 1-m could be omitted at the cost of additional
complexity in the formulas for part a of Steps 2-m
(see Equation ).
Example
Proof
Let basis
for subspace S of be given
and the Gram-Schmidt process be applied
to V yielding
W . For each i such that
, 
is the normalization of 
,
providing 
.
In Step 1,
is an orthogonal set by Prop. .
Also 
, since
otherwise V is dependent.
By Prop. , 
and by
Prop. , 
is an orthonormal set.
From the equations of Steps 1 and 2, we see that
.
In Step 2, 
, since
otherwise V is dependent.
As in Step 1, we find that
,
,
is an orthonormal set,
and
.
Continuing through Step m,
we conclude that W is an orthonormal basis for S.
Proof
Let 
be a given orthonormal basis
for subspace S of with m < n.
By Prop. we can find n-m additional vectors
from
such that
is a basis for .
Apply the Gram-Schmidt process to V, noting that
there will be no change to the first m vectors,
since they are already orthonormal. The basis
produced will be an orthonormal basis for .
Let p be an arbitrary element of 
.
Then
.
Since 
and 
,
by Prop.
.
an arbitrary element of is in 
,
where 
.
Now 
is also an orthonormal set
and is independent by Prop.
T is a basis for .
Proof Left as an exercise.
Proof This follows immediately from Prop. .
Proof
Assume S be a subspace of of dimension k, where 
.
Then has an orthonormal basis of dimension n-k.
Let 
be an orthonormal basis for .
Now 
S is polyhedral.
Proof
Let S and be given nontrivial complementary subspaces
for which 
and let b be an arbitrary element of .
By Prop. , S has a basis from which an orthonormal basis
can be produced by the Gram-Schmidt process
(according to Prop. )
and
by Prop. U can be extended
to an orthonormal basis
for
with
an orthonormal basis for .
Thus we can write
= p + q, where
and
.
The uniqueness of this representation of b follows
from Prop.
Because of the unique representation property in Prop. ,
is said to be the direct sum of any pair of nontrivial
complementary subspaces.
Proof
Let 
be given.
Assume S is a subspace of dimension 1
S has a basis which is a singleton set, say 
,
where 
.
Since S is a subspace and 
,
every vector of the form 
.
Let b be an arbitrary point of S
for some 
(since is a basis for S)
.
Now
and
, a line through the origin.
Let S be a line through the origin
for some 
(according to Prop. ).
Now
is a subspace of (according to Prop. ),
is linearly independent (according to Prop. ),
and
spans
is a basis for subspace
is a subspace of dimension 1.
Proof
Let be given.
Assume S is a subspace of dimension n-1
has dimension 1 (according to Prop. ).
has a basis for some 
(according to Prop. )
(according to Prop. ).
Now 
(according to Prop. )
.
Assume S is a hyperplane through the origin
for some 
.
Since 
is a subspace of dimension 1
(according to Prop. ),
the dimension of 
is n-1
(according to Prop. ).
