Linear Algebra, Theory And Applications, 2012a

294 INNER PRODUCT SPACES
Theorem 12.3.5 Suppose V is a subspace of F n having dimension p ≤ n. Then there exists
a Q ∈L(F n , F n ) such that
QV ⊆ span (e 1 , ··· , e p )
and |Qx| = |x| for all x. Also
Q ∗ Q = QQ ∗ = I.
Proof: By Lemma 12.2.1 there exists an orthonormal basis for V,{v i } p i=1
. By using the
Gram Schmidt process this may be extended to an orthonormal basis of the whole space,
F n ,
{v 1 , ··· , v p , v p+1 , ··· , v n } .
Now define Q ∈L(F n , F n )byQ (v i ) ≡ e i and extend linearly. If ∑ n
i=1 x iv i is an arbitrary
element of F n ,
( n
)∣ ∣ Q ∑ ∣∣∣∣
2 ∣ n∑ ∣∣∣∣
2 ∣
n∑
x i v i =
x i e i = |x i | 2 n∑ ∣∣∣∣
2
=
x i v i .
∣
∣
i=1
It remains to verify that Q ∗ Q = QQ ∗ = I. To do so, let x, y ∈ F n . Then
i=1
i=1
i=1
(Q (x + y) ,Q(x + y)) = (x + y, x + y) .
Thus
|Qx| 2 + |Qy| 2 +2Re(Qx,Qy) =|x| 2 + |y| 2 +2Re(x, y)
and since Q preserves norms, it follows that for all x, y ∈ F n ,
Re (Qx,Qy) =Re(x,Q ∗ Qy) =Re(x, y) .
Thus
Re (x,Q ∗ Qy − y) = 0 (12.7)
for all x, y. Letω be a complex number such that |ω| = 1 and
Then from (12.7),
ω (x,Q ∗ Qy − y) =|(x,Q ∗ Qy − y)| .
0 = Re(ωx,Q ∗ Qy − y) =Reω (x,Q ∗ Qy − y)
= |(x,Q ∗ Qy − y)|
and since x is arbitrary, it follows that for all y,
Q ∗ Qy − y = 0
Thus
Similarly QQ ∗ = I.
I = Q ∗ Q.