Page 153 - Linear Algebra Done Right

Normal Operators on Real Inner-Product Spaces
m
the sum of the squares of the absolute

2
∗
7.20
T e j =
values of the entries of A and B.
j=1
Because T is normal, Te j = T e j for each j (see 7.6); thus 141
∗
m m

2 ∗ 2
Te j = T e j .
j=1 j=1
This equation, along with 7.19 and 7.20, implies that the sum of the
squares of the absolute values of the entries of B must equal 0. In
other words, B must be the matrix consisting of all 0’s. Thus
e 1 ... e m f 1 ... f n
 
e 1
. .  A 0 


.


 
 
7.21 M(T) = e m  
 .

f 1  
 
. .  
.

 0 C 

f n
This representation shows that Tf k is in the span of (f 1 ,...,f n ) for
each k. Because (f 1 ,...,f n ) is a basis of U , this implies that Tv ∈ U ⊥
⊥
whenever v ∈ U . In other words, U is invariant under T, completing
⊥
⊥
the proof of (a).
To prove (b), note that M(T ) has a block of 0’s in the lower left
∗
corner (because M(T), as given above, has a block of 0’s in the upper
right corner). In other words, each T e j can be written as a linear
∗
combination of (e 1 ,...,e m ). Thus U is invariant under T , completing
∗
the proof of (b).
To prove (c), let S = T| U . Fix v ∈ U. Then
Su, v = Tu, v
∗
= u, T v
for all u ∈ U. Because T v ∈ U (by (b)), the equation above shows that
∗
S v = T v. In other words, (T| U ) ∗ = (T )| U , completing the proof
∗
∗
∗
of (c).
To prove (d), note that T commutes with T ∗ (because T is normal)
and that (T| U ) = (T )| U (by (c)). Thus T| U commutes with its adjoint
∗
∗
and hence is normal, completing the proof of (d).