Page 152 - Linear Algebra Done Right

Chapter 7. Operators on Inner-Product Spaces
140
The next result will play a key role in our characterization of the
7.18
Proposition: Suppose T ∈L(V) is normal and U is a subspace
Without normality, an normal operators on a real inner-product space.
easier result also holds: of V that is invariant under T. Then
if T ∈L(V) and U
invariant under T, then (a) U ⊥ is invariant under T;
∗
U ⊥ is invariant under (b) U is invariant under T ;
T ; see Exercise 29 in (c) (T| U ) = (T )| U ;
∗
∗
∗
Chapter 6.
(d) T| U is a normal operator on U;
(e) T| U ⊥ is a normal operator on U .
⊥
Proof: First we will prove (a). Let (e 1 ,...,e m ) be an orthonormal
basis of U. Extend to an orthonormal basis (e 1 ,...,e m ,f 1 ,...,f n ) of V
(this is possible by 6.25). Because U is invariant under T, each Te j is
a linear combination of (e 1 ,...,e m ). Thus the matrix of T with respect
to the basis (e 1 ,...,e m ,f 1 ,...,f n ) is of the form

e 1 ... e m f 1 ... f n
 
e 1
. .  A B 


.


 
 
e m  
M(T) =   ;
f 1  
 
. . .  0 C 


f n  
here A denotes an m-by-m matrix, 0 denotes the n-by-m matrix con-
sisting of all 0’s, B denotes an m-by-n matrix, C denotes an n-by-n
matrix, and for convenience the basis has been listed along the top and
left sides of the matrix.
2
For each j ∈{1,...,m}, Te j equals the sum of the squares of the
absolute values of the entries in the j th column of A (see 6.17). Hence
m

the sum of the squares of the absolute
2
7.19 Te j =
values of the entries of A.
j=1
2
For each j ∈{1,...,m}, T e j equals the sum of the squares of the
∗
absolute values of the entries in the j th rows of A and B. Hence