Page 181 - Linear Algebra Done Right

Chapter 8. Operators on Complex Vector Spaces
170





n
n


 .
M(T ) =M(T) =  λ 1 n . . . n ∗ 
 λ n−1 
 n 
0 λ n
This shows that
n n
T v n = u + λ n v n
for some u ∈ U. To prove 8.14 (still assuming that λ n = 0), suppose
n
v ∈ null T . We can write v in the form
v = ˜ u + av n ,
where ˜ u ∈ U and a ∈ F. Thus
n
n
n
n
n
0 = T v = T ˜ u + aT v n = T ˜ u + au + aλ n v n .
n
n
Because T ˜ u and au are in U and v n ∉ U, this implies that aλ n = 0.
However, λ n = 0, so a = 0. Thus v = ˜ u ∈ U, completing the proof
of 8.14.
Now consider the case where λ n = 0. In this case we will show that
n
8.15 dim null T n = dim null(T| U ) + 1,
which along with 8.13 will complete the proof when λ n = 0.
Using the formula for the dimension of the sum of two subspaces
(2.18), we have
n
n
dim null T n = dim(U ∩ null T ) + dim(U + null T ) − dim U
n
n
= dim null(T| U ) + dim(U + null T ) − (n − 1).
Suppose we can prove that null T n contains a vector not in U. Then
n
n = dim V ≥ dim(U + null T )> dim U = n − 1,
n
which implies that dim(U + null T ) = n, which when combined with
the formula above for dim null T n gives 8.15, as desired. Thus to com-
plete the proof, we need only show that null T n contains a vector not
in U.
Let’s think about how we might find a vector in null T n that is not
in U. We might try a vector of the form
u − v n ,