Page 182 - Linear Algebra Done Right

where u ∈ U. At least we are guaranteed that any such vector is not
in U. Can we choose u ∈ U such that the vector above is in null T ?
Let’s compute: The Characteristic Polynomial n 171
n
n
n
T (u − v n ) = T u − T v n .
To make the above vector equal 0, we must choose (if possible) u ∈ U
n
n
n
n
such that T u = T v n . We can do this if T v n ∈ range(T| U ) . Because
8.11 is the matrix of T with respect to (v 1 ,...,v n ), we see that Tv n ∈ U
(recall that we are considering the case where λ n = 0). Thus
n
n
T v n = T n−1 (Tv n ) ∈ range(T| U ) n−1 = range(T| U ) ,
where the last equality comes from 8.9. In other words, we can indeed
n
choose u ∈ U such that u − v n ∈ null T , completing the proof.
Suppose T ∈L(V). The multiplicity of an eigenvalue λ of T is de- Our definition of
fined to be the dimension of the subspace of generalized eigenvectors multiplicity has a clear
corresponding to λ. In other words, the multiplicity of an eigenvalue λ connection with the
of T equals dim null(T − λI) dim V .If T has an upper-triangular matrix geometric behavior
with respect to some basis of V (as always happens when F = C), then of T. Most texts define
the multiplicity of λ is simply the number of times λ appears on the multiplicity in terms of
diagonal of this matrix (by the last theorem). the multiplicity of the
3
As an example of multiplicity, consider the operator T ∈L(F ) de- roots of a certain
fined by polynomial defined by
determinants. These
8.16 T(z 1 ,z 2 ,z 3 ) = (0,z 1 , 5z 3 ). two definitions turn
out to be equivalent.
You should verify that 0 is an eigenvalue of T with multiplicity 2, that
5 is an eigenvalue of T with multiplicity 1, and that T has no additional
3
eigenvalues. As another example, if T ∈L(F ) is the operator whose
matrix is
 
6 7 7


8.17  0 6 7  ,
0 0 7
then 6 is an eigenvalue of T with multiplicity 2 and 7 is an eigenvalue
of T with multiplicity 1 (this follows from the last theorem).
In each of the examples above, the sum of the multiplicities of the
eigenvalues of T equals 3, which is the dimension of the domain of T.
The next proposition shows that this always happens on a complex
vector space.