Page 198 - Linear Algebra Done Right

Jordan Form
Thus the theorem holds for nilpotent operators.
Now suppose T ∈L(V). Let λ 1 ,...,λ m be the distinct eigenval-
ues of T, with U 1 ,...,U m the corresponding subspaces of generalized 187
eigenvectors. We have
V = U 1 ⊕· · · U m ,
is nilpotent (see 8.23). By the previous para-
where each (T − λ j I)| U j
.
graph, there is a basis of each U j that is a Jordan basis for (T − λ j I)| U j
Putting these bases together gives a basis of V that is a Jordan basis
for T.