Linear Algebra, 2020a

Section IV. Jordan Form 455
The rightmost non-⃗0 vectors ⃗w 1,1 ,...,⃗w k,1 are in the null space of the
associated maps, (t − λ 1 )(⃗w 1,1 )=⃗0, ..., (t − λ i )(⃗w k,1 )=⃗0. Thus the number
of strings associated with t − λ 1 , denoted q in the above diagram, is equal to
the dimension of W ∩ N (t − λ 1 ).
The string basis above is for the space W = R(t − λ 1 ). We now expand it to
make it a basis for all of C n . First, because each of the vectors ⃗w 1,n1 ,..., ⃗w q,nq
is an element of R(t−λ 1 ), each is the image of some vector from C n . Prefix each
string with one such vector, ⃗x 1 ,...,⃗x q , as shown below. Second, the dimension
of W ∩ N (t − λ 1 ) is q, and the dimension of W is r, so the bookkeeping requires
that there be a subspace Y ⊆ N (t − λ 1 ) with dimension r − q whose intersection
with W is {⃗0}. Consequently, pick r − q linearly independent vectors ⃗y 1 , ...,
⃗y r−q ∈ N (t − λ 1 ) and incorporate them into the list of strings, again as shown
below.
⃗x 1
t−λ 1
.
t−λ
↦−−−→
1 t−λ
⃗w1,n1 ↦−−−→ ··· ↦−−−→
1
t−λ 1
⃗w1,1 ↦−−−→ ⃗0
t−λ
⃗x
1
t−λ
q ↦−−−→
1 t−λ
⃗wq,nq ↦−−−→ ··· ↦−−−→
1
t−λ 1
⃗wq,1 ↦−−−→ ⃗0
⃗w q+1,nq+1
t−λ 2
.
⃗w k,nk
t−λ
↦−−−→ ··· ↦−−−→
2
t−λ 2
⃗wq+1,1 ↦−−−→ ⃗0
t−λ
↦−−−→
i t−λ
··· ↦−−−→
i
⃗wk,1
t−λ
↦−−−→
i
⃗0
⃗y 1
t−λ
↦−−−→
1
⃗0
.
⃗y r−q
t−λ
↦−−−→
1
⃗0
We will show that this is the desired basis for C n .
Because of the bookkeeping the number of vectors in the set is equal to the
dimension of the space, so we will be done if we verify that its elements are
linearly independent. Assume that this is equal to ⃗0.
a 1 ⃗x 1 + ···+ a q ⃗x q + b 1,n1 ⃗w 1,n1 + ···+ b k,1 ⃗w k,1 + c 1 ⃗y 1 + ···+ c r−q ⃗y r−q
(∗)
We first show that that all of the a’s are zero. Apply t − λ 1 to (∗). The vectors
⃗y 1 ,...,⃗y r−q go to ⃗0. Each of the vectors ⃗x i is sent to ⃗w i,ni . Astothe⃗w’s,
they have the property that (t − λ k ) ⃗w i,j = ⃗w i,j−1 for some λ k . Write ˆλ for
λ 1 − λ k to get this.
(t − λ 1 )(⃗w i,j )=(t −(λ 1 − ˆλ)+ˆλ)(⃗w i,j )=⃗w i,j−1 + ˆλ · ⃗w i,j (∗∗)
Thus, applying t − λ 1 to (∗) ends with a linear combination of ⃗w’s. These
vectors are linearly independent because they form a basis for W and so in this