1 IMPLICATION OF JORDAN-CHEVALLEY DECOMPOSITION FOR MATRICES 1

MATH 262 Lecture, 28 May 2020

1 Implication of Jordan-Chevalley Decomposition for Matrices
Let V be a finite dimensional vector space over a field F and T an operator on V . As in the
previous lecture, we assume that the characteristic polynomial of T is a product of linear factors
in F [x]:
∆T (x) = (x − λ1 )µ1 (x − λ2 )µ2 . . . (x − λk )µk
By the results of the last lecture, now we know the following: Let Ci = ker((T − λi I)µi ). Then
first of all, we showed that
V = C1 ⊕ C2 ⊕ · · · ⊕ Ck .
The other important result, the Jordan-Chevalley decomposition theorem told us that there
exist unique semisimple and nilpotent operators S and N such that T = S + N , and such that
T, S, N pairwise commute. The proof of that result shows that each Ci above is an invariant
subspace for all three operators T, S, N , and furthermore S restricted to Ci is simply equal to
λi I.
We now want to understand the implications of these results for matrices representing
the operator T in suitable bases. If we select arbitrary bases B1 , B2 , . . . , Bk for the subspa-
ces C1 , C2 , . . . , Ck respectively, then the matrix A representing T with respect to the basis
B = B1 ∪ . . . ∪ Bk is of the block diagonal form
 
A1 0 . . . . . . 0
 0 A2 0 ... 0
 
A = . . . . . . . . . . . . . . . 


 0 . . . 0 Ak−1 0 
0 ... ... 0 Ak

We can do better than this: So far, we selected arbitrary bases for C1 , C2 , . . . , Ck . We can tidy
things up by making more careful choices of bases. The following result is the first improvement
that we can make:

Proposition 1. With the assumptions and notation above, there exists a basis for Ci such that
the matrix Ai for T restricted to Ci is of the form
 
λi ∗ . . . ... ∗
 0 λi ∗ ... ∗ 
 
Ai = . . . . . . . . . . . . . . .
,
 0 ... 0 λi ∗ 
0 ... ... 0 λi

(each ∗ denotes an arbitary element of the field F ), the matrix for S restricted to Ci is λi I, and
the matrix for N restricted to Ci is strictly upper triangular.

Proof: Recall that S restricted to Ci is equal to λi I and N restricted to Ci is equal to T − λi I.

Let 0 6= v1 ∈ Ci ∩ ker(T − λi ). We clearly have T v1 = λi v1 so that v1 is an eigenvector for
T with eigenvalue λi . This will be our first basis vector. To find the second one, look at the
quotient vector space W2 = Ci /Span(v1 ). Since all transformations T, S, N send Span(v1 ) to
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itself, they induce linear operators on W2 . Furthermore, the operator that N induces on W2 is
still nilpotent. Pick v2 ∈ Ci such that its equivalence class in W2 is non-zero and in the kernel
of N . But then this means, there exists a constant a12 ∈ F such that
T v2 = a12 v1 + λi v2 .
Let v2 be the second basis vector. Its independence from v1 is clear.
It is clear how we can proceed inductively: Having chosen v1 , . . . , vm−1 , define the quotient
vector space Wm = Ci /Span(v1 , . . . , vm−1 ). If this vector space is trivial, this means that
{v1 , . . . , vm−1 } is a basis for Ci , so we stop. If not, then N induces a nilpotent operator on it.
So pick vm ∈ Ci such that its equivalence class in Wm is non-zero and belongs to the kernel of
N . Then it is an eigenvector for T in this quotient space. Lifting back to Ci , we see that
T vm = a1m v1 + a2m v2 + . . . + am−1 m vm−1 + λi vm
for some a1m , . . . , am−1 m ∈ F .
Continuing in this way, at some stage we obtain a basis {v1 , . . . , vm } for Ci . It is clear from
the expression for each T vj that the matrix Ai representing T with respect to this basis will be
upper triangular with diagonal elements λi (and the ajl above are going to be the entries above
the diagonal). Since S restricted to Ci is λi I, it is clear that N = T − S will be represented by
Ai − λi I which will be strictly upper triangular. This finishes the proof. 2

2 Jordan Form
Despite the nice structural result proved in the last section that shows that for every linear
operator there is a choice of basis such that the matrix A representing T is of a special form,
there is still some more room for standardization. We will show below that by an even more
careful choice of basis, the representing matrix A can be chosen to be a matrix in so called
Jordan form. We will first define a matrix in Jordan form, and then we will prove this result.
Definition 1. Let F be a field. An m × m matrix Jλ over F is said to be a Jordan block if
it is of the form  
λ 1 0 ... ... 0
0 λ 1 0 ... 0 
 
0 0 λ 1 ... 0 
Jλ = 
. . . . . . . . . . . .

 . . . . . .

 0 ... ... 0 λ 1
0 ... ... ... 0 λ
for some λ ∈ F (namely, (Jλ )ii = λ for all i = 1, . . . , m, (Jλ )i i+1 = 1 for all i = 1, . . . , m − 1
and all other entries are 0.
Definition 2. Let F be a field. An n × n matrix J over F is said to be a matrix in Jordan
form (or a Jordan matrix) if it is in the block form
 
Jλ1 0 . . . . . . 0
 0 Jλ2 0 ... 0 
 
J = ... ... ... ...
 ...
 0 . . . 0 Jλr−1 0 
0 ... ... 0 Jλr
where each Jλi is a Jordan block.
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Remark: The λi ’s don’t need to be distinct. The Jordan blocks can be of various sizes.
Example: The matrices
 
3 1 0 0  
0 0 1 0
3 0 0
A= 0
, B = 0 0 1
0 3 0
0 0 0
0 0 0 −2
are both Jordan matrices. The matrix A has 3 Jordan blocks; the first is 2 × 2 with λ1 = 3,
the second is 1 × 1 with λ = 3 and the third is 1 × 1 with λ = −2. The matrix B has 1 Jordan
block, it is 3 × 3 with λ = 0.
Example: Any diagonal matrix is in Jordan form, whose Jordan blocks are all 1 × 1.
We now want to prove the main result of this lecture:
Theorem 1. Let V be a finite dimensional vector space over a field F . Let T be a linear operator
on V such that its characteristic polynomial is a product of linear factors in F [x]. Then there
exists a basis B of V such that the matrix representing T in this basis is in Jordan form.
Proof: With the notation above, it will be enough to show that we can find a basis for each
subspace Ci such that the corresponding matrix Ai representing the restriction of T to Ci is
in Jordan form. Since N restricted to Ci is nilpotent, there exists a minimal integer m such
that N m = 0 on Ci . If m = 1, then N = 0 on Ci which means that T = λi I on Ci , so it is
in Jordan form (actually, diagonal). Suppose m ≥ 2. Since m is minimal, N m−1 6= 0 so there
exists vm ∈ Ci such that N m−1 vm 6= 0. Let vm−1 = N vm , vm−2 = N 2 vm , . . . , v1 = N m−1 vm .
Notice that for every j we have N j vj = 0 and N j−1 vj 6= 0 by construction. We claim that these
vectors are linearly independent. Indeed, suppose that

c1 v1 + . . . + cj vj = 0

where cj 6= 0. Apply N j−1 to both sides of this equation. This implies that cj N j−1 vj = 0,
which in turn says that cj = 0, and this is a contradiction. Hence {v1 , . . . , vm } is linearly inde-
pendent. Consider the action of N on the subspace Span(v1 , . . . , vm ). Since N v1 = v2 , N v2 =
v3 , . . . , N vm−1 = vm we see that N is represented by the matrix
 
0 1 0 ... 0
0 0 1 ... 0 
 
. . . . . . . . . . . . . . .
 
 0 ... ... 0 1
0 ... ... ... 0

on the subspace Span(v1 , . . . , vm ). Since T = λi I + N , it is represented by a Jordan block on

the same subspace.
Next, we want to pass to the quotient space Ci /Span(v1 , . . . , vm ). If this space is trivial,
then Ci = Span(v1 , . . . , vm ) and there is nothing left to be proven. Suppose this space is not
trivial. The operator N induces a nilpotent operator on Ci /Span(v1 , . . . , vm ), say m0 is the
0
minimal integer such that N m = 0 on this space. Clearly m0 ≤ m. Say that v ∈ Ci is a
0
vector such that N m −1 v 6= 0 where v denotes the equivalence class of v in the quotient space
0
Ci /Span(v1 , . . . , vm ). Lifting back to Ci , the vector N m v must belong to Span(v1 , . . . , vm ) so
we must have
0
N m v = a1 v1 + a2 v2 + . . . + am vm
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for some a1 , . . . , am ∈ F . We claim that am = am−1 = . . . = am−m0 +1 = 0. Suppose not, and

say that j ≥ m − m0 + 1 is the greatest integer such that aj 6= 0. Apply N j−1 to both sides of
the equation above. This will give
0
N m +j−1 v = aj N j−1 vj 6= 0.

But m0 + j − 1 ≥ m so N m v 6= 0, a contradiction. So am = am−1 = . . . = am−m0 +1 = 0. Next,

define the vector

wm0 = v − am−m0 vm − am−m0 −1 vm−1 − . . . − a1 vm0 +1 .

0
It is clear that wm0 = v in Ci /Span(v1 , . . . , vm ). Also, by direct computation we have N m wm0 =
0. Now, set
wm0 −1 = N wm0 , wm0 −2 = N wm0 −1 , . . . , w1 = N w2 .
Then Span(w1 , . . . , wm0 ) is a T -invariant subspace and on this subspace T is represented by a
Jordan block with respect to the basis {w1 , . . . , wm0 }, just like in the previous paragraph.
One proceeds inductively in the same manner by passing to Ci /Span(v1 , . . . , vm , w1 , . . . , wm0 ).
We leave out the details. 2