Download Proof of the Jordan canonical form

PROOF OF THE JORDAN CANONICAL FORM
EDWARD FRENKEL
Let V be a finite-dimensional vector space over R or over C and T a linear operator
V →V.
We will use the definition of generalized eigenvector and the subspace Kλ of generalized
eigenvectors of T on V corresponding to the eigenvalue λ from Chapter 7 of Linear Algebra
by Friedberg, e.a. We will also use the definition of a cycle of generalized eigenvectors of
T , the initial vector, and the end vector from the same chapter.
Theorem 1. Let γ1 , . . . , γq be cycles of generalized eigenvectors of T whose initial vectors
are linearly independent. Then their union is linearly independent as well.
Proof. See the proof of Theorem 7.6.
Theorem 2. Let λ1 , . . . , λk be the distinct roots of the characteristic polynomial of T and
Kλ1 , . . . , Kλk the corresponding generalized eigenspaces. Suppose that we have vectors vi ∈
Kλi such that
v1 + v2 + . . . + vj = 0
for some j ≤ k. Then v1 = v2 = . . . = vj = 0.
Proof. We will prove this by induction on j. For j = 1 there is nothing to prove. Suppose
that the statement is true for j = m < k. Let us prove it for j = m + 1. Consider the linear
operator T − λm+1 I. Note that
(T − λi I)(T − λm+1 I) = (T − λm+1 I)(T − λi I).
Simply open the brakets and use the fact that T commutes with I. Applying this several
times, we obtain
(T − λi I)p (T − λm+1 I) = (T − λm+1 I)(T − λi I)p
for any p = 1, 2, . . ..
If v ∈ Kλi , then (T − λi I)p (v) = 0 for some p. Using the above formula, we then obtain
that (T − λi I)p · (T − λm+1 I)(v) = 0. Therefore Kλi is invariant under (T − λm+1 I) for all
i.
Consider now the restriction of (T − λm+1 I) to Kλi , where i 6= m + 1. Suppose that
v ∈ Kλi is in the null-space of (T − λm+1 I). Then T (v) = λm+1 v, and so we find:
(T − λi I)p (v) = (λm+1 − λi )p v.
Since λi 6= λm+1 by our assumption, (T − λi I)p (v) = 0 if and only if v = 0. Therefore
the null-space of the restriction of (T − λm+1 I) to Kλi with i 6= m + 1 is {0}. Hence the
restriction of (T −λm+1 I) to Kλi with i 6= m+1 is an isomorphism. Therefore the restriction
of (T − λm+1 I)p to Kλi with i 6= m + 1 is also an isomorphism for any p = 1, 2, . . ..
1
2
EDWARD FRENKEL
Now consider the equation
(1)
v1 + v2 + . . . + vm+1 = 0,
where vi ∈ Kλi . Since vm+1 ∈ Kλm+1 , we have (T − λm+1 I)p (vm+1 ) = 0 for some p.
Let us apply (T − λm+1 I)p to the left hand side of formula (1). We obtain
0
v10 + v20 + . . . + vm
= 0,
where vi0 = (T − λm+1 I)p (vi ). By our inductive assumption, this implies that vi0 = 0 for all
i = 1, 2, . . . , m. Since the restriction of (T −λm+1 I)p to Kλi , i = 1, . . . , m is an isomorphism,
this entails vi = 0 for all i = 1, . . . , m. Formula (1) then implies that vm+1 = 0. This
completes the proof.
Theorem 3. There exists a basis of V that is a union of cycles of T .
Proof. We will prove this by induction on the dimension of V . If the dimension of V is equal
to one, then T is the operator of multiplication by a number, and therefore any non-zero
vector of V is an eigenvector. The statement of the theorem follows.
Now suppose we have proved this statement for all vector spaces of dimension less than
the dimension of V . Let us prove it for V .
Let W = R(T − λ1 I). Since λ1 is a root of the characteristic polynomail of T , W is a
subspace of V of dimension strictly less than dim V . Furthermore, W is T -invariant. Hence,
by our inductive assumption, W has a basis β 0 that is a union of cycles of T . We now extend
it to a basis of V that is also a union of cycles.
Let γ10 , . . . , γq0 be the cycles in β 0 corresponding to the eigenvalue λ1 . We have
γi0 = {zi1 , . . . , zi`i },
where zi1 is the initial vector and zi`i is the end vector. Since zi`i ∈ W = R(T − λ1 I), there
exists a vector yi ∈ V such that zi`i = (T − λ1 I)(yi ). Hence
γi = {zi1 , . . . , zi`i , yi },
i = 1, . . . , q,
is a cycle in V (corresponding to the same eigenvalue λ1 ).
Note also that zi1 ∈ N (T − λ1 I) for all i = 1, . . . , q, and these vectors are linearly
independent by our assumption. Denote dim V = n, dim W = r. Then by dimension
theorem, we have dim N (T − λ1 I) = n − r. If q < n − r, then we extend the linearly
independent subset {z11 , . . . , zq1 } of N (T − λ1 I) to a basis of N (T − λ1 I):
{z11 , . . . , zq1 , xq+1 , . . . , xn−r },
and introduce new cycles of length one (note: if a cycle consists of a single vector, then this
vector is necessarily an eigenvector) γq+1 = {xq+1 }, . . . , γn−r = {xn−r }.
Now let β be the union of the above cycles γ1 , γ2 , . . . , γn−r and all the cycles corresponding
to the eigenvalues λ2 , . . . , λk appearing in the basis β 0 of W .
By construction, the initial vectors of the cycles γ1 , γ2 , . . . , γn−r are linearly independent.
Thereofore, by Theorem 1, the union of these cycles is also linearly independent. Furthermore, by Theorem 2, this union is also linearly independent from the union of the cycles
corresponding to the eigenvalues λ2 , . . . , λk appearing in the basis β 0 of W . Thus, the set
β is linearly independent.
PROOF OF THE JORDAN CANONICAL FORM
3
Finally, observe that β contains r + q + (n − r − q) = n elements. Since dim V = n, it
follows that β is a basis of V . The theorem is proved.
Theorem 4. If β is the basis of Theorem 3, then the matrix [T ]β has Jordan canonical
form.
This follows from the fact that each block corresponding to a given cycle in the basis β
is a Jordan block.
Corollary 5. We have
and dim Kλi
V = Kλ1 ⊕ Kλ1 ⊕ . . . ⊕ Kλk ,
= mλi , the algebraic multiplicity of λi in the characteristic polynomial of T .
Proof. By Theorem 3, V has a basis of cycles. But each element of every cycle if a generalized
k
X
eigenvector of T , so it belongs to some Kλi . Therefore V =
Kλi . Theorem 2 together
i=1
with Theorem 5.10(b) from the book then imply that V =
k
M
Kλi .
i=1
Therefore
k
X
dim Kλi = n = dimV .
i=1
On the other hand, dim Kλi ≤ mλi (see Theorem 7.2(a) for a proof). But since
k
X
mλi = n
i=1
(the degree of the characteristic polynomial), this immediately implies that dim Kλi =
mλi .