Linear Algebra and Analysis

7.3 Nilpotent Transformations and Jordan Canonical Form

In general, when you have A ∈ℒ

and the minimum polynomial is ∏ _i=1^pϕ_i^k_i, you can decompose V into a direct sum as

V =p ker(ϕ (A )ki) i=1 i

Then on ker

, it follows by definition that ϕ_i is nilpotent.

The following lemma contains some significant observations about nilpotent transformations.

Proof:  Suppose ∑ _i=0^kc_iNⁱx = 0 where not all c_i = 0. There exists l such that k ≤ l < m and N^l+1x = 0 but N^lx≠0. Then multiply both sides by N^l to conclude that c₀ = 0. Next multiply both sides by N^l−1 to conclude that c₁ = 0 and continue this way to obtain that all the c_i = 0.

Next consider the claim that λ^m is the minimum polynomial. If p

is the minimum polynomial, then by the division algorithm,

m λ = p(λ)l(λ )+ r(λ)

where the degree of r

is less than that of p or else r = 0. The above implies 0 = 0 + r contrary to p being minimum. Hence r = 0 and so p divides λ^m. Hence p = λ^k for k ≤ m. But if k < m, this would contradict the definition of m as being the smallest such that N^m = 0. ■

Note how this lemma implies that if N^kx is a linear combination of the preceeding vectors for k as small as possible, then N^kx = 0.

Now suppose V = ker

where ϕ is irreducible and the minimum polynomial for A on V is ϕ^k as in the above. Letting B = ϕ, we can consider a cyclic basis for V of the form where β_x ≡. From Lemma 7.3.2, B^mx = 0. So what is the matrix of B with respect to this basis? It is block diagonal, the blocks coming from the individual β_xi, the size of the blocks being ×,i = 1,,s.

( ) C1 || .. || ( . ) Cp

Using the useful gimmick and ordering the basis using decreasing powers,¹ one of these blocks is of the form C_k where C_k is the matrix on the right

That is, C_k is the × matrix which has ones down the super diagonal and zeros elsewhere. Note that the size of the blocks is determined by just as in the above rational canonical form. In fact, if you ordered the basis in the opposite way, one of these blocks would just be the companion matrix for B just as in the rational canonical form. This is because of Lemma 7.3.2 which requires B^mx = 0 and so the linear combination of B^mx in terms of B^kx for k < m has all zero coefficients. You would have

( ) 0 0 ⋅⋅⋅ 0 || 1 0 ... ... || || . . . || |( .. .. .. 0 |) 0 0 1 0

because the right column in 7.4 is all zero.

The following convenient notation will be used.

Then with this definition and the above discussion, the following proposition has been proved.

In fact, the matrix of the above proposition is unique. This is a general fact for a nilpotent matrix N.

Proof: Since J and J^′ are similar, it follows that for each k an integer, J^k and J^′k are similar. Hence, for each k, these matrices have the same rank. Now suppose J≠J^′. Note first that

Jr (0)r = 0,Jr(0)r−1 ⁄= 0.

Denote the blocks of J as J_rk

and the blocks of J^′ as J_rk^′. Let k be the first such that J_rk≠J_rk^′. Suppose that r_k > r_k^′. By block multiplication and the above observation, it follows that the two matrices J^r_k−1 and J^′r_k−1 are respectively of the forms

( ) ( ) Mr1 0 Mr ′1 0 || .. || || .. || || . || || . || || Mrk || , || Mr′k || || ∗ || || 0 || || .. || || .. || ( . ) ( . ) 0 ∗ 0 0

where M_rj = M_rj^′ for j ≤ k − 1 but M_rk^′ is a zero r_k^′×r_k^′ matrix while M_rk is a larger matrix which is not equal to 0. For example, M_rk could look like

( ) | 0 ⋅⋅⋅ 1 | Mrk = |( ... ... |) 0 0

Thus this contradicts the requirement that J^k and J^′k have the same rank. ■

The Jordan canonical form is available when the minimum polynomial can be factored in the field of scalars. Thus

∏r p(λ) = (λ− μk)mk k=1

and

⊕r ⊕r V = ker(A − μiI)mi ≡ Vi i=1 i=1

Now here is a useful observation.

So what is the matrix description of the Jordan canonical form? From Proposition 7.3.4, the matrix of A − μ_iI on V _i having dimension d_i is of the form

( J (0) 0 ) | r1 | || Jr2 (0) || |( ... |) 0 J (0) rs

where r₁ +

+ r_s = d_i and we can assume the size of the Jordan blocks decreases from upper left to lower right. It follows then from the above observation that the matrix of A = + μ_iI_di is

( ) Jr1 (μi) 0 || Jr2 (μi) || J (μi) ≡ || . || ( .. ) 0 Jrs (μi)

Then the Jordan form of the matrix A is

( ) J (μ1) 0 || .. || ( . ) 0 J (μr)

Each J

has a size which is uniquely determined by the dimension of ker^k_i which comes from the minimum polynomial. Furthermore, each J is a block diagonal matrix in which the blocks have a certain specified form.

Note that if any of the β_k consists of eigenvectors, then the corresponding Jordan block will consist of a diagonal matrix having λ_k down the main diagonal. This corresponds to m_k = 1. The vectors which are in ker

^m_k which are not in ker are called generalized eigenvectors.

The following is the main result on the Jordan canonical form.

Proof: Suppose there are two, J and J^′. Then these are matrices of A with respect to possibly different bases and so they are similar. Therefore, they have the same minimum polynomials and the generalized eigenspaces ker

^k_i have the same dimension. Thus the size of the matrices J and J^′ defined by the dimension of these generalized eigenspaces, also corresponding to the algebraic multiplicity of λ_k, must be the same. Therefore, they comprise the same set of positive integers. Thus listing the eigenvalues in the same order, corresponding blocks J,J^′ are the same size.

It remains to show that J

and J^′ are not just the same size but also are the same up to order of the Jordan blocks running down their respective diagonals. It is only necessary to worry about the number and size of the Jordan blocks making up J and J^′. Since J,J^′ are similar, so are J −λ_kI and J^′− λ_kI.

Thus the following two matrices are similar

( ) J (λ1)− λkI 0 || ... || || || A ≡ || J (λk)− λkI || |( ... |) 0 J (λ ) − λ I r k

( ) J′(λ1)− λkI 0 || .. || || . || B ≡ || J ′(λk) − λkI || |( ... |) ′ 0 J (λr)− λkI

and consequently, rank

= rank for all k ∈ ℕ. Also, both J − λ_kI and J^′ − λ_kI are one to one for every λ_j≠λ_k. Since all the blocks in both of these matrices are one to one except the blocks J^′ − λ_kI, J − λ_kI, it follows that this requires the two sequences of numbers _m=1^∞ and _m=1^∞ must be the same.

Then

( ) Jk1 (0) 0 || Jk2 (0) || J (λk)− λkI ≡ || ... || ( ) 0 Jkr (0)

and a similar formula holds for J^′

( ) Jl1 (0) 0 || Jl2 (0) || J′(λk)− λkI ≡ || . || ( .. ) 0 Jlp (0)

and it is required to verify that p = r and that the same blocks occur in both. Without loss of generality, let the blocks be arranged according to size with the largest on upper left corner falling to smallest in lower right. Now the desired conclusion follows from Corollary 7.3.6. ■

Note that if any of the generalized eigenspaces ker

^m_k has a basis of eigenvectors, then it would be possible to use this basis and obtain a diagonal matrix in the block corresponding to μ_k. By uniqueness, this is the block corresponding to the eigenvalue μ_k. Thus when this happens, the block in the Jordan canonical form corresponding to μ_k is just the diagonal matrix having μ_k down the diagonal and there are no generalized eigenvectors as fussed over in ordinary differential equations. Recall that these were vectors in ker^m_k but not in ker.

The Jordan canonical form is very significant when you try to understand powers of a matrix. There exists an n×n matrix S² such that

A = S−1JS.

Therefore, A² = S⁻¹JSS⁻¹JS = S⁻¹J²S and continuing this way, it follows

Ak = S−1JkS.

where J is given in the above corollary. Consider J^k. By block multiplication,

( ) Jk1 0 Jk = || ... || . ( k ) 0 Jr

The matrix J_s is an m_s × m_s matrix which is of the form

Js = D + N (7.6)

(7.6)

for D a multiple of the identity and N an upper triangular matrix with zeros down the main diagonal. Thus N^m_s = 0. Now since D is just a multiple of the identity, it follows that DN = ND. Therefore, the usual binomial theorem may be applied and this yields the following equations for k ≥ m_s.

the third equation holding because N^m_s = 0. Thus J_s^k is of the form

( ) αk ⋅⋅⋅ ∗ Jk = || .. .. .. || . s ( . . . ) 0 ⋅⋅⋅ αk

Proof: From 7.7, it follows that for large k, and j ≤ m_s,

( ) k ≤ k(k−-1)⋅⋅⋅(k-−-ms-+-1). j ms!

Therefore, letting C be the largest value of

for 0 ≤ j ≤ m_s,

| | ( ) ||(Jk) || ≤ msC k(k-− 1)-⋅⋅⋅(k−-ms-+-1) |α|k−ms pq ms!

which converges to zero as k →∞. This is most easily seen by applying the ratio test to the series

∑∞ ( ) k-(k-−-1)⋅⋅⋅(k-−-ms +-1) |α|k−ms k=ms ms!

and then noting that if a series converges, then the k^th term converges to zero. ■