7 Jordan Canonical form

We start this chapter with the following theorem which generalizes the Schur Upper triangularization theorem.

Theorem 7.1.1. [Generalized Schur’s Theorem] Let A ∈ M_n(ℂ). Suppose λ₁,…,λ_k are the distinct eigenvalues of A with multiplicities m₁,…,m_k, respectively. Then, there exists a non-singular matrix W such that

⊕k W - 1AW = Ti, where, Ti ∈ Mmi (ℂ ), for 1 ≤ i ≤ k i=1

and T_i’s are upper triangular matrices with constant diagonal λ_i. If A has real entries with real eigenvalues then W can be chosen to have real entries.

Proof. By Schur Upper Triangularization (see Lemma 6.2.12), there exists a unitary matrix U such that U^*AU = T, an upper triangular matrix with diag(T) = (λ₁,…,λ₁,…,λ_k,…,λ_k).

Now, for any upper triangular matrix B, a real number α and i < j, consider the matrix F(B,i,j,α) = E_ij(-α)BE_ij(α), where the matrix E_ij(α) is defined in Definition 2.1.13. Then, for 1 ≤ k,ℓ ≤ n,

Now, using Equation (7.1.1), the diagonal entries of F(T,i,j,α) and T are equal and

Exercise 7.1.2. Apply Theorem 7.1.1 to the matrix given below for better understanding. PICT PICT DRAFT

⌊ | | ⌋ |1 2 3 |4 5 6 |7 8 9| |0 1 2 |3 4 5 |6 7 8| ||0 0 1 |2 3 4 |5 6 7|| ||--------|--------|-------|| ||0 0 0 |2 3 4 |5 6 7|| ||0 0 0 |0 2 3 |4 5 6|| . ||0 0 0 |0 0 2 |3 4 5|| |--------|--------|-------| ||0 0 0 |0 0 0 |3 4 5|| |⌈0 0 0 |0 0 0 |0 3 4|⌉ | | 0 0 0 |0 0 0 |0 0 3

Definition 7.1.3. [Jordan Block and Jordan Matrix]

1.: Let λ ∈ ℂ and k be a positive integer. Then, by the Jordan block J_k(λ) ∈ M_k(ℂ), we understand the matrix $⌊ ⌋ λ 1 || .. .. || | . . | . |⌈ λ 1|⌉ λ$
2.: A Jordan matrix is a direct sum of Jordan blocks. That is, if A is a Jordan matrix having r blocks then there exist positive integers k_i’s and complex numbers λ_i’s (not necessarily distinct), for 1 ≤ i ≤ r such that DRAFT $A = Jk1(λ1 )⊕ ⋅⋅⋅⊕ Jkr(λr).$

Example 7.1.4.

1.: J₁(0) = is the only Jordan matrix of size 1.
2.: J₁(0) ⊕ J₁(0) = and J₂(0) = are Jordan matrices of size 2.
3.: Even though , J₁(0) ⊕ J₂(0) and J₂(0) ⊕ J₁(0) are two Jordan matrices of size 3, we do not differentiate between them as they are similar (use permutations).
4.: J₁(0) ⊕J₁(0) ⊕J₁(0) = , J₂(0) ⊕J₁(0) = and J₃(0) = are Jordan matrices of size 3.
5.: Observe that the number of Jordan matrices of size 4 with 0 on the diagonal are 5.

We now give some properties of the Jordan blocks. The proofs are immediate and hence left for the reader. They will be used in the proof of subsequent results.

Remark 7.1.5. [Jordan blocks] Fix a positive integer k. Then,

1.

J_k(λ) is an upper triangular matrix with λ as an eigenvalue. PICT

DRAFT

2.

J_k(λ) = λI_k + J_k(0).

3.

Alg.Mul_λ(J_k(λ)) = k.

4.

The matrix J_k(0) satisfies the following properties.

(a): Rank((J_k(0)ⁱ) = k - i, for 1 ≤ i ≤ k.
(b): J_k^T(0)J_k(0) = .
(c): J_k(0)^p = 0 whenever p ≥ k.
(d): J_k(0)e_i = e_i-1 for i = 2,…,k.
(e): I - J_k^T(0)J_k(0)x = = ⟨x,e₁⟩e₁.

5.

Thus, using Remark 7.1.5.4d Geo.Mul_λ(J_k(λ)) = 1.

Exercise 7.1.6.

1.

Fix a positive integer k and a complex number λ. Then, prove that

(a): Rank(J_k(λ) - λI_k) = k - 1.
(b): Rank(J_k(λ) - αI_k) = k, whenever α≠λ. Or equivalently, for all α≠λ the matrix J_k(λ) - αI_k is invertible.
(c): for 1 ≤ i ≤ k, Rank((J_k(λ) - λI_k)ⁱ) = k - i.
(d): for α≠λ, Rank((J_k(λ) - αI_k)ⁱ) = k, for all i.

2.

Let J be a Jordan matrix that contains ℓ Jordan blocks for λ. Then, prove that PICT

DRAFT

(a): Rank(J - λI) = n - ℓ.
(b): J has ℓ linearly independent eigenvectors for λ.
(c): Rank(J - λI) ≥Rank((J - λI)²) ≥Rank((J - λI)³) ≥.

3.

Let A ∈ M_n(ℂ). Then, prove that AJ_n(λ) = J_n(λ)A if and only if AJ_n(0) = J_n(0)A.

Definition 7.1.7. [Index of an Eigenvalue] Let J be a Jordan matrix containing J_t(λ), for some positive integer t and some complex number λ. Then, the smallest value of k for which Rank((J - λI)^k) stops decreasing is the order of the largest Jordan block J_k(λ) in J. This number k is called the index of the eigenvalue λ.

Lemma 7.1.8. Let A ∈ M_n(ℂ) be strictly upper triangular. Then, A is similar to a direct sum of Jordan blocks. That is, there exists a non-singular matrix S and integers n₁ ≥… ≥ n_m ≥ 1 such that

A = S- 1(J (0)⊕ ⋅⋅⋅⊕ J (0))S. n1 nm

If A ∈ M_n(ℝ) then S can be chosen to have real entries.

Proof. We will prove the result by induction on n. For n = 1, the statement is trivial. So, let the result be true for matrices of size ≤ n - 1 and let A ∈ M_n(ℂ) be strictly upper triangular. Then, A = [ T]
0 a
0 A1

. By induction hypothesis there exists an invertible matrix S₁ such that

So, let us now assume that ⟨a₁,e₁⟩≠0. Then, writing α = ⟨a₁,e₁⟩, we have

DRAFT

Exercise 7.1.9. Convert ⌊ ⌋
0 1 1
|⌈0 0 1|⌉

0 0 0 to J₃(0) and ⌊ ⌋
0 1 2
|⌈ 0 0 0|⌉

0 0 0 to J₂(0) ⊕ J₁(0).

Corollary 7.1.10. A ∈ M_n(ℂ). Then, A is similar to J, a Jordan matrix.

Proof. Let λ₁,…,λ_k be the distinct eigenvalues of A with algebraic multiplicities m₁,…,m_k. By Theorem 7.1.1, there exists a non-singular matrix S such that S^-1AS = ⊕ _i=1^kT_i, where T_i is an upper triangular with diagonal (λ_i,…,λ_i). Thus T_i - λ_iI_{m_i} is a strictly upper triangular matrix. Thus, by Theorem 7.1.8, there exist a non-singular matrix S_i such that

Let A ∈ M_n(ℂ). Suppose λ ∈ σ(A) and J is a Jordan matrix that is similar to A. Then, for each fixed i,1 ≤ i ≤ n, by ℓ_i(λ), we denote the number of Jordan blocks J_k(λ) in J for which k ≥ i. Then, the next result uses Exercise 7.1.6 to determine the number ℓ_i(λ).

Remark 7.1.11. Let A ∈ M_n(ℂ). Suppose λ ∈ σ(A) and J is a Jordan matrix that is similar to A. Then, for 1 ≤ k ≤ n, PICT PICT DRAFT

k-1 k ℓk(λ) = Rank(A - λI) - Rank(A - λI) .

Proof. In view of Exercise 7.1.6, we need to consider only the Jordan blocks J_k(λ), for different values of k. Hence, without loss of generality, let us assume that J = ⊕ _i=1ⁿa_iJ_i(λ), where a_i’s are non-negative integers and J contains exactly a_i copies of the Jordan block J_i(λ), for 1 ≤ i ≤ n. Then, by definition and Exercise 7.1.6, we observe the following:

Lemma 7.1.12. [Similar Jordan Matrices] Let J and J′ be two similar Jordan matrices of size n. Then, J is a block permutation of J′.

Proof. For 1 ≤ i ≤ n, let ℓ_i and ℓ_i′ be, respectively, the number of Jordan blocks of J and J′ of size at least i corresponding to λ. Since J and J′ are similar, the matrices (J -λI)ⁱ and (J′-λI)ⁱ are similar for all i,1 ≤ i ≤ n. Therefore, their ranks are equal for all i ≥ 1 and hence, ℓ_i = ℓ_i′ for all i ≥ 1. Thus the required result follows. _

We now state the main result of this section which directly follows from Lemma 6.2.12, Theorem 7.1.1 and Corollary 7.1.10 and hence the proof is omitted.

Theorem 7.1.13. [Jordan Canonical Form Theorem] Let A ∈ M_n(ℂ). Then, A is similar to a Jordan matrix J, which is unique up to permutation of Jordan blocks. If A ∈ M_n(ℝ) and has real eigenvalues then the similarity transformation matrix S may be chosen to have real entries. This matrix J is called the the Jordan canonical form of A, denoted Jordan CF(A).

Example 7.1.14. Let us use the idea from Lemma 7.1.11 to find the Jordan Canonical Form of the following matrices.

1.

Let A = J₄(0)² = ⌊ ⌋
0 0 1 0
||0 0 0 1||
|| ||
⌈0 0 0 0⌉
0 0 0 0

Solution: Note that ℓ₁ = 4 -Rank(A - 0I) = 2. So, there are two Jordan blocks.

Also, ℓ₂ = Rank(A - 0I) -Rank((A - 0I)²) = 2. So, there are at least 2 Jordan blocks of size 2. As there are exactly two Jordan blocks, both the blocks must have size 2. Hence, Jordan CF(A) = J₂(0) ⊕ J₂(0). PICT PICT DRAFT

2.

Let A₁ =

⌊ ⌋
1 1 0 1
||0 1 1 1||
|| ||
⌈0 0 1 1⌉
0 0 0 1

Solution: Let B = A₁ - I. Then, ℓ₁ = 4 -Rank(B) = 1. So, B has exactly one Jordan block and hence A₁ is similar to J₄(1).

3.

A₂ =

⌊ ⌋
|1 1 0 1|
|0 1 1 1|
||0 0 1 0||
⌈ ⌉
0 0 0 1

Solution: Let C = A₂ - I. Then, ℓ₁ = 4 -Rank(C) = 2. So, C has exactly two Jordan blocks. Also, ℓ₂ = Rank(C) -Rank(C²) = 1 and ℓ₃ = Rank(C²) -Rank(C³) = 1. So, there is at least 1 Jordan blocks of size 3.

Thus, we see that there are two Jordan blocks and one of them is of size 3. Also, the size of the matrix is 4. Thus, A₂ is similar to J₃(1) ⊕ J₁(1).

4.

Let A = J₄(1)² ⊕ A₁ ⊕ A₂, where A₁ and A₂ are given in the previous exercises.

Solution: One can directly get the answer from the previous exercises as the matrix A is already in the block diagonal form. But, we compute it again for better understanding.

Let B = A - I. Then, ℓ₁ = 16 -Rank(B) = 5, ℓ₂ = Rank(B) -Rank(B²) = 11 - 7 = 4, ℓ₃ = Rank(B²) -Rank(B³) = 7 - 3 = 4 and ℓ₄ = Rank(B³) -Rank(B⁴) = 3 - 0 = 3.

Hence, J₄(1) appears thrice (as ℓ₄ = 3 and ℓ₅ = 0), J₃(1) also appears once (as ℓ₃-ℓ₄ = 1), J₂(1) does not appear as (as ℓ₂ - ℓ₃ = 0) and J₁(1) appears once (as ℓ₁ - ℓ₂ = 1). Thus, the required result follows.

Remark 7.1.15. [Observations about Jordan CF(A)]

1.: What are the steps to find Jordan CFA?
Łet λ₁,…,λ_k be the distinct eigenvalues of A. Now, apply the Schur Upper Triangularization Lemma (see Lemma 6.2.12) to get an upper triangular matrix, say T such that the diagonal entries of T are λ₁,…,λ₁,λ₂,…,λ₂,…,λ_k,…,λ_k. Now, apply similarity transformations (see Theorem 7.1.1) to get T = ⊕ _i=1^kT_i, where each diagonal entry of T_i is λ_i. Then, for each DRAFT i,1 ≤ i ≤ k, use Theorem 7.1.8 to get an invertible matrix S_i such that S_i^-1(T_i-λ_iI)S_i = , a Jordan matrix. Thus, we obtain a Jordan matrix J_i = + λ_iI = S_i^-1T_iS_i, where each diagonal entry of J_i is λ_i. Hence, S = ⊕ _i=1^kS_i converts T = ⊕ _i=1^kT_i into the required Jordan matrix.
2.: Let A ∈ M_n(ℂ) be a diagonalizable matrix. Then, by definition, A is similar to ⊕ _i=1ⁿλ_i, where λ_i ∈ σ(A), for 1 ≤ i ≤ n. Thus, Jordan CF(A) = ⊕ _i=1ⁿλ_i, up to a permutation of λ_i’s.
3.: In general, the computation of Jordan CF(A) is not numerically stable. To understand this, let A_ϵ = . Then, A_ϵ is diagonalizable as A has distinct eigenvalues. So, Jordan CF(A_ϵ) = .
Whereas, for A = , we know that Jordan CF(A) = ≠lim_ϵ→0Jordan CF(A_ϵ). Thus, a small change in the entries of A may change Jordan CF(A) significantly.
4.: Let A ∈ M_n(ℂ) and ϵ > 0 be given. Then, there exists an invertible matrix S such that S^-1AS = ⊕ _i=1^kJ_{n_i}(λ_i,ϵ), where J_{n_i}(λ_i,ϵ) is obtained from J_{n_i}(λ_i) by replacing each off diagonal entry 1 by an ϵ. To get this, define Di(ϵ) = diag(1,ϵ,ϵ²,…,ϵ^n_i-1), for 1 ≤ i ≤ k. Now compute ⊕ _i=1^k.
5.: Let Jordan CF(A) contain ℓ Jordan blocks for λ. Then, A has ℓ linearly independent eigenvectors for λ.
For if, A has at least ℓ + 1 linearly independent eigenvectors for λ, then dim(Null(A - λI)) > ℓ. So, Rank(A - λI) < n - ℓ. But, the number of Jordan blocks for λ in A is ℓ. Thus, we must have Rank(J - λI) = n - ℓ, a contradiction.
6.: Let λ ∈ σ(A). Then, by Remark 7.1.5.5, Geo.Mul_λ(A) = the number of Jordan blocks J_k(λ) in Jordan CF(A).
7.: Let λ ∈ σ(A). Then, by Remark 7.1.5.3, Alg.Mul_λ(A) = the sum of the sizes of all Jordan blocks J_k(λ) in Jordan CF(A).
8.: Let λ ∈ σ(A). Then, Jordan CF(A) does not get determined by Alg.Mul_λ(A) and Geo.Mul_λ(A). For example, ⊕⊕ and ⊕⊕ DRAFT are different Jordan CFs but they have the same algebraic and geometric multiplicities.
9.: Let A ∈ M_n(ℂ). Suppose that, for each λ ∈ σ(A), the values of Rank(A - λI)^k, for k = 1,…,n are known. Then, using Remark 7.1.11, Jordan CF(A) can be computed. But, note here that finding rank is numerically unstable as has rank 1 but it converges to which has a different rank.

Theorem 7.1.16. [A is similar to A^T] Let A ∈ M_n(ℂ). Then, A is similar to A^T.

Proof. Let K_n = ⌊ ⌋
| 1|
|⌈ ... |⌉

1

. Then, observe that K^-1 = K and KJ_n(a)K = J_n(a)^T, as the (i,j)-th entry of A goes to (n - i + 1,n - j + 1)-th position in KAK. Hence,

7.2 Minimal polynomial

We start this section with the following definition. Recall that a polynomial p(x) = a₀ + a₁x + ⋅⋅⋅

+ a_nxⁿ with a_n = 1 is called a monic polynomial.

PICT PICT DRAFT Definition 7.2.1. [Companion Matrix] Let P(t) = tⁿ + a_n-1t^n-1 + ⋅⋅⋅ + a₀ be a monic polynomial in t of degree n. Then, the n×n matrix A = ⌊ ⌋
| 0 0 0 ⋅⋅⋅ 0 - a0 |
| 1 0 0 ⋅⋅⋅ 0 - a1 |
|| 0 1 0 ⋅⋅⋅ 0 - a2 ||
|| . . . . ||
|| 0 0 .. .. .. .. ||
|⌈ 0 0 0 ⋅⋅⋅ 0 - a |⌉
n-2
0 0 0 1 - an-1 , denoted A(n : a₀,…,a_n-1) or Companion(P), is called the companion matrix of P(t).

Definition 7.2.2. [Annihilating Polynomial] Let A ∈ M_n(ℂ). Then, the polynomial P(t) is said to annihilate (destroy) A if P(A) = 0.

Let P(x) be the characteristic polynomial of A. Then, by the Cayley-Hamilton Theorem, P(A) = 0. So, if f(x) = P(x)g(x), for any multiple of g(x), then f(A) = P(A)g(A) = 0g(A) = 0. Thus, there are infinitely many polynomials which annihilate A. In this section, we will concentrate on a monic polynomial of least positive degree that annihilates A.

Definition 7.2.3. [Minimal polynomial] Let A ∈ M_n(ℂ). Then, the minimal polynomial of A, denoted m_A(x), is a monic polynomial of least positive degree satisfying m_A(A) = 0.

Theorem 7.2.4. Let A be the companion matrix of the monic polynomial P(t) = tⁿ + a_n-1t^n-1 + ⋅⋅⋅ + a₀. Then, P(t) is both the characteristic and the minimal polynomial of A.

We will now show that P(t) is the minimal polynomial of A. To do so, we first observe that Ae₁ = e₂,…,Ae_n-1 = e_n. That is,

Now, Suppose we have a monic polynomial Q(t) = t^m + b_m-1t^m-1 + ⋅⋅⋅

+ b₀, with m < n, such that Q(A) = 0. Then, using Equation (7.2.1), we get

The next result gives us the existence of such a polynomial for every matrix A. To do so, recall that the well-ordering principle implies that if S is a subset of natural numbers then it contains a least element.

Lemma 7.2.5. [Existence of the Minimal Polynomial] Let A ∈ M_n(ℂ). Then, there exists a unique monic polynomial m(x) of minimum (positive) degree such that m(A) = 0. Further, if f(x) is any polynomial with f(A) = 0 then m(x) divides f(x).

Proof. Let P(x) be the characteristic polynomial of A. Then, deg(P(x)) = n and by the Cayley-Hamilton Theorem, P(A) = 0. So, consider the set

Also, without loss of generality, we can assume that m(x) is monic and unique (non-uniqueness will lead to a polynomial of smaller degree in S).

Now, suppose there is a polynomial f(x) such that f(A) = 0. Then, by division algorithm, there exist polynomials q(x) and r(x) such that f(x) = m(x)q(x) + r(x), where either r(x) is identically the zero polynomial of deg(r(x)) < M = deg(m(x)). As

Corollary 7.2.6. [Minimal polynomial divides the Characteristic Polynomial] Let m_A(x) and P_A(x) be, respectively, the minimal and the characteristic polynomials of A ∈ M_n(ℂ).

1.: Then, m_A(x) divides P_A(x).
2.: Further, if λ is an eigenvalue of A then m_A(λ) = 0.

Proof. The first part following directly from Lemma 7.2.5. For the second part, let (λ,x) be an eigen-pair. Then, f(A)x = f(λ)x, for any polynomial of f, implies that

Lemma 7.2.7. Let A and B be two similar matrices. Then, they have the same minimal polynomial.

Proof. Since A and B are similar, there exists an invertible matrix S such that A = S^-1BS. Hence, f(A) = F(S^-1BS) = S^-1f(B)S, for any polynomial f. Hence, m_A(A) = 0 if and only if m_A(B) = 0 and thus the required result follows. _

PICT PICT DRAFT Theorem 7.2.8. Let A ∈ M_n(ℂ) and let λ₁,…,λ_k be the distinct eigenvalues of A. If n_i is the size of the largest Jordan block for λ_i in J = Jordan CFA then

∏k n mA (x) = (x- λi) i. i=1

Proof. Using 7.2.6, we see that m_A(x) = ∏ _i=1^k(x-λ_i)^α_i, for some α_i’s with 1 ≤ α_i ≤ Alg.Mul_{λ_i}(A). As m_A(A) = 0, using Lemma 7.2.7 we have m_A(J) = ∏ _i=1^k (J - λiI)

^α_i = 0. But, observe that for the Jordan block J_{n_i}(λ_i), one has

Thus ∏ _i=1^k(J - λ_iI)^n_i = 0 and ∏_i=1^k(x - λ_i)^n_i divides ∏_i=1^k(x - λ_i)^α_i = m_A(x) and ∏ _i=1^k(x - λ_i)^n_i is a monic polynomial, the result follows. _

As an immediate consequence, we also have the following result which corresponds to the converse of the above theorem.

Theorem 7.2.9. Let A ∈ M_n(ℂ) and let λ₁,…,λ_k be the distinct eigenvalues of A. If the minimal polynomial of A equals ∏ _i=1^k(x - λ_i)^n_i then n_i is the size of the largest Jordan block for λ_i in J = Jordan CFA.

Theorem 7.2.10. Let A ∈ M_n(ℂ). Then, the following statements are equivalent. PICT PICT DRAFT

1.: A is diagonalizable.
2.: Every zero of m_A(x) has multiplicity 1.
3.: Whenever m_A(α) = 0, for some α, then m_A(x)_x=α≠0.

Proof. Part 1 ⇒ Part 2. If A is diagonalizable, then each Jordan block in J = Jordan CFA has size 1. Hence, by Theorem 7.2.8, m_A(x) = ∏ _i=1^k(x - λ_i), where λ_i’s are the distinct eigenvalues of A.

Part 2 ⇒ Part 3. Let m_A(x) = ∏ _i=1^k(x-λ_i), where λ_i’s are the distinct eigenvalues of A. Then, m_A(x) = 0 if and only if x = λ_i, for some i,1 ≤ i ≤ k. In that case, it is easy to verify that d
---
dx

m_A(x)≠0, for each λ_i.

Part 3 ⇒ Part 1. Suppose that for each α satisfying m_A(α) = 0, one has d
---
dx

m_A(α)≠0. Then, it follows that each zero of m_A(x) has multiplicity 1. Also, using Corollary 7.2.6, each zero of m_A(x) is an eigenvalue of A and hence by Theorem 7.2.8, the size of each Jordan block is 1. Thus, A is diagonalizable. _

Remark 7.2.11.

1.

Let f(x) be a monic polynomial and A = Companion(f) be the companion matrix of f. Then, by Theorem 7.2.4) f(A) = 0 and no monic polynomial of smaller degree annihilates A. Thus P_A(x) = m_A(x) = f(x), where P_A(x) is the characteristic polynomial and m_A(x), the minimal polynomial of A.

2.

Let A ∈ M_n(ℂ). Then, A is similar to Companion(f), for some monic polynomial f if and only if m_A(x) = f(x).

Proof. Let B = Companion (f). Then, using Lemma 7.2.7, we see that m_A(x) = m_B(x). But, by Remark 7.2.11.1, we get m_B(x) = f(x) and hence the required result follows.

Conversely, assume that m_A(x) = f(x). But, by Remark 7.2.11.1, m_B(x) = f(x) = P_B(x), the characteristic polynomial of B. Since m_A(x) = m_B(x), the matrices A and B have the same largest Jordan blocks for each eigenvalue λ. As P_B = m_B, we know that for each λ, there is only one Jordan block in Jordan CFB. Thus, Jordan CFA = Jordan CFB and hence A is similar to Companion (f). _

DRAFT

Exercise 7.2.12. The following are some facts and questions.

1.: Let A ∈ M_n(ℂ). If P_A(x) is the minimal polynomial of A then A is similar to Companion (P_A) if and only if A is nonderogatory. T/F?
2.: Let A,B ∈ M₃(ℂ) with eigenvalues 1,2,3. Is it necessary that A is similar to B?
3.: Let A,B ∈ M₃(ℂ) with eigenvalues 1,1,3. Is it necessary that A is similar to B?
4.: Let A,B ∈ M₄(ℂ) with the same minimal polynomial. Is it necessary that A is similar to B?
5.: Let A,B ∈ M₃(ℂ) with the same minimal polynomial. Is it necessary that A is similar to B?
6.: Let A ∈ M_n(ℂ) be idempotent and let J = Jordan CFA. Thus, J² = J and hence conclude that J must be a diagonal matrix. Hence, every idempotent matrix is diagonalizable.
7.: Let A ∈ M_n(ℂ). Suppose that m_A(x)|x(x - 1)(x - 2)(x - 3). Must A be diagonalizable?
8.: Let A ∈ M₉(ℂ) be a nilpotent matrix such that A⁵≠0 but A⁶ = 0. Determine P_A(x) and m_A(x).
9.: Recall that for A,B ∈ M_n(ℂ), the characteristic polynomial of AB and BA are the same. That is, P_AB(x) = P_BA(x). However, they need not have the same minimal polynomial. Take A = and B = to verify that m_AB(x)≠m_BA(x).

We end this section with a method to compute the minimal polynomial of a given matrix.

PICT PICT DRAFT Example 7.2.13. [Computing the Minimal Polynomial] Let λ₁,…,λ_k be the distinct eigenvalues of A ∈ M_n(ℂ).

7.3 Applications of Jordan Canonical Form

In the last section, we say that the matrices if A is a square matrix then A and A^T are similar. In this section, we look at some more applications of the Jordan Canonical Form.

7.3.1 Coupled system of linear differential equations

Consider the first order Initial Value Problem (IVP) x′(t) = ⌊ ⌋
x ′1(t)
|| . ||
⌈ .. ⌉
x ′(t)
n

= A

= Ax(t), with x(0) = 0. If A is not a diagonal matrix then the system is called coupled and is hard to solve. Note that if A can be transformed to a nearly diagonal matrix, then the amount of coupling among x_i’s can be reduced. So, let us look at J = Jordan CF(A) = S^-1AS. Then, using S^-1A = JS^-1. verify that the initial problem x′(t) = Ax(t) is equivalent to the equation S^-1x′(t) = S^-1Ax(t) which in turn is equivalent to y′(t) = Jy(t), where S^-1x(t) = y(t) with y(0) = S^-1x(0) = 0. Therefore, if y is a solution to the second equation then x(t) = Sy is a solution to the initial problem.

When J is diagonalizable then solving the second is as easy as solving y_i′(t) = λ_iy_i(t) for which the required solution is given by y_i(t) = y_i(0)e^λ_it.

7.3.2 Commuting matrices

Let P(x) be a polynomial and A ∈ M_n(ℂ). Then, P(A)A = AP(A). What about the converse? That is, suppose we are given that AB = BA for some B ∈ M_n(ℂ). Does it necessarily imply that B = P(A), for some nonzero polynomial P(x)? The answer is No as I commutes with A for every A. We start with a set of remarks.

Theorem 7.3.1. Let A ∈ M_n(ℂ) and B ∈ M_m(ℂ). Then, the linear system AX -XB = 0, in the variable matrix X of size n×m, has a unique solution, namely X = 0 (the trivial solution), if and only if σ(A) and σ(B) are disjoint.

Since σ(A) and σ(B) are disjoint, the matrix P_B(A) = ( )
∏
[λI - A ]
λ∈σ(B)

, obtained by evaluating A at the characteristic polynomial, P_B(t), of B, is invertible. So, let us look at the implication of the condition AX = XB. This condition implies that A²X = AXB = XBB = XB² and hence, P(A)X = XP(B), for any polynomial P(t). In particular, P_B(A)X = XP_B(B) = X0 = 0. As P_B(A) is invertible, we get X = 0.

Now, conversely assume that AX - XB = 0 has only the trivial solution X = 0. Suppose on the contrary λ is a common eigenvalue of both A and B. So, choose nonzero vectors x ∈ ℂⁿ and y ∈ ℂ^m such that (λ,x) is an eigen-pair of A and (λ,y) is a left eigen-pair of B. Now, define X₀ = xy^T. Then, X₀ is an n × m nonzero matrix and

Corollary 7.3.2. Let A ∈ M_n(ℂ),B ∈ M_m(ℂ) and C be an n × m matrix. Also, assume that PICT PICT DRAFT σ(A) and σ(B) are disjoint. Then, it can be easily verified that the system AX - XB = C, in the variable matrix X of size n × m, has a unique solution, for any given C.

Proof. Consider the linear transformation T : M_n,m(ℂ) → M_n,m(ℂ), defined by T(X) = AX - XB. Then, by Theorem 7.3.1, Null(T) = {0}. Hence, by the rank-nullity theorem, T is a bijection and the required result follows. _

Definition 7.3.3. [Toeplitz Matrix] A square matrix A is said to be of Toeplitz type if each (super/sub)-diagonal of A consists of the same element. For example, A = ⌊ ⌋
b1 b2 b3 b4
|| a1 b1 b2 b3||
|| ||
⌈ a2 a1 b1 b2⌉
a3 a2 a1 b1 is a 4 × 4 Toeplitz type matrix. and the matrix B = ⌊b b b b ⌋
| 1 2 3 4|
|| 0 b1 b2 b3||
|⌈ 0 0 b1 b2|⌉

0 0 0 b1 is an upper triangular Toeplitz type matrix.

Exercise 7.3.4. Let J_n(0) ∈ M_n(ℂ) be the Jordan block with 0 on the diagonal.

1.

Further, if A ∈ M_n(ℂ) such that AJ_n(0) = J_n(0)A then prove that A is an upper Toeplitz type matrix.

2.

Further, if A,B ∈ M_n(ℂ) are two upper Toeplitz type matrices then prove that

(a): there exists a_i ∈ ℂ,1 ≤ i ≤ n, such that A = a₀I + a₁J_n(0) + + a_nJ_n(0)^n-1.
(b): P(A) is a Toeplitz matrix for any polynomial P(t).
(c): AB is a Toeplitz matrix.
(d): if A is invertible then A^-1 is also an upper Toeplitz type matrix.

DRAFT

To proceed further, recall that a matrix A ∈ M_n(ℂ) is called non-derogatory if Geo.Mul_α(A) = 1, for each α ∈ σ(A) (see Definition 6.2.4).

Theorem 7.3.5. Let A ∈ M_n(ℂ) be a non-derogatory matrix. Then, the matrices A and B commute if and only if B = P(A), for some polynomial P(t) of degree at most n - 1.

Proof. If B = P(A), for some polynomial P(t), then A and B commute. Conversely, suppose that AB = BA, σ(A) = {λ₁,…,λ_k} and let J = Jordan CFA = S^-1AS be the Jordan matrix of A. Then, J = ⌊ ⌋
Jn1(λ1)
|| .. ||
⌈ . ⌉
Jnk(λk)

. Now, write B = S^-1BS = ⌊ -- -- ⌋
B11 ⋅⋅⋅ B1k
|| .. .. ..||
⌈-- . . -- .⌉
Bk1 ⋅⋅⋅ Bkk

, where B is partitioned conformally with J. Note that AB = BA gives JB = BJ. Thus, verify that

To proceed further, for 1 ≤ i ≤ k, define F_i(t) = ∏ _j≠i(t-λ_j)^n_j. Then, F_i(t) is a polynomial with deg(F_i(t)) = n - n_i and F_i(J_{n_j}(λ_j)) = 0 if j≠i. Also, note that F_i(J_{n_i}(λ_i)) is a nonsingular upper triangular Toeplitz type matrix. Hence, its inverse has the same form and using Exercise 7.3.4.1, the matrix F_i(J_{n_i}(λ_i))^-1B_ii is also a Toeplitz type upper triangular matrix. Hence,