Lecture 4: Jordan Canonical Forms This lecture introduces the Jordan canonical form of a matrix — we prove that every square matrix is equivalent to a (essentially) unique Jordan matrix and we give a method to derive the latter. We also introduce the notion of minimal polynomial and we point out how to obtain it from the Jordan canonical form. Finally, we make an encounter with companion matrices. 1 Jordan form and an application Definition 1. A Jordan block is a matrix of the form J1(λ) = λ 2 C when k = 1 and 2λ 1 0 ··· 039 > 6 . 7> 60 λ 1 0 . 7> 6 7=> 6 .. .. .. 7 Jk(λ) = 60 . 07 k when k ≥ 2: 6 7> 6 . .. .. 7> 4 . λ 15> > 0 ··· 0 0 λ ; | {z } k Theorem 2. For any A 2 Mn, there is an invertible matrix S 2 Mn such that 2 3 Jn1 (λ1) 0 ··· 0 . 6 .. 7 6 0 Jn2 (λ2) . 7 −1 −1 A = S 6 7 S =: SJS ; 6 . .. .. 7 4 . 0 5 0 ··· 0 Jnk (λk) where n1 + n2 + ··· + nk = n. The numbers λ1; λ2; : : : ; λk are the (not necessarily distinct) eigenvalues of A. The Jordan matrix J is unique up to permutation of the blocks. Observation: two matrices close to one another can have Jordan forms far apart, for instance " # " # " # 1 0 1 " 1 1 and ∼ : 0 1 0 1 0 1 Thus, any algorithm determining the Jordan canonical form is inevitably unstable! Try typing help jordan in MATLAB... The Jordan form can be useful when solving a system of ordinary differential equations in 0 −1 the form [x = Ax; x(0) = x0]. If A = SJS is the Jordan canonical form, then the change of −1 0 unknown functions xe = S x transforms the original system to [xe = Jx;e xe(0) = xe0]. Writing 2 3 2 3 Jn1 (λ1) 0 ··· 0 u(1) . 6 .. 7 6u 7 6 0 Jn2 (λ2) . 7 6 (2)7 J = 6 7 and x = 6 . 7 ; 6 . .. .. 7 e 6 . 7 4 . 0 5 4 . 5 u 0 ··· 0 Jnk (λk) (k) 1 0 the new system decouples as u(`) = Jn` (λ`)u(`), ` 2 [1 : k]. Each of these k systems has the form 2 3 2 u0 3 λ 1 ··· 0 2 u 3 1 . 1 6 u0 7 60 λ .. 7 6 u 7 6 2 7 6 7 6 2 7 6 7 = 6 . 7 6 7 : 4···5 6 . .. .. 17 4···5 0 4 5 um 0 ··· 0 λ um This can be solved by backward substitution: first, use um(t) = λum(t) to derive λt um(t) = um(0)e : d Then, u0 (t) = λu (t) + u (t) reads [u (t)e−λt] = [u0 (t) − λu (t)]e−λt = u (0), m−1 m−1 m dt m−1 m−1 m−1 m so that λt um−1(t) = [um−1(0) + um(0)t]e : d Next, u0 (t) = λu (t)+u (t) reads [u (t)e−λt] = [u0 (t)−λu (t)]e−λt = u (0)+ m−2 m−2 m−1 dt m−2 m−2 m−2 m−1 um(0)t, so that 2 λt um−2(t) = [um−2(0) + um−1(0)t + um(0)t ]e ; > etc. The whole vector [u1; : : : ; um] can be determined in this fashion. 2 Proof of Theorem 2 Uniqueness: The argument is based on the observation that, for 20 1 0 ··· 039 > 6 .7> 60 0 1 0 .7> 6 7=> 6 .. .. .. 7 Nk := 60 . 07 k; 6 7> 6. .. .. 7> 4. 0 15> > 0 ··· 0 0 0 ; | {z } k ` we have Nk = 0 if ` ≥ k and ` ! 20 ··· 1 ··· 03 6 .. .7 60 0 . 1 .7 6 7 ` 6 .. .. .. 7 Nk = 60 . 17 has rank k − ` if ` < k. 6 7 6. .. .. .7 4. 0 .5 0 ··· 0 0 0 2 Let µ1 < : : : < µ` be the distinct values of λ1; : : : ; λk. After permutation, 2 3 Jn1;1 (µ1) 6 .. 7 6 . 7 6 7 6 Jn (µ1) 7 6 1;h1 7 6 . 7 A ∼ 6 .. 7 : 6 7 6 7 6 Jn`;1 (µ`) 7 6 7 6 .. 7 4 . 5 Jn (µ ) `;h` ` We are going to prove that for each µi (say, for µ1) and for each m, the number of blocks Jm(µi) of size m is uniquely determined by A. From 2 3 Nn1;1 6 .. 7 6 . 7 6 7 6 Nn 7 A − µ I ∼ 6 1;h1 7 ; 1 6 . 7 6 .. 7 6 7 6 7 4 5 Full Rank we obtain, for each m ≥ 0, 2 m 3 Nn1;1 6 .. 7 6 . 7 6 m 7 6 Nn 7 (A − µ I)m ∼ 6 1;h1 7 : 1 6 . 7 6 .. 7 6 7 6 7 4 5 Full Rank Therefore, we derive, for each m ≥ 1, ! ! 1 if m ≤ n 1 if m ≤ n rank (A − µ I)m−1 − rank ((A − µ I)m) = 1;1 + ··· + 1;h1 1 1 0 otherwise 0 otherwise = [number of Jordan blocks of size ≥ m] =: j≥m: Since the numbers j≥m are completely determined by A, so are the numbers jm = j≥m −j≥m+1, which represent the numbers of Jordan blocks of size = m. Let us emphasize that the previous argument is also the basis of a method to find the Jordan canonical form. We illustrate the method on the example 2 3 1 2 −1 6 7 A = 40 3 −15 : 0 4 −1 3 The eigenvalues of A are the roots of det(xI − A) = (1 − x)3 — calculation left to the reader — hence 1 is the unique eigenvalue of A. Therefore, there are three possibilities for the Jordan canonical form J of A: 2 3 2 3 2 3 1 0 0 1 1 0 1 1 0 6 7 6 7 6 7 J1 = 40 1 05 ;J2 = 40 1 05 ;J3 = 40 1 15 : 0 0 1 0 0 1 0 0 1 The observation that rank(J − I) = rank(A − I) = 1 — calculation left to the reader — shows that J = J2 (since rank(J1 − I) = 0, rank(J2 − I) = 1, rank(J3 − I) = 2). Existence: Let µ1 < ··· < µ` be the distinct eigenvalues of A. We know (see Lecture 1) that 2 3 2 3 T1 0 ··· 0 µi x ··· x 6 .. 7 6 .. 7 6 0 T2 . 7 6 0 µi . 7 A ∼ 6 7 ; where Ti = 6 7 : 6 . .. .. 7 6 . .. .. 7 4 . 0 5 4 . x 5 0 ··· 0 T` 0 ··· 0 µi It is now enough to show that each matrix Ti is equivalent to a Jordan matrix, i.e., that 2 3 2 3 µ x ··· x Jm1 (µ) 0 ··· 0 . 6 .. 7 6 .. 7 60 µ . 7 6 0 Jm2 (µ) . 7 T = 6 7 2 Mm =) T ∼ J = 6 7 ; 6 . .. .. 7 6 . .. .. 7 4 . x5 4 . 0 5 0 ··· 0 µ 0 ··· 0 Jmk (µ) with m1 + m2 + ··· + mk = m, or equivalently that 2 3 2 3 0 x ··· x Nm1 0 ··· 0 . 6 .. .7 6 .. 7 60 0 .7 6 0 Nm2 . 7 T = 6 7 2 Mm =) T ∼ M = 6 7 : 6. .. .. 7 6 . .. .. 7 4. x5 4 . 0 5 0 ··· 0 0 0 ··· 0 Nmk In other words, we want to find vectors m1−1 mk−1 x1;T (x1); ··· ;T (x1); : : : ; xk;T (xk); ··· ;T (xk) m m1 m forming a basis of C and with T (x1) = 0;:::;T k (xk) = 0. This is guaranteed by the following lemma, stated at the level of linear transformations. Lemma 3. If V is a vector space of dimension m and if T : V ! V is a linear map with T p = 0 for some integer p ≥ 1, then there exist integers m1; : : : ; mk ≥ 1 with m1 + ··· + mk = m and m1−1 m −1 vectors v1; : : : ; vk 2 V such that (v1;T (v1);:::;T (v1); : : : ; vk;T (vk);:::;T k (vk)) forms a m1 m basis for V and that T (v1) = 0;:::;T k (vk) = 0. 4 Proof. We proceed by induction on n ≥ 1. For n = 1, there is nothing to do. Let us now suppose the result true up to m − 1, m ≥ 2, and let us prove that it holds for the integer m, too. Note that U := ran T is a vector space of dimension n < m (otherwise T would be surjective, hence bijective, and T p = 0 would be impossible). Consider the restriction of T to U, i.e., the linear map Te : x 2 U 7! T x 2 U, and note that Tep = 0. Applying the induction hypothesis, there exist integers n1; : : : ; n` ≥ 1 with n1 + ··· + n` = n and vec- n1−1 n −1 tors u1; : : : ; u` 2 U such that (u1; Te(u1);:::; Te (u1); : : : ; u`; Te(u`);:::; Te ` (u`)) forms a ba- n1 n sis for U and that Te (u1) = 0;:::; Te ` (u`) = 0. Since u1; : : : ; u` 2 U = ran T , there exist n +1 n1 n v1; : : : ; v` 2 V such that ui = T vi. Note that T i (vi) = 0, so that (T (v1);:::;T ` (v`)) is a linearly independent system of ` vectors in ker T , which has dimension m − n. Com- plete this system with vectors v`+1; : : : ; vm−n 2 ker T to form a basis for ker T . Now consider n1 n the system (v1;T (v1);:::;T (v1); : : : ; v`;T (v`);:::;T ` (v`); v`+1; ··· ; vm−n). Observe that this is a linearly independent system. Observe also that the number of vectors in this system is n1 + 1 + ··· + n` + 1 + m − n − ` = n1 + ··· + n` + m − n = m, so that the system is in fact a basis n1+1 n +1 for V .