NOTES II FOR 130A JACOB STERBENZ

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1 NOTES II FOR 130A JACOB STERBENZ Abstract. Here are some notes on the Jordan canonical form as it was covered in class. Contents 1. Polynomials 1 2. The Minimal Polynomial and the Primary Decomposition 3 3. When the Characteristic Polynomial Factors 4 4. The Structure of Nilpotent Operators 5 5. The Real and Complex Jordan Forms 7 1. Polynomials As will become apparent shortly, the theory of invariant subspaces of a linear transformation T L(V ) turns on the issue of factoring polynomials into prime powers. Accordingly we begin here with a basic review of polynomial arithmetic. First some notation. Let F denote a field of numbers. For our purposes one can take this to be either real numbers R or complex numbers C. However, everything we say here is equally valid for any field (e.g. rational numbers Q). We denote by: F[x] = { p(x) d p(x) = a i x i, where a i F }, for various values of d. For any particular p F[x] we define: deg(p) = min{d d p(x) = a i x i and a d 0}. i=0 Note that this is just the usual notion of degree of a polynomial (the highest power of x you see in p(x)). We also say that p F[x] is monic if deg(p) = d and a d = 1. This is simply a normalization condition that is useful in the statement of several results to follow. Perhaps the most basic result about polynomials in one variable is the following: Theorem 1.1 (Division with remainder). Let p, q F[x] be any two polynomials. Then there exists g, r F[x] such that q = pg + r where deg(r) < deg(p). Proof. This follows from long division of polynomials just as one learns in high school algebra. For starters we may assume deg(p) deg(q) because otherwise we can simply choose g = 0 and r = q. Then we have: q(x) = d 1 i=0 i=0 a i x i, p(x) = with d 1 d 2, and a d1 0 and b d2 0. Let g 1 = a d 1 b d2 x d1 d2, and then set q 1 = q g 1 p. We have deg(q 1 ) < deg(q). If deg(q 1 ) < deg(p) we are done by setting r = q 1. Otherwise repeat this process to produce q 2 = q 1 g 2 p with deg(q 2 ) < deg(q 1 ). Again check if deg(q 2 ) < deg(p) or not, and divide again if this condition has not yet been achieved. Eventually this process will produce some q k with deg(q k ) < deg(p) and q k = q (g 1 + g g k )p. Finally set g = k i=1 g i and r = q k are we are done. 1 d 2 i=0 b i x i,

2 For collections of polynomials we have the following importion notion: Definition 1.2. A collection of polynomials I F[x] is called an ideal if: i) For any p, q I we have that p + q I. ii) For any p I and any other q F[x] we have pq I. An example of an ideal would be the set I = {p F[x] a0 = 0}. In other words this ideal consists of all polynomials of the form xg(x) for some (not fixed) g F[x]. Division with remainder shows that in some sense this is the general picture for polynomial ideals: Theorem 1.3 (Generation of ideals). Let I F[x] be any nonzero ideal. Then there exists a unique monic p F[x] such that I = {q F[x] q = pg for some g F[x]}. In this case we write I = (p) and say I is the ideal generated by p. Proof. Let d be the minimum degree over all nonzero polynomials in I, and let p I be a monic polynomial with deg(p) = d. Let q I be anything else. Then we have q = gp + r for some deg(r) < deg(p) polynomial. But q gp I by rules i) and ii) for ideals, so r I as well. If r 0 then it would be a nonzero polynomial in I with degree less that p, a contradiction. Thus r = 0 and we have q = pg as desired. Note that uniqueness follows because if deg(p) = deg(q) and q = gp for some g F[x], we must have g = a for some a F (this is simply because deg(gp) = deg(g) + deg(p)). Thus p is the only monic polynomial that can generate I. Definition 1.4. Next, given two polynomials p, q F[x] we say p divides q of there exists a g F[x] with q = pg. In this case we write p q. We say a polynomial p is prime if deg(p) 1 and if q p for some q F[x] we must have either p = aq for some a F or deg(q) = 0 (in the latter case q = a 0 for some a 0 F). Given a (finite) collection of polynomials p i F[x] we say p i are relatively prime if q p i for all i implies that deg(q) = 0. In other words the only common factors of all the p i are constants. It is important to understand that the notion of being prime really depends on F. The polynomial p = x is prime when F = R but not when F = C. The fundamental theorem of algebra states that the only prime polynomials if C[x] are those of the form p = a(x λ). On the other hand the only prime polynomials in R[x] are of the form p = a(x λ) or p(x) = ax 2 + bx + c where b 2 4ac < 0. This is because if a polynomial has real coefficients then its roots are either real or come in complex conjugate pairs. Also, note that a collection of non-constant polynomials p i C[x] are relatively prime iff they have no root in common. The same is true for p i R[x] as long as one takes into account complex roots (this is necessary because p 1 = x and p 2 = x 4 + 2x have no real roots in common but they are also not relatively prime in R[x]). From the previous result on generation of ideals we now have the following theorem which will be our main tool in the next Section: Theorem 1.5. Let q i F[x] be a finite collection of nonzero relatively prime polynomials. Then there exists g i F[x] such that i q ig i = 1. Proof. First define an ideal that is generated by the q i : I = {q F[x] q = i g i q i for some g i F[x]}. One can check immediately that this collection of polynomials satisfies i) and ii) above, and since the q i are nonzero I itself is also nonzero. Thus there exists a unique monic p F[x] with I = (p). In particular p q i for all i, and because the q i are relatively prime this means deg(p) = 0. The only monic polynomial of degree 0 is p = 1. Thus 1 I, and so by the construction of I there must exists g i F[x] with i g iq i = 1. We end with a prime factorization theorem for polynomials which mirrors what one knows about integers: Theorem 1.6. Let q F[x]. Then one can write uniquely q = a k i=1 pri i, where a F, where p i F[x] are distinct prime monic polynomials, and r i N. If F = C then each deg(p i ) = 1, and if F = R then each 1 deg(p i ) 2. 2

3 Proof. The existence of some prime factorization of q F[x] follows by inductively splitting q into smaller factors. For example either q is already prime or q = q 1 q 2 where each 1 deg(q i ) < deg(q). Then if at least one of q 1, q 2 is not prime we split further and continue this process until all factors are prime. The main issue then is the uniqueness of the prime factorization. This is a consequence of the following claim: If p is prime and p q 1 q 2, then either p q 1 or p q 2. The claim follows because if p, q 1 are relatively prime then one can find f, g F[x] with fp + gq 1 = 1. Then fq 2 p + gq 1 q 2 = q 2. But p q 1 q 2 implies q 1 q 2 = gp so rearranging things we have (fq 2 + g)p = q 2. In other words p q 2. By repeatedly applying the previous fact to both sides of the following equation we see that if: q = a k i=1 p ri i = a then we must have a = a, k = k, r i = r i, and p i = p i. (p i) r i, 2. The Minimal Polynomial and the Primary Decomposition Now let V be a vector space over the real or complex numbers F. Let T L(V ). The primary purpose of this section is to associate a certain polynomial m T F[x] with T in such a way that the prime factorization of m T gives a great deal of information about the invariant subspaces of T. Recall that finding the invariant subspaces of T is the key step to finding all of the solutions to ẋ = T x. To set things up we first define an ideal in F[x] associated with T. This is: k i=1 I(T ) = {q F[x] q(t ) = 0}. Here we mean that if q(x) = d i=0 a ix i then we define q(t ) = d i=0 a it i with the proviso that T 0 = I for any T. Note that q(t ) = 0 simply means q(t )x = 0 for all x V. It is not hard to check that I(T ) is indeed an ideal (basically zero plus zero is zero, and zero times anything is zero). A little bit less obvious is that I(T ) is not the zero ideal. To see this note that if x V then the collection of vectors {x, T x, T 2 x,..., T n x}, where n = dim(v ), must be linearly dependent. Thus there exists q F[x] such that q(t )x = 0. Constructing such q i for a basis B = {x 1,..., x n } and then setting q = q i gives a nonzero polynomial with q(t )x = 0 for all x V. Definition 2.1. From Theorem 1.3 in the previous section we must have I(T ) = (m T ) for some unique monic m T F[x]. In other words there exists a polynomial m T with m T q for all other q F[x] with q(t ) = 0. We call m T the minimal polynomial of T. Using the prime factorization of m T we have the fundamental result: Theorem 2.2 (Primary Decomposition Theorem). Let T L(V ) and m T its minimal polynomial. Then if m T = k i=1 pri i is the prime factorization of m T there exists a T invariant direct sum decomposition V = k i=1 V i where for each i = 1,..., k one has m T Vi = p ri i and moreover each V i = ker ( p ri i (T )). Proof. We ll break the proof down into a number of steps. Step 1:(Construction of V i ) This is the key part of the proof. Let m T = k m T into distinct prime powers and for each i = 1,..., k define: q i = j i p rj j = m T p ri i. i=1 pri i be the factorization of Then by construction the q i are relatively prime so by Theorem 1.5 there exists g i F[x] with i g iq i = 1. Using these polynomials define E i L(V ) by the formulas E i = g i (T )q i (T ). The first thing that is immediate is that [E i, T ] = E i T T E i = 0, and also i E i = I. Now define: V i = ran(e i ) = {y V y = Ei x for some x V }. Note that span(v 1,..., V k ) = V because x = i E ix for each x V. Therefore to show V = k i=1 V i we only need V i V j = { 0} for all i j. This will follow if E i E j = 0 for all i j. But m T g i q i g j q j whenever i j because q i q j for i j contains every prime factor p r l l for all l = 1,..., k. Thus E i E j = g i (T )q i (T )g j (T )q j (T ) = 0. 3

4 To complete this part of the proof we just need to show the V i are T invariant. By E i E j = 0 we have x = E i x for all x V i. Thus if x V i we have T x = T E i x = E i T x ran(e i ) = V i. Step 2:(Show that V i = ker ( p ri i (T )) ) First let x V i. Then x = g i (T )q i (T )x, so multiplying both sides by p ri i (T ) and using pri i (T )q i(t ) = m T (T ) = 0 we have p ri i (T )x = 0. Thus V i ker ( p ri i (T )). On the other hand suppose p ri i (T )x = 0. Since x = x j where each x j V j, and since T (V j ) V j we must also have p ri i (T )x j V j. But the V j are linearly independent so this implies p ri i (T )x j = 0 for each j. Now for j i we have p ri i g jq j, so there exists g F[x] with gp ri i = g j q j. Multiplying the previous equation by g(t ) gives E j x j = 0 for all i j. In other words x j = 0 for all i j. Thus x = x i so we have shown ker ( p ri i (T )) V i. Step 3:(Show that the minimal polynomial of T Vi is p ri i ) Since V i = ker ( p ri i (T )) we have p ri i (T V i ) = 0. Thus m T Vi p ri i, so we must have m T Vi = p r i i for some r i r i. Now define m = p r i i q i = p r i i j i prj j. Since each p rj j (T )V j = { 0} and p r i i (T )V i = { 0} we have m(t ) = 0 on all of V. Thus m T m which implies r i r i and we are done. 3. When the Characteristic Polynomial Factors We now bring in the connection between the minimal polynomial m T and the characteristic polynomial p T. Recall that the latter is defined by the equation p T (x) = det(xi T ), and λ F is an eigenvalue of T iff p T (λ) = 0. First we give an indication that m T is in fact closely related to p T : Lemma 3.1. Given T L(T ) the polynomials m T and p T have exactly the same roots. Moreover, m T factors over F iff p T factors over F. Proof. First of all let p I(T ) be any polynomial such that p(t ) = 0. Then if T x = λx for some x 0 a direct computation shows p(t )x = p(λ)x. Thus p(λ) = 0. Therefore the roots of p T are roots of m T as well. The other direction is a bit more work. Let λ F be a root of m T. Then by prime factorization m T = (x λ) r q(x) where (x λ) and q are relatively prime. Let V = V λ V q be the corresponding invariant decomposition of V according to Theorem 2.2. Then the minimal polynomial of T Vλ is exactly (x λ) r. In other words (T Vλ λi) r = 0 but (T Vλ λi) r 1 0. Let y = (T Vλ λi) r 1 x with y 0. Then (T Vλ λi)y = 0 so y V λ V is an eigenvector for T (on all of V ). Finally, the part about factorization is trivial if F = C (in this case every polynomial factors). On the other hand if V is a real vector space, by extending T to real operator on the complexification V C, we just need to see that the minimal polynomial of the extension is the same as the original. Since p(t ) = p(t ) for any p = C[x], we see that if p(t ) = 0 then R(p)(T ) = I(p)(T ) = 0, where R(p) and I(p) are the real polynomials with p = R(p) + ii(p). This implies that the minimal polynomial of a real operator is real, and so the minimal polynomials of T on both V and V C are the same. Before stating our main result we need one more idea. First a definition: Definition 3.2. An operator N L(V ) is called nilpotent if N k = 0 for some k 1. If N is nilpotent we call k 0 = min{k N k = 0} the order of N. Now: Lemma 3.3. Let N L(V ) be nilpotent. Then the order of N is dim(v ). Moreover for any λ F wth λ 0 one has λi + N is invertible. Proof. Let k 0 be the order of N, and choose some x V with y = N k0 1 x 0. Consider the collection of vectors {x, Nx, N 2 x,..., N k0 1 x}, and suppose k 0 1 i=0 a in i x = 0 for some a i F. Applying N k0 1 to both sides gives a 0 y = 0. Hence a 0 = 0. Applying N k0 2 gives a 1 y = 0, so a 1 = 0. Continuing in this way applying N k0 j for j = 3,..., k 0 (in order) gives a j 1 = 0. This shows that {x, Nx, N 2 x,..., N k0 1 x} is a linearly independent collection of k 0 vectors, and so we must have k 0 dim(v ). Finally, the part about the inverse follows from the direct calculation (λi + N) 1 = 1 k0 1 λ i=0 ( 1 λ N)i. We can now state the Main Theorem of these notes: 4

5 Theorem 3.4 (Abstract Jordan Form). Let V be a vector space over F and let T L(V ). Suppose that the characteristic polynomial of T factors over F, that is p T (x) = k i=1 (x λ i) di for some λ i F. Then there is a T invariant direct sum decomposition V = k i=1 V i where V i = ker ( ) (T λ i I) di. Moreover V i = ker ( ) (T λ i I) ri are exactly the subspaces associated with the factorization of the minimal polynomial m T (x) = k i=1 (x λ i) ri, and we have the relation r i d i = dim(v i ). In particular m T p T. As a consequence we can write T = S+N where S is diagonalizable with eigenvalues λ i, and N is nilpotent with [S, N] = 0. Moreover there is only one such pair S, N which yields this decomposition. Therefore, we have that T itself is diagonalizable iff N = 0, which is equivalent to r i = 1 for every prime factor of the minimal polynomial. Proof. Assuming that p T (x) = k i=1 (x λ i) di for some λ i F and d i N implies by Lemma 3.1 that m T (x) = k i=1 (x λ i) ri for the same λ i but possibly different r i. Now let V = k i=1 V i be the direct sum decomposition according to Theorem 2.2, so that by construction V i = ker ( ) (T λ i I) ri. Since the Vi are T invariant subspaces we must have p T = k i=1 p T Vi. On the other hand since m T Vi = (x λ i ) ri and p T Vi have the same roots then p T Vi = (x λ i ) di where d i = dim(v i ) and d i is the same as in the factorization of p T. Also, because T Vi λ i I is nilpotent of order r i, by Lemma 3.3 we have r i d i. Next we show that V i = ker ( ) (T λ i I) di. Since we already know that Vi = ker ( ) (T λ i I) ri and r i d i the containment V i ker ( ) (T λ i I) di is immediate. On the other hand if x = j x j, where x j V j, is such that x ker ( ) (T λ i I) di, then we know from invariance and independence that individually x j ker ( ) (T λ i I) di. For j i we can write T Vj λ i I = λi +N where λ = λ j λ i 0 and N = T Vj λ j I is nilpotent. Thus by Lemma 3.3 T Vj λ i I is invertible when i j, so ker(t Vj λ i I) di = { 0}. Thus x j = 0 for all j i which shows x V i. Finally, to get the T = S + N decomposition we just need to construct S and N in terms of blocks for each V i. Let S Vi = λ i I, and let N Vi = T Vi λ i I. Then it is clear that S is semisimple and N is nilponent with [S, N]. Now let T = S + N be any other such decomposition. By the condition [S, T ] = 0 we see that [S, E i ] = 0 for each of the projections constructed in Theorem 2.2, so S (V i ) V i for each i. Therefore V = k i=1 V i is also an invariant direct sum decomposition for S, and hence for N as well. Now if S x = λx, where x = i x i and x i V i we must have Sx i = λx i by independence. Thus, if B = {e 1,..., e n } is a basis for V consisting of eigenvectors of S, then S i = {E i e 1,..., E i e n } must be a collection of eigenvectors of S Vi which spans V i. In other words S Vi is also diagonalizable for each i. Now let x V i be any nonzero vector with S x = λx, and let k x 1 be the smallest power such that (N ) kx x = 0. Then if we set y = (N ) kx 1 x have 0 y V i and T y = (S + N )y = λy. Thus, we have shown that any eigenvalue of S Vi must also be an eigenvalue of T Vi, so S Vi = λ i I. Now N Vi = T Vi λ i I and we are done. We ll end this section we a few additional definitions regarding the subspaces V i constructed in the previous theorem. Definition 3.5. Let T L(V ) and assume that its characteristic polynomial factors as p T (x) = (x λ) d λ q(x) where (x λ), q(x) are relatively prime. Then we call V λ = ker(t λi) d λ the generalized eigenspace associated to λ. Note that d λ = dim(v λ ). We call d λ the algebraic multiplicity of eigenvalue λ, and we call e λ = dim ( ker(t λi) ) the geometric multiplicity of eigenvalue λ. Note that we always have 1 e λ d λ, and that e λ is exactly the number of linearly independent eigenvectors of T in V λ (which all must have eigenvalue λ). In particular T is diagonalizable iff p T (x) factors and e λ = d λ for all eigenvalues λ. Finally, note it is possible for e λ to be much smaller than d λ. For example if r λ = d λ, where r λ is the power of (x λ) in the minimal polynomial m T, then we must have e λ = The Structure of Nilpotent Operators In light of Theorem 3.4 we see that to get the simplest form for a general T L(V ) is suffices to study a general nilpotent operator N L(V ). It turns out that the key notion in this regard is the following: Definition 4.1. Let T L(V ) be any linear operator, and let x V be a fixed vector. Then we define: Z(x, T ) = span{x, T x, T 2 x,..., T dim(v ) 1 x}. 5

6 We call Z(x, T ) the T -cyclic subspace of V generated by x. We call dim(z(x, T )) = O(x, T ) the T -order of x. Note that by the properties of the characteristic polynomial we have T dim(v )+j x Z(x, T ) for all j 0. Our main theorem now is the following: Theorem 4.2. Let N L(V ) be a nilpotent operator of order k 0. Then there exists a collection of vectors x 1,..., x k V with O(x 1, N) O(x 2, N)... O(x k, N) = k 0 and V = k i=1 Z(x i, N). This decomposition is unique in the sense that if V = k i=1 Z(y i, N) is any other such decomposition we must have k = k and O(x i, N) = O(y i, N) for all i = 1,..., k. In particular there is a basis B consisting of vectors of the form N i x j such that: (1) [N] B = N l1 N l2 Nlk, where each l i = O(x i, N), and N l M(l l) is the elementary nilpotent matrix of order l : (2) N l = Any such form of N is unique with the proviso that l 1 l 2... l k. Proof. The proof of the cyclic decomposition V = k i=1 Z(x i, N) we give here is the same as in Appendix III of the text. The proof is by induction on the dimension of V. If dim(v ) = 1 then the only nilpotent operator is N = 0 and so V = Z(x, N) for any x 0. Now let N, V be arbitrary, and set W = N(V ) V. Since N is nilpotent we must have ker(n) { 0} so W V. Now N(W ) W, and dim(w ) < dim(v ) so by induction we can decompose W = k i=1 Z(y i, N W ) for some y i W. Choose any x i V with y i = Nx i. First suppose that there exists p i F[x] with i p i(n)x i = 0 and each deg(p i ) < O(x i, N) = O(y i, N)+1. Applying N to this and using the independence of Z(y i, N W ) we see that for each i either p i = 0 or p i 0 (as a polynomial) and p i (N)y i = 0. In other words p i = 0 or else p i (x) = a i x O(yi,N). But this also implies that p a i 0 in O(yi,N) 1 y i = 0, so a i = 0 for each term in that sum. This shows that the Z(x i, N) themselves are independent. It remains to show V = k i=1 Z(x i, N) L where L = k j=k+1 Z(x j, N) for some other collection of vectors x j. Since N( k i=1 Z(x i, N)) = N(V ), we must have V = span ( k i=1 Z(x i, N), ker(n) ). Let ker(n) = L L where L = ker(n) k i=1 Z(x i, N) and L is any other complimentary subspace. Then N(L) = { 0} L, so V = k i=1 Z(x i, N) L is an invariant direct sum decomposition. If {x k+1,..., x k } is any basis for L we have L = k j=k+1 span(x j), and automatically span(x j ) = Z(x j, N), so we are done. Notice that this inductive procedure also establishes uniqueness in terms of the orders O(x i, N) because if the orders of O(y i, N W ) are uniquely determined for i = 1... k, then so are the orders O(x i, N) = O(y i, N W ) + 1. And the number of remaining vectors, which all have order 1, is also determined by the number dim(v ) k i=1 O(x i, N). Finally, to see the canonical form (1) is valid suppose that V = span(x, Nx,..., N l 1 x) where l = dim(v ) = O(x, N). Then in the basis B = { e 1 = N l 1 x, e 2 = N l 2 x,..., e l = x } we have Ne 1 = 0 and Ne i = e i 1 for all i = 2,..., l. Thus [N] B = N l where N l is defined by (2) above. 6

7 Remark 4.3. Note that if V = span(x, Nx,..., N l 1 x) where l = dim(v ) = O(x, N), and instead we use the basis B = { e 1 = x, e 2 = Nx,..., e l = N l 1 x } we will get: [N] B = N l = This is the basic form for nilpotent matrices used in the text. I have chosen to employ the upper triangular form here which is a little more common by modern standards. 5. The Real and Complex Jordan Forms Combining what we have done so far yields the so called Jordan canonical form of a matrix: Theorem 5.1 (Jordan form). Let V be a vector space over F and let T L(V ) be such that it characteristic polynomial p T (x) factors over F as p T (x) = k i=1 (x λ i) di. In addition suppose the minimal polynomial of T factors as m T (x) = k i=1 (x λ i) ri. Then there exists a basis B for V such that: J l11 (λ 1 ) J l12 (λ 1 ) [T ] B = (λ1) Jl1n1 Jlk1 (λk)... (λk) Jlknk where l ij l i j+1 for each ij and l ini l ij l ij Jordan matrix of the form: = r i, and n i j=1 l ij = d i, and each elementary block J lij (λ i ) is an J l (λ) = λi + N l = λ 1 λ λ Proof. This follows more or less directly by combining Theorems 3.4 and 4.2. First note that T takes a block form along the primary direct sum decomposition V = k i=1 V i, and on each V i we have T Vi = λ i I + N i for some nilpotent operator N i L(V i ) or order r i. By applying Theorem 4.2 to N i we are done. The previous Theorem is not quite satisfactory when we want to compute solutions to ẋ = T x on a real vectors space V when the characteristic polynomial T L(V ) does not factor over R. However, the only obstruction here is complex conjugate pairs of roots in the factorization of p T (x). Using the material from the previous notes on complexifications and the previous Theorem we have: Theorem 5.2 (Real Jordan form). Let V be a real vector space and let T L(V ) be such that it characteristic polynomial p T (x) factors over C as p T (x) = k r i=1 (x λ i) di k c j=1 (x µ j) ej (x µ j ) ej, where λ i R and µ j = a j + 1b j and b j 0. In addition suppose the minimal polynomial of T factors as m T (x) = 7

8 kr i=1 (x λ i) ri k c j=1 (x µ j) tj (x µ j ) tj. Then there exists a (real) basis B for V such that: [T ] B = J l11 (λ 1 ) Jlkr (λ nkr kr ) K s11 (µ 1 ) Kskc mkc (µkc) where l ij l i j+1 for each ij and l ini = r i, and n i j=1 l ij = d i, and each real elementary Jordan block J l (λ) is the same as before. Furthermore s ij s i j+1 for each ij and s imi = t i while m i j=1 s ij = e i and each matrix K sij (µ i ), where µ i = a i + 1b i, is a 2s ij 2s ij block matrix of the form: a b I b a a b (3) K s (µ) = b a. I a b b a Proof. Using Theorem 3.4 we can reduce to the invariant factor associates to a single pair of conjugate complex eigenvalues. In other words we may assume p T (x) = (x µ) e (x µ) e and m T (x) = (x µ) t (x µ) t where µ = a + ib with b 0. Then using the complex Jordan normal form we know that V C = V µ V µ where each factor is the invariant subspace associated to eigenvalues µ and µ respectively. Let σ be the complex conjugation on V C. The key observation now is that σ(v µ ) = V µ. To see this recall that V µ = ker(t C µi) e, and an easy calculation shows that σ ( ker(t C µi) e) = ker(t C µi) e thanks to σt C = T C σ. Because of this we further have T C = S + N where S = µe µ + µσe µ σ where σe µ σ = E µ and the E are projections onto the generalized eigenspaces. This implies that both σs = Sσ and σn = Nσ, so both S and N in the semisimple/nilpotent decomposition of T are real operators. Now suppose V µ = k j=1 Z(z j, N). Then using the mirror symmetry σ(v µ ) = V µ and the fact N is real we must have V µ = k j=1 Z(σ(z j), N). In particular we can pair cyclic spaces to get the real decomposition: V = k j=1(z(z j, N) Z(σ(z j ), N)) R. For each fixed value of j let s j = O(z j, N) denote the (complex) dimension of the cyclic factor Z(z j, N), and choose the real basis of (Z(z j, N) Z(σ(z j ), N)) R to be B j = {y 1, x 1,..., y sj, x sj } where x l + iy l = N sj l z j. Then one can readily check that in this basis the matrix of T restricted to the factor (Z(z j, N) Z(σ(z j ), N)) R is exactly K sj (µ) from line (3). Department of Mathematics, University of California, San Diego (UCSD) 9500 Gilman Drive, Dept 0112 La Jolla, CA USA address: jsterben@math.ucsd.edu 8