NOTES II FOR 130A JACOB STERBENZ

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 2 +... + 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,

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 2 + 1 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 2 + 1 and p 2 = x 4 + 2x 2 + 1 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

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

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

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 = 1. 4. 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

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 = 0 1 0 1 0... 1 0. 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

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: 0 1 0 [N] B = N l = 1 0. 1 0 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...... 1. 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

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 92093-0112 USA E-mail address: jsterben@math.ucsd.edu 8