PhL on Jordan Form

PhL on Jordan Form PhL 1.4.05 Several authors have contributed bits and pieces to my understanding of this subject, but no author has given a clean linear presentation, so I will here attempt my own integration. 1. The Jordan Block. Consider a Jordan Block of dimension 3 x 3, J = = + = + 1 = C + 1 I We have isolated a matrix called C here. As shown on page 33 of Matthews (apart from transpose), this is the companion matrix for the minimum polynomial f(x) = x3 . That is, we have that C3 = 0. As you keep raising the power of C, you keep shifting the row of 1's to the upper right until it finally shifts completely away and you have a completely zero matrix left. Here is a more detailed example with k = 4 C = C2 = C3 = C4 = To see why this is happening, note that you can write Cjk = j+1,k to represent the initial diagonal of ones in the first off diagonal. Then we have (C2 )jk = m Cjm Cmk = m j+1,m m+1,k = j+2,k (C3 )jk = m C2jm Cmk = m j+2,m m+1,k = j+3,k and so on Recall the way you are supposed to find the min poly of a matrix: keep raising the power until you get a linear combination of lower powers. Normally when you do this with some arbitrary matrix C, you get something like this at some point, Ck = -a0C0 - a1C1 + ... - ak-1Ck-1 with some coefficients ai . We can rewrite this as (think ak = 1) a0C0 + a1C1 + ... + ak-1Ck-1 + Ck = 0 or f(C) = 0 where f(x) = xk + ak-1xk-1 + .... + a0 and then f(x) is the min poly for matrix C. In our special case, we get this: Ck = 0 k = 4 f(x) = xk Notice that the lower powers C, C2, C3 cannot be expressed as lin coms of previous powers, because the ones are in the wrong place. That is WHY the min poly is Ck . So far, then, we have shown this fact: Theorem 1: The min poly mC() of the k x k matrix (J - 1I) = C is given by mC() = k, and mC(C) = Ck = 0. Theorem 2: The min poly mJ() of the k x k matrix J is given by ( - 1)k. Proof: We know that min poly of C is m(C) = Ck which we can write as (J - 1I)k. We can then expand this thing to get Jk - 1Jk-1 + ... = 0, and this then must be the min poly for the matrix J. Any lower power of J (below Jk) cannot be expressed as a lincom of lower powers. So mJ(J) = (J - 1I)k. But now write this in terms of a scalar argument to get mJ() = ( - 1)k . Theorem 3: The char poly chJ(x) of the k x k matrix J is given by chJ() = ( - 1)k, so we can regard k as the algebraic multiplicity of the eigenvalue 1 of the matrix J. Proof: chJ() = det( J - I) = ( - 1)k , just manually do the det, the off diagonal 1's have no effect. Comment: The Jordan Block is constructed just to get this very result: the min poly = the char poly. Usually the min poly has a lower exponent, but not here! Theorem 4: The matrix J has only one eigenvector (x,0,0,0...) and thus has geometric multiplicity = 1. Proof: The eigenvector problem is this: Jv = 1v, which is the same as Cv = 0 which tells us that vi+1 = 0 for i = 0,1,2... Thus, our eigenvector is (v0, 0, 0, 0....). We can regard this as the one eigenvector of the nullspace of matrix C = (J - 1I). The nullity of this space is then 1. Conclusions reached concerning the Jordan block J of dimension k with diagonal elements 1 : (1) J has char poly = min poly = ( - 1)k (2) J has one eigenvalue 1 and that eigenvalue has an algebraic multiplicity of k (3) J has a geometric multiplicity of 1 and its only eigenvector is (v0, 0, 0, 0....) 2. Direct Sum of Jordan Blocks with same eigenvalue 1 Suppose we now construct a larger matrix (dim k = k1+ k2) by doing a direct sum of 2 Jordan blocks, both of which have the SAME eigenvalue 1 : A = Jk1 Jk2 The first block has min poly ( - 1)k1, the second has ( - 1)k2. What is the min poly for the larger matrix? The result is derived on page 36 of Matthews. If we let C = AB, then he shows that mC = lcm(mA, mB) ' lcm = least common multiple' We need to pause to prove this important result: Theorem 5: If C = AB, then mC = lcm(mA, mB). Proof: Assume that mC is some polynomial f(x). Then we know that: f(C) = 0 = f(AB) = f(A) f(B) which => f(A) = 0 and f(B) = 0. This says that mA | f and mB | f . It says mA | f, for example, because mA is the lowest degree poly that satisfies mA(A) = 0, so if f(A) is some other poly with f(A) = 0, then mA must divide into f. Now, we want to know the lowest degree poly f(x) that can be divided by both mA(x) and mB(x). This is exactly what lcm(mA(x), mB(x)) is. And if the f(x) so found is in fact the poly of lowest degree, then it must be the min poly of C. There can be no poly of lower degree that satisfies the requirement mA | f and mB | f. Now we apply this theorem to our direct product situation. We have mA = ( - 1)k1 and mB = ( - 1)k2. The answer is that lcm(mA, mB) = ( - 1)MAX(k1,k2). For example, if we have mA = ( - 1)2 and we have mB = ( - 1)3, then ( - 1)3 can be divided by both mA and mB. So we can easily extend this idea to show the following Theorem 6: Suppose we construct a matrix by combining multiple Jordan Blocks of various sizes, but which all have the same eigenvalue 1 , A = Jk1 Jk2 Jk3 .... Then the min poly for A is given by mA = ( - 1)MAX(k1,k2, k3.....) . This conclusion is reinforced on page 37 of Matthews. Comment: If you are staring at the matrix A just described, you can easily see which submatrix has the largest dimension, call it kmax = MAX(k1, k2, ....). Later, when we combine matrices Ai for different eigenvalues i to make a still larger matrix B, and when we talk about the overall min poly for this larger matrix B, it will have a factor ( - 1)kmax for our particular submatrix A. So, the size of largest Jordan block in each manifold equals the exponent of the overall min poly for the matrix A. Theorem 7: For the same matrix A shown in Theorem 6, the char poly is given by ( - 1)k1 + k2 + ... . Proof: When we have any block diagonal form, we know that we can multiply the sub-determinants to get the total determinant, so in this case we can multiply the char polys of the sub-matrices, and that then leads to a sum of exponents as shown. 3. Primary Decomposition Suppose we are given some matrix T and we have computed its char poly and min poly to be: chT() = ( - 1)a1 ( - 2)a2 mT() = ( - 1)b1 ( - 2)b2 where we know that ai bi because we know that mT() | chT(). Then: Theorem 8: According to the Primary Decomposition Theorem on page 61 of Matthews, we can decompose our vector space V, in which this matrix T acts, into a pair of subspaces, V = V1 V2 where V1 = N [ (T - 1I)b1] with = a1 N[X] means "Nullspace of matrix X " V2 = N [ (T - 2I)b2] with = a2 where the exponent bi is the kmax mentioned earlier for each subspace i = 1 and 2. So bi will be the size of the largest Jordan block in each of the two subspaces. Remember: bi is the min poly exponent. The ai are the algebraic multiplicities of each subspace. Nothing is said here about geometric multiplicity, but we know that geo mult is the number of Jordan blocks in each subspace. We also know these facts: N [ (T - 1I)1] has = geo mult 1 N [ (T - 2I)1] has = geo mult 2 These nullspaces are the "eigenmanifolds" of Stakgold. Here we have unity exponents. 4. The Cyclic Subspace Business Following Matthews page 54, we imagine some "vector specific" min poly mT,v(T)v = 0 rather than the more general min poly mT(T) = 0. This specific thing has to divide the normal min poly. Theorem 9: If f is any polynomial and v is some vector in V (the space in which T acts), then the set of all vectors of the form f(T)v forms a subspace of V. This is called the T-cyclic subspace generated by the vector v, and this subspace is a T-invariant subspace, and it is represented as CT,v. Proof: I think the subspace part follows from usual linearity stuff. That is, f1(T) and f2(T) make a sum that is in the same space, so it is closed under addition, etc. Now suppose w is in CT, v. Then w = f(T)v for some polynomial f. Then Tw = Tf(T)v = g(T)v which is also in the space since g is just some other polynomial. Theorem 10: In the situation described in Theorem 9, assume that for vector v we have some vector-specific min poly mT,v(T) of degree k. Recall that v is the generator vector of the subspace CT,v. Then the following vectors form a basis in CT,v: { v, Tv, T2v, .... Tk-1v } This is the T-cyclic basis. This theorem has a proof on page 55,56 of Matthews which I have not read. Theorem 11: Again in the situation of the previous two theorems, if we work in the T-cyclic basis, then the matrix T is none other than the companion matrix of the min poly mT,v. This is shown on page 56 Matthews. Application of the Previous Theorems. Suppose mT',v() = k . This is a poly of degree k, and we can form the T-cyclic basis of theorem 10: CT',v : { v, T'v, T'2v, .... T'k-1v } where we use T' everywhere. Moreover, the matrix T' in this cyclic basis is given (according to theorem 11) by the companion matrix which corresponds to mT',v() = k, and we know from earlier work that this matrix is exactly the matrix C we discussed in section 1 above, the matrix with all zeros except for the first off-diagonal of ones. Now if we define T as the shift T' = (T - 1I), we can do what we did in section 1 above and claim that mT,v() = ( - 1)k is the v-specific min poly for operator T. Furthermore, the T-cyclic basis is now CT,v: { v, (T - 1I)v, (T - 1I)2v, .... (T - 1I)k-1v } The matrix T = T' + 1I is then none other than J, the k x k Jordon Block matrix with eigenvalue 1. Comments: So, we have now learned that the Jordan Block arises when you take as your basis vectors the strange set { v, (T - 1I)v, (T - 1I)2v, .... (T - 1I)k-1v }. Status: At this stopping point, we have "come down from above" with the primary decomposition to get a set of subspaces V1 and V2 , and we have "come up from below" with some strange cyclic subspace gizmo to create one Jordan block. It remains to connect these two worlds. Status 1/7/05. Things are now pretty well connected, I think I am done with this subject for now, after spending quite a few days on it. Results are all in a binder and in a directory.