Jordan Form Theorem: Let V = Cn (or a vector space over an any algebraically closed field).Any linear transformation T : V → V has a basis vi,j (with two subscripts i = 1,...,s and j = 1,...,ni) and eigenvalues λ1,...,λs (a list possibly with repeat entries), such that T acts by:T(vi,1) = λi vi,1 and T(vi,j) = λi vi,j + vi, j-1 for j > 1 .Equivalently, any square matrix M can be written in the form M = P J P-1 where J is a block-diagonal matrix with blocks Jn1(λ1), ... , Jns(λs), where Jn(λ) is the n×n Jordan block:λ | 1 | 0 | ... | 0 |
0 | λ | 1 | ... | 0 |
0 | 0 | λ | ... | 0 |
: | : | : | | : |
0 | 0 | 0 | ... | 1 |
0 | 0 | 0 | ... | λ |
This form is unique except for permuting the order of the Jordan blocks.Different Jordan blocks Jni(λi) may have the same the same eigenvalue: λi = λj for i ≠ j.
T is diagonalizable (has a basis of eigenvectors) if and only if all ni = 1, in which case J is a diagonal matrix.Algorithm to compute the Jordan form of a transformation T on Cn, which has standard basis e = {e1,...,en}. Preliminaries:
- All vectors will be denoted as n×1 column vectors in standard coordinates.
- A subspace W ⊂ Cn is always specified by a basis w = {w1,...,wm}, which can also be considered as an n×m matrix of column vectors. Using column reduction, we may require that some submatrix of w consisting of rows (i1,...,im) forms an m×m identity matrix.
- If T(W) ⊂ W, we say W is a T-invariant subspace, and we let T|W denote the restricted transformation T acting on W . Below we will find the invariant subspaces Wi = Span(vi,1,...,vi,ni) corresponding to the Jordan blocks.
- Procedure: Given a T-invariant subspace W, compute the the nullspace Ker(T|W). Form the n×m matrix of column vectors M = [T(w1),...,T(wm)].Perform row reduction on M to find its kernel vectors u1,...,ut ∈ Cm, then matrix multiply w*ui to get a basis of Ker(T|W).
For extra efficiency, we could instead work with the m×m matrix M' consisting of rows (i1,...,im) of M. The vectors u ∈ Ker(M') ⊂ Cm are the same as those for M, and we again take w*u. - Procedure: Given two spaces W ⊂ V ⊂ Cn, find a complementary space W' such that V = W ⊕ W' ; that is, V = Span(W,W'), and W ∩ W' = 0. Letting m = dim(W), p = dim(V), form the n × (m+p) matrix of the basis vectors of W followed by the basis vectors of V : M := [w1,...,wm,v1,...,vp]. Reduce the last p columns by adding multiples of the first m columns (for efficiency, concentrate on rows (i1,...,im) ). Finally, column-reduce the last p columns, obtaining p-m non-zero column vectors, a basis of W'.
Now, given T, we compute the basis vectors vi,j.
- First find the eigenvalues of T, the roots x = λk of the characteristic polynomial f(x) = det(M-xI) . The rational root test might be useful here, as well as other methods of solving polynomials.
- For each distinct λk , use row reduction to compute Ker(T-λkI)q for q = 1,2,..., which increases until we reach a q with: Ker (T-λkI)q-1 ≠ Ker (T-λkI)q = Ker (T-λkI)q+1 .Let Vk := Ker (T-λkI)q , the generalized eigenspace of T corresponding to all the Jordan blocks with the eigenvalue λk .We thus have a decomposition V = ⊕ Vk into T-invariant subspaces.
- Let Tk := (T-λkI)|Vk, a nilpotent transformation on Vk with Tkq = 0 but Tkq-1 ≠ 0 .We want to find a Jordan basis for this Tk . First, we find a complement Vk,1 to Ker Tkq-1 ⊂ Ker Tkq = Vk . We repeat this, finding complementary spaces Vk,j so that:Vk = Ker Tkq = Ker Tkq-1 ⊕ Vk,1
Ker Tkq-1 = Ker Tkq-2 ⊕ Tk(Vk,1) ⊕ Vk,2
Ker Tkq-2 = Ker Tkq-3 ⊕ Tk2(Vk,1) ⊕ Tk(Vk,2) ⊕ Vk,3
...We get a decomposition Vk = ⊕ Tkp(Vk,j), summed over all j and p. - Now we take v1,q to be a basis vector of V1,1, and v1,q-p := T1pv1,q. We repeat this with another basis vector v2,q of V1,1, until we fill out V1,1. We go on with a basis vector vi,q-1 of V1,2, and continue in this way until we have a basis of V1 .Repeating with V2 , V3 , ... produces the Jordan basis.
Exercises:- Show that if we let D and N be the diagonal and off-diagonal parts of J (respectively), then we get the decomposition M = P D P-1 + P N P-1, where the first term is diagonalizable, the second term is nilpotent (some power equals zero), and the two summands commute with each other under matrix multiplication.
- Extra Credit: Give another proof of the Jordan Form Theorem by showing that the vectors vi,j found in the algorithm are actually a basis of V. The main point is that the spaces Vk are linearly independent and span V, and the spaces Tkp(Vk,j) are linearly independent and span Vk .
- Consider the following matrices J, P, and M = PJP-1, respectively:
2 | 1 | 0 | | 1 | 0 | 0 | | 0 | 2 | -1 |
0 | 2 | 0 | | 1 | 1 | 1 | | -2 | 4 | -1 |
0 | 0 | 2 | | 0 | 1 | 2 | | 0 | 0 | 2 |
Use the above (improved) algorithm to recover the Jordan form J of M, along with the Jordan basis vi,j given by the columns of P.