T. S. Blyth & E. F. Robertson · First-year university · Study material generated by AI from this edition. Not affiliated with the publisher.
Chapter 3 · pp. 41–53
This chapter shows how to bring a linear map to a nearly diagonal shape called the Jordan canonical form. We start with nilpotent maps, whose kernel chains let us build a special basis. Then we glue the pieces together using generalised eigenspaces, read off the form from the characteristic and minimum polynomials, find the basis in practice, and finish with applications.
A map is nilpotent of index if but . The kernels of its powers form a strictly increasing chain: (p. 42). The reason is that if two neighbouring kernels were equal, every later kernel would equal them too. The chain would then stop growing before reaching , and would have to be 0. That contradicts the index being . Example: for on , has dimension 2 and (p. 42).
To build a good basis we need to push vectors down the chain. Choose bases of the kernels by successive extension, each basis extending the previous one. Take a basis of and add the extra vectors that complete a basis of . Applying to those extra vectors lands in . The basis of together with these images is linearly independent. To see why, suppose a combination of them is 0. Then the combination of the new vectors lies in , so by independence its coefficients are all 0. The remaining coefficients must then be 0 too (p. 42).
An elementary Jordan matrix for a scalar is either the matrix or a square matrix with on the diagonal, 1 on the superdiagonal and 0 elsewhere. For and size 3 the rows are , , (p. 43).
A Jordan block matrix for is a block diagonal matrix whose diagonal blocks are all elementary Jordan matrices for that same . The sizes may differ. For example, a and a elementary matrix for combine into a Jordan block (p. 43).
If is nilpotent of index on a non-zero finite-dimensional space , there is a basis of in which the matrix of is a Jordan block with 0 on the diagonal. The diagonal must be 0 because a nilpotent map has only the eigenvalue 0 (pp. 43–44).
The construction goes as follows (p. 44).
Worked example: is nilpotent of index 2. Here . Take the top-level vectors and . Their images are and . Order the basis as . The matrix then has 1s in positions (1,2) and (3,4), so it consists of two elementary matrices with 0 on the diagonal (pp. 45–46).
Every column of the basis table ends in the bottom row, which is a basis of . So the number of elementary Jordan matrices equals . At least one has size , where is the index (p. 46). Example: on a 4-dimensional space with and index 2, the only possibility is two blocks.
More precisely, let and . The number of blocks of size exactly is . The number of blocks of size exactly is . The reason is that counts blocks of size at least , and we subtract those of size (p. 46).
Suppose the minimum polynomial is . Let be the generalised eigenspace and the restriction of to . Then is nilpotent of index . The minimum polynomial of is , so and no smaller power vanishes (p. 46). We can therefore apply the nilpotent result on each .
If all eigenvalues of lie in the field , there is an ordered basis in which the matrix of is block diagonal. Each diagonal block is a Jordan block for one eigenvalue. This is the Jordan canonical form. On each , adding back to the nilpotent Jordan block puts on the diagonal, and joining the bases of the gives the whole matrix. Over this applies to every linear map (p. 47).
The form is unique up to the order of the Jordan blocks. The number of blocks, their sizes and the number of elementary matrices in each are determined by . Matrices with the same Jordan form up to ordering are similar (p. 47).
Let and . In the Jordan form (p. 47):
Example: and forces block sizes 2 and 1.
If and , eigenvalue 1 has sizes 2+1 and eigenvalue 2 has sizes 3+1. Nothing else fits, so the form is unique (p. 48).
The minimum polynomial does not always decide the form. If instead , eigenvalue 2 occurs four times with largest block exactly 2. The sizes can be 2+2 or 2+1+1, so two forms are possible (p. 48).
If , there are six possible minimum polynomials with and , and each gives a possible form. For eigenvalue 3: gives 1+1+1, gives 2+1, and gives one block. For eigenvalue 2: gives one block and gives two blocks (pp. 48–49).
A common misconception is that any matrix with a repeated eigenvalue can be diagonalised. In fact diagonalisation works only when every Jordan block is , which happens exactly when every factor of the minimum polynomial has power 1. If , the eigenvalue 2 has a block with a 1 above the diagonal, so the matrix is not diagonalisable (pp. 47–48, 50).
A Jordan basis for is a basis in which the matrix of is a Jordan canonical matrix . Equivalently, there is an invertible with , where is the matrix of in a fixed basis. The columns of are the Jordan basis vectors written in the fixed basis (p. 49).
For an elementary block of size for , a chain must satisfy and for . So is an eigenvector lying in both and . Then solve for , then , and so on. Each produces a 1 above the diagonal. All vectors must be linearly independent (pp. 49–50).
Example: has and . We get . Take , which is in the image and the kernel. Then satisfies . Finally is in the kernel and independent of the others. The Jordan form has a block and a block for eigenvalue 2 (p. 50).
Similarity to the transpose. Every square matrix over is similar to its transpose. It is enough to check an elementary Jordan matrix. If is a Jordan basis, reversing it to gives the transpose, with the 1s below the diagonal. Block diagonal matrices then follow block by block (p. 51).
Differential equations. To solve with constant coefficients, choose with and put . Since is constant, . Solve for , then set . The Jordan form is nearly diagonal, so the equations can be solved one after another from the bottom up (pp. 51–52).
Example: , , . Here . So has the block and one block for eigenvalue 2 (p. 52). The reduced system is , , . Solving upward gives and . Then gives . The factor appears because of the block. Multiplying by gives the general solution (p. 53).
