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An m×m matrix is of the Jordan block form if it has a constant on the principal diagonal and 1's for all the elements next to the principal diagonal on the right. All other elements are zero. For example, suppose m=3. Then
|λ | 1 | 0| |
|0 | λ | 1| |
|0 | 0 | λ| |
is a Jordan block. Such blocks can be represented as λI+H, where H is a square matrix of zeroes except for elements of 1 to the immediate right of the principal diagonal.
For m=1 a Jordan block is just a constant, say λ, a 1×1 matrix.
Suppose an n×n matrix A of complex values has k eigenvalues of {λ1, λ2, …, λk} of mulitiplicities {m1, m2, …, mk}, respectively, and Σmj=n.
Let Λk be the Jordan block for λk and mk. Let J be an n×n matrix with the principal diagonals of the Λk's aligned along its principal diagonal and zeros everywhere else.
| Λ1 | 0 | … | 0 | |
| 0 | Λ2 | … | 0 | |
| 0 | … | … | 0 | |
| 0 | … | 0 | Λk| |
This is a Jordan canonical form of the matrix A. There may be a number of such canonical forms because the ordering of the eigenvalues is arbitrary. If all of the eigenvalues are different then the canonical form is a diagonal matrix with the eigenvalues on the principal diagonal.
Let J be an n×n Jordan canonical form matrix. The elements Jjk=0 if k<j. Let Jjj be designated as λj and Jj,j+1 as δj, which can be either 0 or 1. The elements Jjk for k>j+1 are equal to zero.
Now consider the computation of K=J·J=J². The element for the j-th row and j-th column of K comes from multiplying the j-th row of J times the j-th column. The only non-zero product is the square of the element in j-th row and j-th column of J; i.e., Kjj=λj².
It is helpful for the computation of Kjk to write the k-th column of J as a row vector below the j-th row of J.
For k=j this is
… | 0 | λj | δj | 0 … |
… | δj-1 | λj | 0 | 0 … |
For k=j-1 it is
… | 0 | 0 | λj | δj | 0 … | |
… | δj-2 | λj-1 | 0 | 0 | 0 | … |
Thus the element Kj,j-1 is equal to zero.
For k=j+1
… | 0 | 0 | λj | δj | 0 … | |
… | 0 | 0 | δj | λj+1 | 0 | … |
Thus the element Kj,j+1 is equal to (λj+λj+1)δj.
For k=j+2
… | 0 | 0 | λj | δj | 0 … | ||
… | 0 | 0 | 0 | δj+1 | λj+2 | 0 | … |
Thus the element Kj,j+1 is equal to δj·δj+1.
For all other cases Kjk is equal to zero.
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