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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 {λ_{1}, λ_{2}, …, λ_{k}} of mulitiplicities {m_{1}, m_{2}, …, m_{k}}, respectively, and Σm_{j}=n.
Let Λ_{k} be the Jordan block for λ_{k} and m_{k}. Let J be an n×n matrix with the principal diagonals of the Λ_{k}'s aligned along its principal diagonal and zeros everywhere else.
 Λ_{1}  0  …  0  
 0  Λ_{2}  …  0  
 0  …  …  0  
 0  …  0  Λ_{k} 
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 J_{jk}=0 if k<j. Let J_{jj} be designated as λ_{j} and J_{j,j+1} as δ_{j}, which can be either 0 or 1. The elements J_{jk} for k>j+1 are equal to zero.
Now consider the computation of K=J·J=J². The element for the jth row and jth column of K comes from multiplying the jth row of J times the jth column. The only nonzero product is the square of the element in jth row and jth column of J; i.e., K_{jj}=λ_{j}².
It is helpful for the computation of K_{jk} to write the kth column of J as a row vector below the jth row of J.
For k=j this is
…  0  λ_{j}  δ_{j}  0 … 
…  δ_{j1}  λ_{j}  0  0 … 
For k=j1 it is
…  0  0  λ_{j}  δ_{j}  0 …  
…  δ_{j2}  λ_{j1}  0  0  0  … 
Thus the element K_{j,j1} is equal to zero.
For k=j+1
…  0  0  λ_{j}  δ_{j}  0 …  
…  0  0  δ_{j}  λ_{j+1}  0  … 
Thus the element K_{j,j+1} is equal to (λ_{j}+λ_{j+1})δ_{j}.
For k=j+2
…  0  0  λ_{j}  δ_{j}  0 …  
…  0  0  0  δ_{j+1}  λ_{j+2}  0  … 
Thus the element K_{j,j+1} is equal to δ_{j}·δ_{j+1}.
For all other cases K_{jk} is equal to zero.
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