﻿ The Square of a Jordan Canonical Form Matrix
San José State University

applet-magic.com
Thayer Watkins
Silicon Valley
USA

 The Square of a Jordan Canonical Form Matrix

## Jordan Block Matrices

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

#### |λ10| |0λ1| |00λ|

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.

## The Jordan Canonical Form of a 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.

#### | Λ10…0  | | 0Λ2…0  | | 0……0  | | 0…0Λk|

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.

## The Square of a Jordan Canonical Form Matrix

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., Kjjj².

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 (λjj+1j.

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.