Editing Jordan normal form (section)

=== Uniqueness ===

It can be shown that the Jordan normal form of a given matrix ''A'' is unique up to the order of the Jordan blocks.

Knowing the algebraic and geometric multiplicities of the eigenvalues is not sufficient to determine the Jordan normal form of ''A''. Assuming the algebraic multiplicity ''m''(''λ'') of an eigenvalue ''λ'' is known, the structure of the Jordan form can be ascertained by analyzing the ranks of the powers {{math|1=(''A'' − ''λI'')<sup>''m''(''λ'')</sup>}}. To see this, suppose an ''n'' × ''n'' matrix ''A'' has only one eigenvalue ''λ''. So ''m''(''λ'') = ''n''. The smallest integer ''k''<sub>1</sub> such that

:<math>(A - \lambda I)^{k_1} = 0</math>

is the size of the largest Jordan block in the Jordan form of ''A''. (This number ''k''<sub>1</sub> is also called the '''index''' of ''λ''. See discussion in a following section.) The rank of

:<math>(A - \lambda I)^{k_1 - 1}</math>

is the number of Jordan blocks of size ''k''<sub>1</sub>. Similarly, the rank of

:<math>(A - \lambda I)^{k_1 - 2}</math>

is twice the number of Jordan blocks of size ''k''<sub>1</sub> plus the number of Jordan blocks of size ''k''<sub>1</sub> − 1. The general case is similar.

This can be used to show the uniqueness of the Jordan form. Let ''J''<sub>1</sub> and ''J''<sub>2</sub> be two Jordan normal forms of ''A''. Then ''J''<sub>1</sub> and ''J''<sub>2</sub> are similar and have the same spectrum, including algebraic multiplicities of the eigenvalues. The procedure outlined in the previous paragraph can be used to determine the structure of these matrices. Since the rank of a matrix is preserved by similarity transformation, there is a bijection between the Jordan blocks of ''J''<sub>1</sub> and ''J''<sub>2</sub>. This proves the uniqueness part of the statement.