Editing Jordan normal form (section)

== Overview ==
=== Notation ===
Some textbooks have the ones on the [[subdiagonal]]; that is, immediately below the main diagonal instead of on the superdiagonal. The eigenvalues are still on the main diagonal.<ref>{{harvtxt|Cullen|1966|p=114}}</ref><ref>{{harvtxt|Franklin|1968|p=122}}</ref>

=== Motivation ===
An ''n'' × ''n'' matrix ''A'' is [[diagonalizable matrix|diagonalizable]] if and only if the sum of the dimensions of the eigenspaces is ''n''. Or, equivalently, if and only if ''A'' has ''n'' [[linearly independent]] [[eigenvectors]]. Not all matrices are diagonalizable; matrices that are not diagonalizable are called [[defective matrix|defective]] matrices. Consider the following matrix:

: <math>A =
  \left[\begin{array}{*{20}{r}}
     5 &  4 &  2 &  1 \\[2pt]
     0 &  1 & -1 & -1 \\[2pt]
    -1 & -1 &  3 &  0 \\[2pt]
     1 &  1 & -1 &  2
  \end{array}\right].
</math>

Including multiplicity, the eigenvalues of ''A'' are ''λ'' = 1, 2, 4, 4. The [[Hamel dimension|dimension]] of the eigenspace corresponding to the eigenvalue 4 is 1 (and not 2), so ''A'' is not diagonalizable. However, there is an invertible matrix ''P'' such that ''J'' = ''P''<sup>−1</sup>''AP'', where

:<math>J = \begin{bmatrix}
  1 & 0 & 0 & 0 \\[2pt]
  0 & 2 & 0 & 0 \\[2pt]
  0 & 0 & 4 & 1 \\[2pt]
  0 & 0 & 0 & 4
\end{bmatrix}.</math>

The matrix <math>J</math> is almost diagonal. This is the Jordan normal form of ''A''. The section [[#Example|''Example'']] below fills in the details of the computation.