Editing Jordan normal form (section)

=== Example: Obtaining the normal form ===

This example shows how to calculate the Jordan normal form of a given matrix.

Consider the matrix
:<math>A =
\left[ \begin{array}{rrrr}
 5 &  4 &  2 &  1 \\
 0 &  1 & -1 & -1 \\
-1 & -1 &  3 &  0 \\ 
 1 &  1 & -1 &  2
\end{array} \right] </math>
which is mentioned in the beginning of the article.

The [[characteristic polynomial]] of ''A'' is
:<math> \begin{align} \chi(\lambda) & = \det(\lambda I - A) \\ &  = \lambda^4 - 11 \lambda^3 + 42 \lambda^2 - 64 \lambda + 32  \\ & = (\lambda-1)(\lambda-2)(\lambda-4)^2. \, \end{align} </math>
This shows that the eigenvalues are 1, 2, 4 and 4, according to algebraic multiplicity. The eigenspace corresponding to the eigenvalue 1 can be found by solving the equation {{math|1=''Av'' = 1''v''.}} It is spanned by the column vector ''v'' = (−1, 1, 0, 0)<sup>T</sup>. Similarly, the eigenspace corresponding to the eigenvalue 2 is spanned by ''w'' = (1, −1, 0, 1)<sup>T</sup>. Finally, the eigenspace corresponding to the eigenvalue 4 is also one-dimensional (even though this is a double eigenvalue) and is spanned by ''x'' = (1, 0, −1, 1)<sup>T</sup>. So, the [[geometric multiplicity]] (that is, the dimension of the eigenspace of the given eigenvalue) of each of the three eigenvalues is one. Therefore, the two eigenvalues equal to 4 correspond to a single Jordan block, and the Jordan normal form of the matrix ''A'' is the [[Matrix addition#Direct sum|direct sum]]
:<math> J = J_1(1) \oplus J_1(2) \oplus J_2(4) = 
\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 4 & 1 \\ 0 & 0 & 0 & 4 \end{bmatrix}. </math>
There are three [[Generalized eigenvector#Jordan chains|Jordan chains]]. Two have length one: {''v''} and {''w''}, corresponding to the eigenvalues 1 and 2, respectively. There is one chain of length two corresponding to the eigenvalue 4. To find this chain, calculate
: <math>\ker(A-4I)^2 = \operatorname{span} \, \left\{ \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}, \left[ \begin{array}{r} 1 \\ 0 \\ -1 \\ 1 \end{array} \right] \right\}</math>

where {{mvar|I}} is the {{val|4| × |4}} identity matrix. Pick a vector in the above span that is not in the kernel of {{math|''A'' − 4''I'';}} for example, ''y'' = (1,0,0,0)<sup>T</sup>. Now, {{math|1=(''A'' − 4''I'')''y'' = ''x''}} and {{math|1=(''A'' − 4''I'')''x'' = 0}}, so {''y'', ''x''} is a chain of length two corresponding to the eigenvalue 4.

The transition matrix ''P'' such that ''P''<sup>−1</sup>''AP'' = ''J'' is formed by putting these vectors next to each other as follows
:<math> P = \left[\begin{array}{c|c|c|c} v & w & x & y \end{array}\right] = 
\left[ \begin{array}{rrrr}
-1 &  1 &  1 &  1 \\
 1 & -1 &  0 &  0 \\ 
 0 &  0 & -1 &  0 \\
 0 &  1 &  1 &  0
\end{array} \right]. </math>
A computation shows that the equation ''P''<sup>−1</sup>''AP'' = ''J'' indeed holds.

:<math>P^{-1}AP=J=\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 2 & 0 & 0 \\
0 & 0 & 4 & 1 \\
0 & 0 & 0 & 4 \end{bmatrix}.</math>

If we had interchanged the order in which the chain vectors appeared, that is, changing the order of ''v'', ''w'' and {''x'', ''y''} together, the Jordan blocks would be interchanged. However, the Jordan forms are equivalent Jordan forms.