Editing Jordan normal form (section)

== Complex matrices ==

In general, a square complex matrix ''A'' is [[similar (linear algebra)|similar]] to a [[block diagonal matrix]]

:<math>J = \begin{bmatrix}
J_1 & \;     & \; \\
\;  & \ddots & \; \\ 
\;  & \;     & J_p\end{bmatrix}</math>

where each block ''J<sub>i</sub>'' is a square matrix of the form

:<math>J_i = 
\begin{bmatrix}
\lambda_i & 1            & \;     & \;  \\
\;        & \lambda_i    & \ddots & \;  \\
\;        & \;           & \ddots & 1   \\
\;        & \;           & \;     & \lambda_i       
\end{bmatrix}.</math>

So there exists an invertible matrix ''P'' such that ''P''<sup>−1</sup>''AP'' = ''J'' is such that the only non-zero entries of ''J'' are on the diagonal and the superdiagonal. ''J'' is called the '''Jordan normal form''' of ''A''. Each ''J''<sub>''i''</sub> is called a [[Jordan block]] of ''A''. In a given Jordan block, every entry on the superdiagonal is 1.

Assuming this result, we can deduce the following properties:

* Counting multiplicities, the eigenvalues of ''J'', and therefore of ''A'', are the diagonal entries.
* Given an eigenvalue ''λ''<sub>''i''</sub>, its '''[[geometric multiplicity]]''' is the dimension of {{math|ker(''A'' &minus; ''λ''<sub>''i'' </sub>''I''),}} where {{mvar|I}} is the [[identity matrix]], and it is the number of Jordan blocks corresponding to ''λ''<sub>''i''</sub>.<ref name="HJp321">{{harvtxt|Horn|Johnson|1985|loc=§3.2.1}}</ref>
* The sum of the sizes of all Jordan blocks corresponding to an eigenvalue ''λ''<sub>''i''</sub> is its '''[[algebraic multiplicity]]'''.<ref name="HJp321" />
* ''A'' is diagonalizable if and only if, for every eigenvalue ''λ'' of ''A'', its geometric and algebraic multiplicities coincide. In particular, the Jordan blocks in this case are {{val|1|×|1}} matrices; that is, scalars.
* The Jordan block corresponding to ''λ'' is of the form {{math|''λI'' + ''N''}}, where ''N'' is a [[nilpotent matrix]] defined as ''N''<sub>''ij''</sub> = ''δ<sub>i</sub>''<sub>,''j''&minus;1</sub> (where δ is the [[Kronecker delta]]). The nilpotency of ''N'' can be exploited when calculating ''f''(''A'') where ''f'' is a complex analytic function. For example, in principle the Jordan form could give a closed-form expression for the exponential exp(''A'').
* The number of Jordan blocks corresponding to ''λ''<sub>''i''</sub> of size at least ''j'' is {{math|dim ker(''A'' − ''λ''<sub>''i''</sub>''I'')<sup>''j''</sup> − dim ker(''A'' − ''λ''<sub>''i''</sub>''I'')<sup>''j''−1</sup>.}} Thus, the number of Jordan blocks of size ''j'' is
*:<math>2 \dim \ker (A - \lambda_i I)^j - \dim \ker (A - \lambda_i I)^{j+1} - \dim \ker (A - \lambda_i I)^{j-1}</math>
* Given an eigenvalue ''λ''<sub>''i''</sub>, its multiplicity in the minimal polynomial is the size of its largest Jordan block.

=== Example ===
Consider the matrix <math>A</math> from the example in the previous section. The Jordan normal form is obtained by some [[Matrix similarity|similarity transformation]]:

:<math>P^{-1}AP = J;</math> that is, <math>AP = PJ.</math>

Let <math>P</math> have column vectors <math>p_i</math>, <math>i = 1, \ldots, 4</math>, then

: <math>A \begin{bmatrix} p_1 & p_2 & p_3 & p_4 \end{bmatrix} = \begin{bmatrix} p_1 & p_2 & p_3 & p_4 \end{bmatrix}
\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 2 & 0 & 0 \\ 
0 & 0 & 4 & 1 \\
0 & 0 & 0 & 4 \end{bmatrix} = \begin{bmatrix} p_1 & 2p_2 & 4p_3 & p_3+4p_4 \end{bmatrix}.</math>

We see that
:<math> (A - 1 I) p_1 = 0 </math>
:<math> (A - 2 I) p_2 = 0 </math>
:<math> (A - 4 I) p_3 = 0 </math>
:<math> (A - 4 I) p_4 = p_3. </math>

For <math>i = 1,2,3</math> we have <math>p_i \in \ker(A-\lambda_{i} I)</math>, that is, <math>p_i</math> is an eigenvector of <math>A</math> corresponding to the eigenvalue <math>\lambda_i</math>. For <math>i=4</math>, multiplying both sides by <math>(A-4I)</math> gives
:<math> (A-4I)^2 p_4 = (A-4I) p_3. </math>
But <math>(A-4I)p_3 = 0</math>, so
:<math> (A-4I)^2 p_4 = 0. </math>
Thus, <math>p_4 \in \ker(A-4 I)^2.</math>

Vectors such as <math>p_4</math> are called [[generalized eigenvector]]s of ''A''.

=== 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.