In general, a square complex matrix
A is
similar to a
block diagonal matrix :J = \begin{bmatrix} J_1 & \; & \; \\ \; & \ddots & \; \\ \; & \; & J_p\end{bmatrix} where each block
Ji is a square matrix of the form :J_i = \begin{bmatrix} \lambda_i & 1 & \; & \; \\ \; & \lambda_i & \ddots & \; \\ \; & \; & \ddots & 1 \\ \; & \; & \; & \lambda_i \end{bmatrix}. So there exists an invertible matrix
P such that
P−1
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
Ji 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
λi, its
geometric multiplicity is the dimension of where is the
identity matrix, and it is the number of Jordan blocks corresponding to
λi. • The sum of the sizes of all Jordan blocks corresponding to an eigenvalue
λi is its
algebraic multiplicity. •
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 matrices; that is, scalars. • The Jordan block corresponding to
λ is of the form , where
N is a
nilpotent matrix defined as
Nij =
δi,
j−1 (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
λi of size at least
j is Thus, the number of Jordan blocks of size
j is • :2 \dim \ker (A - \lambda_i I)^j - \dim \ker (A - \lambda_i I)^{j+1} - \dim \ker (A - \lambda_i I)^{j-1} • Given an eigenvalue
λi, its multiplicity in the minimal polynomial is the size of its largest Jordan block.
Example Consider the matrix A from the example in the previous section. The Jordan normal form is obtained by some
similarity transformation: :P^{-1}AP = J; that is, AP = PJ. Let P have column vectors p_i, i = 1, \ldots, 4, then : 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}. We see that : (A - 1 I) p_1 = 0 : (A - 2 I) p_2 = 0 : (A - 4 I) p_3 = 0 : (A - 4 I) p_4 = p_3. For i = 1,2,3 we have p_i \in \ker(A-\lambda_{i} I), that is, p_i is an eigenvector of A corresponding to the eigenvalue \lambda_i. For i=4, multiplying both sides by (A-4I) gives : (A-4I)^2 p_4 = (A-4I) p_3. But (A-4I)p_3 = 0, so : (A-4I)^2 p_4 = 0. Thus, p_4 \in \ker(A-4 I)^2. Vectors such as p_4 are called
generalized eigenvectors of
A.
Example: Obtaining the normal form This example shows how to calculate the Jordan normal form of a given matrix. Consider the matrix :A = \left[ \begin{array}{rrrr} 5 & 4 & 2 & 1 \\ 0 & 1 & -1 & -1 \\ -1 & -1 & 3 & 0 \\ 1 & 1 & -1 & 2 \end{array} \right] which is mentioned in the beginning of the article. The
characteristic polynomial of
A is : \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} 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 It is spanned by the column vector
v = (−1, 1, 0, 0)T. Similarly, the eigenspace corresponding to the eigenvalue 2 is spanned by
w = (1, −1, 0, 1)T. 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)T. 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
direct sum : 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}. There are three
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 : \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\} where is the identity matrix. Pick a vector in the above span that is not in the kernel of for example,
y = (1,0,0,0)T. Now, and , so {
y,
x} is a chain of length two corresponding to the eigenvalue 4. The transition matrix
P such that
P−1
AP =
J is formed by putting these vectors next to each other as follows : 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]. A computation shows that the equation
P−1
AP =
J indeed holds. :P^{-1}AP=J=\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 4 & 1 \\ 0 & 0 & 0 & 4 \end{bmatrix}. 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. == Generalized eigenvectors ==