In linear algebra, a Jordan normal form, also known as a Jordan canonical form
or JCF,
is an upper triangular matrix of a particular form called a Jordan matrix representing a linear operator on a finite-dimensional vector space with respect to some Basis (linear algebra), basis. Such a matrix has each non-zero off-diagonal entry equal to 1, immediately above the main diagonal (on the superdiagonal), and with identical diagonal entries to the left and below them.
Let ''V'' be a vector space over a field (mathematics), field ''K''. Then a basis with respect to which the matrix has the required form exists if and only if all eigenvalues of the matrix lie in ''K'', or equivalently if the characteristic polynomial of the operator splits into linear factors over ''K''. This condition is always satisfied if ''K'' is algebraically closed (for instance, if it is the field of complex numbers). The diagonal entries of the normal form are the eigenvalues (of the operator), and the number of times each eigenvalue occurs is called the algebraic multiplicity of the eigenvalue.
If the operator is originally given by a square matrix ''M'', then its Jordan normal form is also called the Jordan normal form of ''M''. Any square matrix has a Jordan normal form if the field of coefficients is extended to one containing all the eigenvalues of the matrix. In spite of its name, the normal form for a given ''M'' is not entirely unique, as it is a block diagonal matrix formed of Jordan blocks, the order of which is not fixed; it is conventional to group blocks for the same eigenvalue together, but no ordering is imposed among the eigenvalues, nor among the blocks for a given eigenvalue, although the latter could for instance be ordered by weakly decreasing size.
The Jordan–Chevalley decomposition is particularly simple with respect to a basis for which the operator takes its Jordan normal form. The diagonal form for diagonalizable matrices, for instance normal matrix, normal matrices, is a special case of the Jordan normal form.
The Jordan normal form is named after Camille Jordan, who first stated the Jordan decomposition theorem in 1870.Brechenmacher

"Histoire du théorème de Jordan de la décomposition matricielle (1870-1930). Formes de représentation et méthodes de décomposition"

Thesis, 2007

^{−1}''AP'', where
:$J\; =\; \backslash begin\; 1\; \&\; 0\; \&\; 0\; \&\; 0\; \backslash \backslash [2pt]\; 0\; \&\; 2\; \&\; 0\; \&\; 0\; \backslash \backslash [2pt]\; 0\; \&\; 0\; \&\; 4\; \&\; 1\; \backslash \backslash [2pt]\; 0\; \&\; 0\; \&\; 0\; \&\; 4\; \backslash end.$
The matrix $J$ is almost diagonal. This is the Jordan normal form of ''A''. The section #Example, ''Example'' below fills in the details of the computation.

_{i}'' is a square matrix of the form
:$J\_i\; =\; \backslash begin\; \backslash lambda\_i\; \&\; 1\; \&\; \backslash ;\; \&\; \backslash ;\; \backslash \backslash \; \backslash ;\; \&\; \backslash lambda\_i\; \&\; \backslash ddots\; \&\; \backslash ;\; \backslash \backslash \; \backslash ;\; \&\; \backslash ;\; \&\; \backslash ddots\; \&\; 1\; \backslash \backslash \; \backslash ;\; \&\; \backslash ;\; \&\; \backslash ;\; \&\; \backslash lambda\_i\; \backslash end.$
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 ''J''_{''i''} 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''_{''i''}, its geometric multiplicity is the dimension of Ker(''A'' − ''λ''_{''i'' }''I''), where ''I'' is the identity matrix, and it is the number of Jordan blocks corresponding to ''λ''_{''i''}.
* The sum of the sizes of all Jordan blocks corresponding to an eigenvalue ''λ''_{''i''} 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 ''1 × 1'' matrices; that is, scalars.
* The Jordan block corresponding to ''λ'' is of the form ''λI'' + ''N'', where ''N'' is a nilpotent matrix defined as ''N''_{''ij''} = ''δ_{i}''_{,''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 ''λ'' of size at least ''j'' is dim Ker(''A'' − ''λI'')^{''j''} − dim Ker(''A'' - ''λI'')^{''j''}^{−1}. Thus, the number of Jordan blocks of size ''j'' is
*:$2\; \backslash dim\; \backslash ker\; (A\; -\; \backslash lambda\_i\; I)^j\; -\; \backslash dim\; \backslash ker\; (A\; -\; \backslash lambda\_i\; I)^\; -\; \backslash dim\; \backslash ker\; (A\; -\; \backslash lambda\_i\; I)^$
* Given an eigenvalue ''λ''_{''i''}, its multiplicity in the minimal polynomial is the size of its largest Jordan block.

^{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 Matrix addition#Direct sum, direct sum
:$J\; =\; J\_1(1)\; \backslash oplus\; J\_1(2)\; \backslash oplus\; J\_2(4)\; =\; \backslash begin\; 1\; \&\; 0\; \&\; 0\; \&\; 0\; \backslash \backslash \; 0\; \&\; 2\; \&\; 0\; \&\; 0\; \backslash \backslash \; 0\; \&\; 0\; \&\; 4\; \&\; 1\; \backslash \backslash \; 0\; \&\; 0\; \&\; 0\; \&\; 4\; \backslash end.$
There are three Generalized eigenvector#Jordan chains, Jordan chains. Two have length one: and , 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
: $\backslash ker^2\; =\; \backslash operatorname\; \backslash ,\; \backslash left\backslash $
where ''I'' is the 4 × 4 identity matrix. Pick a vector in the above span that is not in the kernel of ''A'' − 4''I''; for example, ''y'' = (1,0,0,0)^{T}. Now, (''A'' − 4''I'')''y'' = ''x'' and (''A'' − 4''I'')''x'' = 0, so 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
:$P\; =\; \backslash left[\backslash begin\; v\; \&\; w\; \&\; x\; \&\; y\; \backslash end\backslash right]\; =\; \backslash begin\; -1\; \&\; 1\; \&\; 1\; \&\; 1\; \backslash \backslash \; 1\; \&\; -1\; \&\; 0\; \&\; 0\; \backslash \backslash \; 0\; \&\; 0\; \&\; -1\; \&\; 0\; \backslash \backslash \; 0\; \&\; 1\; \&\; 1\; \&\; 0\; \backslash end.$
A computation shows that the equation ''P''^{−1}''AP'' = ''J'' indeed holds.
:$P^AP=J=\backslash begin\; 1\; \&\; 0\; \&\; 0\; \&\; 0\; \backslash \backslash \; 0\; \&\; 2\; \&\; 0\; \&\; 0\; \backslash \backslash \; 0\; \&\; 0\; \&\; 4\; \&\; 1\; \backslash \backslash \; 0\; \&\; 0\; \&\; 0\; \&\; 4\; \backslash end.$
If we had interchanged the order in which the chain vectors appeared, that is, changing the order of ''v'', ''w'' and together, the Jordan blocks would be interchanged. However, the Jordan forms are equivalent Jordan forms.

_{r}'', of the chain is a generalized eigenvector such that (''A'' − λ I)^{''r''}''p''_{''r''} = 0, where ''r'' is the size of the Jordan block. The vector ''p''_{1} = (''A'' − λ I)^{''r''−1}''p''_{''r''} is an eigenvector corresponding to λ. In general, ''p''_{''i''} is a preimage of ''p''_{''i''−1} under ''A'' − λ I. So the lead vector generates the chain via multiplication by (''A'' − λ I).
Therefore, the statement that every square matrix ''A'' can be put in Jordan normal form is equivalent to the claim that there exists a basis consisting only of eigenvectors and generalized eigenvectors of ''A''.

_{''i''} be such that
:$\backslash ;\; (A\; -\; \backslash lambda\; I)\; q\_i\; =\; p\_i\; \backslash mbox\; i\; =\; r-s+1,\; \backslash ldots,\; r.$
Clearly no non-trivial linear combination of the ''q''_{''i''} can lie in Ker(''A'' − λ ''I''), for _{''i=r−s+1, ..., r''} are linearly independent. Furthermore, no non-trivial linear combination of the ''q''_{''i''} can be in Ran(''A'' − λ I), for that would contradict the inductive hypothesis that each ''p_{i}'' is a lead vector in a Jordan chain. The set , being preimages of the linearly independent set under ''A'' − λ I, is also linearly independent.
Finally, we can pick any linearly independent set whose projection spans
:$\backslash ;\; \backslash mathrm(A\; -\; \backslash lambda\; I)\; /\; Q.$
By construction, the union of the three sets , , and is linearly independent. Each vector in the union is either an eigenvector or a generalized eigenvector of ''A''. Finally, by the rank–nullity theorem, the cardinality of the union is ''n''. In other words, we have found a basis that consists of eigenvectors and generalized eigenvectors of ''A'', and this shows ''A'' can be put in Jordan normal form.

^{''m''(λ)}. To see this, suppose an ''n'' × ''n'' matrix ''A'' has only one eigenvalue λ. So ''m''(λ) = ''n''. The smallest integer ''k''_{1} such that
:$(A\; -\; \backslash lambda\; I)^\; =\; 0$
is the size of the largest Jordan block in the Jordan form of ''A''. (This number ''k''_{1} is also called the index of λ. See discussion in a following section.) The rank of
:$(A\; -\; \backslash lambda\; I)^$
is the number of Jordan blocks of size ''k''_{1}. Similarly, the rank of
:$(A\; -\; \backslash lambda\; I)^$
is twice the number of Jordan blocks of size ''k''_{1} plus the number of Jordan blocks of size ''k''_{1}−1. The general case is similar.
This can be used to show the uniqueness of the Jordan form. Let ''J''_{1} and ''J''_{2} be two Jordan normal forms of ''A''. Then ''J''_{1} and ''J''_{2} 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''_{1} and ''J''_{2}. This proves the uniqueness part of the statement.

^{−1}''AP'' = ''J'' is a real block diagonal matrix with each block being a real Jordan block. A real Jordan block is either identical to a complex Jordan block (if the corresponding eigenvalue $\backslash lambda\_i$ is real), or is a block matrix itself, consisting of 2×2 blocks (for non-real eigenvalue $\backslash lambda\_i\; =\; a\_i+ib\_i$ with given algebraic multiplicity) of the form
:$C\_i\; =\; \backslash begin\; a\_i\; \&\; -b\_i\; \backslash \backslash \; b\_i\; \&\; a\_i\; \backslash \backslash \; \backslash end$
and describe multiplication by $\backslash lambda\_i$ in the complex plane. The superdiagonal blocks are 2×2 identity matrices and hence in this representation the matrix dimensions are larger than the complex Jordan form. The full real Jordan block is given by
:$J\_i\; =\; \backslash begin\; C\_i\; \&\; I\; \&\; \&\; \backslash \backslash \; \&\; C\_i\; \&\; \backslash ddots\; \&\; \backslash \backslash \; \&\; \&\; \backslash ddots\; \&\; I\; \backslash \backslash \; \&\; \&\; \&\; C\_i\; \backslash \backslash \; \backslash end.$
This real Jordan form is a consequence of the complex Jordan form. For a real matrix the nonreal eigenvectors and generalized eigenvectors can always be chosen to form complex conjugate pairs. Taking the real and imaginary part (linear combination of the vector and its conjugate), the matrix has this form with respect to the new basis.

^{''k''} for 1 ≤ ''k'' ≤ ''m'', where ''m'' is the algebraic multiplicity of the eigenvalue λ, allows one to determine the Jordan form of ''M''. We may view the underlying vector space ''V'' as a ''K''[''x'']-module (mathematics), module by regarding the action of ''x'' on ''V'' as application of ''M'' and extending by ''K''-linearity. Then the polynomials (''x'' − λ)^{''k''} are the elementary divisors of ''M'', and the Jordan normal form is concerned with representing ''M'' in terms of blocks associated to the elementary divisors.
The proof of the Jordan normal form is usually carried out as an application to the ring (mathematics), ring ''K''[''x''] of the structure theorem for finitely generated modules over a principal ideal domain, of which it is a corollary.

_{1}, ..., λ_{''n''}, then for any polynomial ''p'', ''p''(''A'') has eigenvalues ''p''(λ_{1}), ..., ''p''(λ_{''n''}).

_{1}, ..., λ_{''q''} be the distinct eigenvalues of ''A'', and ''s''_{''i''} be the size of the largest Jordan block corresponding to λ_{''i''}. It is clear from the Jordan normal form that the minimal polynomial of ''A'' has degree ''s''_{''i''}.
While the Jordan normal form determines the minimal polynomial, the converse is not true. This leads to the notion of elementary divisors. The elementary divisors of a square matrix ''A'' are the characteristic polynomials of its Jordan blocks. The factors of the minimal polynomial ''m'' are the elementary divisors of the largest degree corresponding to distinct eigenvalues.
The degree of an elementary divisor is the size of the corresponding Jordan block, therefore the dimension of the corresponding invariant subspace. If all elementary divisors are linear, ''A'' is diagonalizable.

_{''i''} corresponds to an invariant subspace ''X''_{''i''}. Symbolically, we put
:$\backslash mathbb^n\; =\; \backslash bigoplus\_^k\; X\_i$
where each ''X''_{''i''} is the span of the corresponding Jordan chain, and ''k'' is the number of Jordan chains.
One can also obtain a slightly different decomposition via the Jordan form. Given an eigenvalue λ_{''i''}, the size of its largest corresponding Jordan block ''s''_{i} is called the index of λ_{''i''} and denoted by ''ν''(λ_{''i''}). (Therefore, the degree of the minimal polynomial is the sum of all indices.) Define a subspace ''Y''_{''i''} by
:$Y\_i\; =\; \backslash operatorname\; (\backslash lambda\_i\; I\; -\; A)^.$
This gives the decomposition
:$\backslash mathbb^n\; =\; \backslash bigoplus\_^l\; Y\_i$
where ''l'' is the number of distinct eigenvalues of ''A''. Intuitively, we glob together the Jordan block invariant subspaces corresponding to the same eigenvalue. In the extreme case where ''A'' is a multiple of the identity matrix we have ''k'' = ''n'' and ''l'' = 1.
The projection onto ''Y_{i}'' and along all the other ''Y_{j}'' ( ''j'' ≠ ''i'' ) is called the spectral projection of ''A'' at λ_{''i''} and is usually denoted by ''P''(λ_{''i''} ; ''A''). Spectral projections are mutually orthogonal in the sense that ''P''(λ_{''i''} ; ''A'') ''P''(λ_{''j''} ; ''A'') = 0 if ''i'' ≠ ''j''. Also they commute with ''A'' and their sum is the identity matrix. Replacing every λ_{''i''} in the Jordan matrix ''J'' by one and zeroing all other entries gives ''P''(λ_{''i''} ; ''J''), moreover if ''U J U''^{−1} is the similarity transformation such that ''A'' = ''U J U''^{−1} then ''P''(λ_{''i''} ; ''A'') = ''U P''(λ_{''i''} ; ''J'') ''U''^{−1}. They are not confined to finite dimensions. See below for their application to compact operators, and in holomorphic functional calculus for a more general discussion.
Comparing the two decompositions, notice that, in general, ''l'' ≤ ''k''. When ''A'' is normal, the subspaces ''X''_{''i''}'s in the first decomposition are one-dimensional and mutually orthogonal. This is the spectral theorem for normal operators. The second decomposition generalizes more easily for general compact operators on Banach spaces.
It might be of interest here to note some properties of the index, ''ν''(''λ''). More generally, for a complex number λ, its index can be defined as the least non-negative integer ''ν''(λ) such that
:$\backslash mathrm(\backslash lambda\; -\; A)^\; =\; \backslash operatorname\; (\backslash lambda\; -\; A)^m,\; \backslash ;\; \backslash forall\; m\; \backslash geq\; \backslash nu(\backslash lambda)\; .$
So ''ν''(λ) > 0 if and only if λ is an eigenvalue of ''A''. In the finite-dimensional case, ''ν''(λ) ≤ the algebraic multiplicity of λ.

^{n}'':
:$\backslash begin\; \backslash lambda\_1\; \&\; 1\; \&\; 0\; \&\; 0\; \&\; 0\; \backslash \backslash \; 0\; \&\; \backslash lambda\_1\; \&\; 1\; \&\; 0\; \&\; 0\; \backslash \backslash \; 0\; \&\; 0\; \&\; \backslash lambda\_1\; \&\; 0\; \&\; 0\; \backslash \backslash \; 0\; \&\; 0\; \&\; 0\; \&\; \backslash lambda\_2\; \&\; 1\; \backslash \backslash \; 0\; \&\; 0\; \&\; 0\; \&\; 0\; \&\; \backslash lambda\_2\; \backslash end^n\; =\backslash begin\; \backslash lambda\_1^n\; \&\; \backslash tbinom\backslash lambda\_1^\; \&\; \backslash tbinom\backslash lambda\_1^\; \&\; 0\; \&\; 0\; \backslash \backslash \; 0\; \&\; \backslash lambda\_1^n\; \&\; \backslash tbinom\backslash lambda\_1^\; \&\; 0\; \&\; 0\; \backslash \backslash \; 0\; \&\; 0\; \&\; \backslash lambda\_1^n\; \&\; 0\; \&\; 0\; \backslash \backslash \; 0\; \&\; 0\; \&\; 0\; \&\; \backslash lambda\_2^n\; \&\; \backslash tbinom\backslash lambda\_2^\; \backslash \backslash \; 0\; \&\; 0\; \&\; 0\; \&\; 0\; \&\; \backslash lambda\_2^n\; \backslash end,$
where the binomial coefficients are defined as $\backslash binom=\backslash prod\_^k\; \backslash frac$. For integer positive ''n'' it reduces to standard definition
of the coefficients. For negative ''n'' the identity $\backslash binom=\backslash left(-1\backslash right)^k\backslash binom$ may be of use.

_{''i''} be the function that is 1 in some open neighborhood of λ_{''i''} and 0 elsewhere. By property 3 of the functional calculus, the operator
:$e\_i(T)$
is a projection. Moreover, let ''ν_{i}'' be the index of λ_{''i''} and
:$f(z)=\; (z\; -\; \backslash lambda\_i)^.$
The spectral mapping theorem tells us
:$f(T)\; e\_i\; (T)\; =\; (T\; -\; \backslash lambda\_i)^\; e\_i\; (T)$
has spectrum . By property 1, ''f''(''T'') can be directly computed in the Jordan form, and by inspection, we see that the operator ''f''(''T'')''e_{i}''(''T'') is the zero matrix.
By property 3, ''f''(''T'') ''e''_{''i''}(''T'') = ''e''_{''i''}(''T'') ''f''(''T''). So ''e''_{''i''}(''T'') is precisely the projection onto the subspace
:$\backslash operatorname\; e\_i\; (T)\; =\; \backslash mathrm(T\; -\; \backslash lambda\_i)^.$
The relation
:$\backslash sum\_i\; e\_i\; =\; 1$
implies
:$\backslash mathbb^n\; =\; \backslash bigoplus\_i\; \backslash ;\; \backslash operatorname\; e\_i\; (T)\; =\; \backslash bigoplus\_i\; \backslash ;\; \backslash mathrm(T\; -\; \backslash lambda\_i)^$
where the index ''i'' runs through the distinct eigenvalues of ''T''. This is the invariant subspace decomposition
:$\backslash mathbb^n\; =\; \backslash bigoplus\_i\; Y\_i$
given in a previous section. Each ''e_{i}''(''T'') is the projection onto the subspace spanned by the Jordan chains corresponding to λ_{''i''} and along the subspaces spanned by the Jordan chains corresponding to λ_{''j''} for ''j'' ≠ ''i''. In other words, ''e_{i}''(''T'') = ''P''(λ_{''i''};''T''). This explicit identification of the operators ''e_{i}''(''T'') in turn gives an explicit form of holomorphic functional calculus for matrices:
:For all ''f'' ∈ Hol(''T''),
:$f(T)\; =\; \backslash sum\_\; \backslash sum\_^\; \backslash frac\; (T\; -\; \backslash lambda\_i)^k\; e\_i\; (T).$
Notice that the expression of ''f''(''T'') is a finite sum because, on each neighborhood of λ_{''i''}, we have chosen the Taylor series expansion of ''f'' centered at λ_{''i''}.

_{''T''} defined by
:$R\_T(\backslash lambda)\; =\; (\backslash lambda\; -\; T)^$
has a pole (complex analysis), pole of order ν at λ.
We will show that, in the finite-dimensional case, the order of an eigenvalue coincides with its index. The result also holds for compact operators.
Consider the annular region ''A'' centered at the eigenvalue λ with sufficiently small radius ''ε'' such that the intersection of the open disc ''B_{ε}''(λ) and ''σ''(''T'') is . The resolvent function ''R''_{''T''} is holomorphic on ''A''.
Extending a result from classical function theory, ''R''_{''T''} has a Laurent series representation on ''A'':
:$R\_T(z)\; =\; \backslash sum\; \_\; ^\; a\_m\; (\backslash lambda\; -\; z)^m$
where
:$a\_\; =\; -\; \backslash frac\; \backslash int\_C\; (\backslash lambda\; -\; z)\; ^\; (z\; -\; T)^\; d\; z$ and ''C'' is a small circle centered at λ.
By the previous discussion on the functional calculus,
:$a\_\; =\; -(\backslash lambda\; -\; T)^\; e\_\; (T)$ where $e\_$ is 1 on $B\_(\backslash lambda)$ and 0 elsewhere.
But we have shown that the smallest positive integer ''m'' such that
:$a\_\; \backslash neq\; 0$ and $a\_\; =\; 0\; \backslash ;\; \backslash ;\; \backslash forall\; \backslash ;\; l\; \backslash geq\; m$
is precisely the index of λ, ''ν''(λ). In other words, the function ''R''_{''T''} has a pole of order ''ν''(λ) at λ.

''Jordan Canonical Form'' article at mathworld.wolfram.com

{{Matrix classes Linear algebra Matrix theory Matrix normal forms Matrix decompositions

"Histoire du théorème de Jordan de la décomposition matricielle (1870-1930). Formes de représentation et méthodes de décomposition"

Thesis, 2007

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.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: : $A\; =\; \backslash left[\backslash begin\; 5\; \&\; 4\; \&\; 2\; \&\; 1\; \backslash \backslash [2pt]\; 0\; \&\; 1\; \&\; -1\; \&\; -1\; \backslash \backslash [2pt]\; -1\; \&\; -1\; \&\; 3\; \&\; 0\; \backslash \backslash [2pt]\; 1\; \&\; 1\; \&\; -1\; \&\; 2\; \backslash end\backslash right].$ 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''Complex matrices

In general, a square complex matrix ''A'' is similar (linear algebra), similar to a block diagonal matrix :$J\; =\; \backslash begin\; J\_1\; \&\; \backslash ;\; \&\; \backslash ;\; \backslash \backslash \; \backslash ;\; \&\; \backslash ddots\; \&\; \backslash ;\; \backslash \backslash \; \backslash ;\; \&\; \backslash ;\; \&\; J\_p\backslash end$ where each block ''JExample

Consider the matrix $A$ from the example in the previous section. The Jordan normal form is obtained by some similarity transformation: :$P^AP\; =\; J;$ that is, $AP\; =\; PJ.$ Let $P$ have column vectors $p\_i$, $i\; =\; 1,\; \backslash ldots,\; 4$, then : $A\; \backslash begin\; p\_1\; \&\; p\_2\; \&\; p\_3\; \&\; p\_4\; \backslash end\; =\; \backslash begin\; p\_1\; \&\; p\_2\; \&\; p\_3\; \&\; p\_4\; \backslash end\; \backslash begin\; 1\; \&\; 0\; \&\; 0\; \&\; 0\; \backslash \backslash \; 0\; \&\; 2\; \&\; 0\; \&\; 0\; \backslash \backslash \; 0\; \&\; 0\; \&\; 4\; \&\; 1\; \backslash \backslash \; 0\; \&\; 0\; \&\; 0\; \&\; 4\; \backslash end\; =\; \backslash begin\; p\_1\; \&\; 2p\_2\; \&\; 4p\_3\; \&\; p\_3+4p\_4\; \backslash end.$ 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 $p\_i\; \backslash in\; \backslash operatorname(A-\backslash lambda\_\; I)$, that is, $p\_i$ is an eigenvector of $A$ corresponding to the eigenvalue $\backslash lambda\_i$. 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, $p\_4\; \backslash in\; \backslash operatorname(A-4\; I)^2.$ 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 :$A\; =\; \backslash begin\; 5\; \&\; 4\; \&\; 2\; \&\; 1\; \backslash \backslash \; 0\; \&\; 1\; \&\; -1\; \&\; -1\; \backslash \backslash \; -1\; \&\; -1\; \&\; 3\; \&\; 0\; \backslash \backslash \; 1\; \&\; 1\; \&\; -1\; \&\; 2\; \backslash end$ which is mentioned in the beginning of the article. The characteristic polynomial of ''A'' is :$\backslash begin\; \backslash chi(\backslash lambda)\; \&\; =\; \backslash det(\backslash lambda\; I\; -\; A)\; \backslash \backslash \; \&\; =\; \backslash lambda^4\; -\; 11\; \backslash lambda^3\; +\; 42\; \backslash lambda^2\; -\; 64\; \backslash lambda\; +\; 32\; \backslash \backslash \; \&\; =\; (\backslash lambda-1)(\backslash lambda-2)(\backslash lambda-4)^2.\; \backslash ,\; \backslash end$ 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 ''Av'' = ''λ v''. It is spanned by the column vector ''v'' = (−1, 1, 0, 0)Generalized eigenvectors

Given an eigenvalue λ, its corresponding Jordan block gives rise to a Jordan chain. The generator, or lead vector, say ''pA proof

We give a proof by induction that any complex-valued matrix A may be put in Jordan normal form. The 1 × 1 case is trivial. Let ''A'' be an ''n'' × ''n'' matrix. Take any eigenvalue λ of ''A''. The range of a function, range of ''A'' − λ I, denoted by Ran(''A'' − λ I), is an invariant subspace of ''A''. Also, since λ is an eigenvalue of ''A'', the dimension of Ran(''A'' − λ I), ''r'', is strictly less than ''n''. Let ''A' '' denote the restriction of ''A'' to Ran(''A'' − λ I), by inductive hypothesis, there exists a basis (linear algebra), basis such that ''A' '', expressed with respect to this basis, is in Jordan normal form. Next consider the kernel (linear algebra), kernel, that is, the linear subspace, subspace Ker(''A'' − λ I). If :$\backslash mathrm(A\; -\; \backslash lambda\; I)\; \backslash cap\; \backslash mathrm(A\; -\; \backslash lambda\; I)\; =\; \backslash ,$ the desired result follows immediately from the rank–nullity theorem. (This would be the case, for example, if ''A'' were Hermitian matrix, Hermitian.) Otherwise, if :$Q\; =\; \backslash mathrm(A\; -\; \backslash lambda\; I)\; \backslash cap\; \backslash mathrm(A\; -\; \backslash lambda\; I)\; \backslash neq\; \backslash ,$ let the dimension of ''Q'' be ''s'' ≤ ''r''. Each vector in ''Q'' is an eigenvector of ''A' ''corresponding to eigenvalue ''λ''. So the Jordan form of ''A' '' must contain ''s'' Jordan chains corresponding to ''s'' linearly independent eigenvectors. So the basis must contain ''s'' vectors, say , that are lead vectors in these Jordan chains from the Jordan normal form of ''A. We can "extend the chains" by taking the preimages of these lead vectors. (This is the key step of argument; in general, generalized eigenvectors need not lie in Ran(''A'' − λ I).) Let ''q''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 (''A'' − λ ''I'')Real matrices

If ''A'' is a real matrix, its Jordan form can still be non-real. Instead of representing it with complex eigenvalues and 1's on the superdiagonal, as discussed above, there exists a real invertible matrix ''P'' such that ''P''Matrices with entries in a field

Jordan reduction can be extended to any square matrix ''M'' whose entries lie in a field (mathematics), field ''K''. The result states that any ''M'' can be written as a sum ''D'' + ''N'' where ''D'' is semisimple operator, semisimple, ''N'' is nilpotent matrix, nilpotent, and ''DN'' = ''ND''. This is called the Jordan–Chevalley decomposition. Whenever ''K'' contains the eigenvalues of ''M'', in particular when ''K'' is algebraically closed, the normal form can be expressed explicitly as the direct sum of Jordan blocks. Similar to the case when ''K'' is the complex numbers, knowing the dimensions of the kernels of (''M'' − λ''I'')Consequences

One can see that the Jordan normal form is essentially a classification result for square matrices, and as such several important results from linear algebra can be viewed as its consequences.Spectral mapping theorem

Using the Jordan normal form, direct calculation gives a spectral mapping theorem for the functional calculus, polynomial functional calculus: Let ''A'' be an ''n'' × ''n'' matrix with eigenvalues λCharacteristic polynomial

The characteristic polynomial of is $p\_A(\backslash lambda)=\backslash det\; (\backslash lambda\; I-A)$. Matrix similarity, Similar matrices have the same characteristic polynomial. Therefore, $p\_A(\backslash lambda)=p\_J(\backslash lambda)=\backslash prod\_i\; (\backslash lambda-\backslash lambda\_i)^$, where $\backslash lambda\_i$ is the ''i''th root of $p\_J$ and $m\_i$ is its multiplicity, because this is clearly the characteristic polynomial of the Jordan form of ''A''.Cayley–Hamilton theorem

The Cayley–Hamilton theorem asserts that every matrix ''A'' satisfies its characteristic equation: if is the characteristic polynomial of , then $p\_A(A)=0$. This can be shown via direct calculation in the Jordan form, since if $\backslash lambda\_i$ is an eigenvalue of multiplicity $m$, then its Jordan block $J\_i$ clearly satisfies $(J\_i-\backslash lambda\_i\; I)^=0$. As the diagonal blocks do not affect each other, the ''i''th diagonal block of $(A-\backslash lambda\_i\; I)^$ is $(J\_i-\backslash lambda\_i\; I)^=0$; hence $p\_A(A)=\backslash prod\_i\; (A-\backslash lambda\_i\; I)^=0$. The Jordan form can be assumed to exist over a field extending the base field of the matrix, for instance over the splitting field of ; this field extension does not change the matrix in any way.Minimal polynomial

The Minimal polynomial (linear algebra), minimal polynomial P of a square matrix ''A'' is the unique monic polynomial of least degree, ''m'', such that ''P''(''A'') = 0. Alternatively, the set of polynomials that annihilate a given ''A'' form an ideal ''I'' in ''C''[''x''], the principal ideal domain of polynomials with complex coefficients. The monic element that generates ''I'' is precisely ''P''. Let λInvariant subspace decompositions

The Jordan form of a ''n'' × ''n'' matrix ''A'' is block diagonal, and therefore gives a decomposition of the ''n'' dimensional Euclidean space into invariant subspaces of ''A''. Every Jordan block ''J''Plane (flat) normal form

The Jordan form is used to find a normal form of matrices up to conjugacy such that normal matrices make up an algebraic variety of a low fixed degree in the ambient matrix space. Sets of representatives of matrix conjugacy classes for Jordan normal form or rational canonical forms in general do not constitute linear or affine subspaces in the ambient matrix spaces. Vladimir Arnold posed a problem: Find a canonical form of matrices over a field for which the set of representatives of matrix conjugacy classes is a union of affine linear subspaces (flats). In other words, map the set of matrix conjugacy classes injectively back into the initial set of matrices so that the image of this embedding—the set of all normal matrices, has the lowest possible degree—it is a union of shifted linear subspaces. It was solved for algebraically closed fields by Peteris Daugulis. The construction of a uniquely defined plane normal form of a matrix starts by considering its Jordan normal form.Matrix functions

Iteration of the Jordan chain motivates various extensions to more abstract settings. For finite matrices, one gets matrix functions; this can be extended to compact operators and the holomorphic functional calculus, as described further below. The Jordan normal form is the most convenient for computation of the matrix functions (though it may be not the best choice for computer computations). Let ''f''(''z'') be an analytical function of a complex argument. Applying the function on a ''n''×''n'' Jordan block ''J'' with eigenvalue ''λ'' results in an upper triangular matrix: :$f(J)\; =\backslash begin\; f(\backslash lambda)\; \&\; f\text{'}(\backslash lambda)\; \&\; \backslash tfrac\; \&\; \backslash cdots\; \&\; \backslash tfrac\backslash \backslash \; 0\; \&\; f(\backslash lambda)\; \&\; f\text{'}(\backslash lambda)\; \&\; \backslash cdots\; \&\; \backslash tfrac\; \backslash \backslash \; \backslash vdots\; \&\; \backslash vdots\; \&\; \backslash ddots\; \&\; \backslash ddots\; \&\; \backslash vdots\; \backslash \backslash \; 0\; \&\; 0\; \&\; 0\; \&\; f(\backslash lambda)\; \&\; f\text{'}(\backslash lambda)\; \backslash \backslash \; 0\; \&\; 0\; \&\; 0\; \&\; 0\; \&\; f(\backslash lambda)\; \backslash end,$ so that the elements of the ''k''-th superdiagonal of the resulting matrix are $\backslash tfrac$. For a matrix of general Jordan normal form the above expression shall be applied to each Jordan block. The following example shows the application to the power function ''f''(''z'')=''zCompact operators

A result analogous to the Jordan normal form holds for compact operators on a Banach space. One restricts to compact operators because every point ''x'' in the spectrum of a compact operator ''T'' is an eigenvalue; The only exception is when ''x'' is the limit point of the spectrum. This is not true for bounded operators in general. To give some idea of this generalization, we first reformulate the Jordan decomposition in the language of functional analysis.Holomorphic functional calculus

Let ''X'' be a Banach space, ''L''(''X'') be the bounded operators on ''X'', and ''σ''(''T'') denote the spectrum (functional analysis), spectrum of ''T'' ∈ ''L''(''X''). The holomorphic functional calculus is defined as follows: Fix a bounded operator ''T''. Consider the family Hol(''T'') of complex functions that is Holomorphic function, holomorphic on some open set ''G'' containing ''σ''(''T''). Let Γ = be a finite collection of Jordan curves such that ''σ''(''T'') lies in the ''inside'' of Γ, we define ''f''(''T'') by : $f(T)\; =\; \backslash frac\; \backslash int\_\; f(z)(z\; -\; T)^\; dz.$ The open set ''G'' could vary with ''f'' and need not be connected. The integral is defined as the limit of the Riemann sums, as in the scalar case. Although the integral makes sense for continuous ''f'', we restrict to holomorphic functions to apply the machinery from classical function theory (for example, the Cauchy integral formula). The assumption that ''σ''(''T'') lie in the inside of Γ ensures ''f''(''T'') is well defined; it does not depend on the choice of Γ. The functional calculus is the mapping Φ from Hol(''T'') to ''L''(''X'') given by : $\backslash ;\; \backslash Phi(f)\; =\; f(T).$ We will require the following properties of this functional calculus: # Φ extends the polynomial functional calculus. # The ''spectral mapping theorem'' holds: ''σ''(''f''(''T'')) = ''f''(''σ''(''T'')). # Φ is an algebra homomorphism.The finite-dimensional case

In the finite-dimensional case, ''σ''(''T'') = is a finite discrete set in the complex plane. Let ''e''Poles of an operator

Let ''T'' be a bounded operator λ be an isolated point of ''σ''(''T''). (As stated above, when ''T'' is compact, every point in its spectrum is an isolated point, except possibly the limit point 0.) The point λ is called a pole of operator ''T'' with order ν if the Resolvent formalism, resolvent function ''R''Numerical analysis

If the matrix ''A'' has multiple eigenvalues, or is close to a matrix with multiple eigenvalues, then its Jordan normal form is very sensitive to perturbations. Consider for instance the matrix :$A\; =\; \backslash begin\; 1\; \&\; 1\; \backslash \backslash \; \backslash varepsilon\; \&\; 1\; \backslash end.$ If ''ε'' = 0, then the Jordan normal form is simply :$\backslash begin\; 1\; \&\; 1\; \backslash \backslash \; 0\; \&\; 1\; \backslash end.$ However, for ''ε'' ≠ 0, the Jordan normal form is :$\backslash begin\; 1+\backslash sqrt\backslash varepsilon\; \&\; 0\; \backslash \backslash \; 0\; \&\; 1-\backslash sqrt\backslash varepsilon\; \backslash end.$ This condition number, ill conditioning makes it very hard to develop a robust numerical algorithm for the Jordan normal form, as the result depends critically on whether two eigenvalues are deemed to be equal. For this reason, the Jordan normal form is usually avoided in numerical analysis; the stable Schur decomposition or pseudospectrum, pseudospectraSee Golub & Van Loan (2014), §7.9 are better alternatives.See also

* Canonical basis * Canonical form * Frobenius normal form * Jordan matrix * Jordan–Chevalley decomposition * Matrix decomposition * Modal matrix * Weyr canonical formNotes

References

* * * * * * * * * * * * * * * *''Jordan Canonical Form'' article at mathworld.wolfram.com

{{Matrix classes Linear algebra Matrix theory Matrix normal forms Matrix decompositions