Jordan Normal Form

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Introduction

After learning the basics of matrix operations we will now look more in depth into the representation of matrix transformations. Some of the topics covered below (like eigenvalues and eigenvectors) are covered in the second year but the majority is not part of the university course at Surrey and has been added for extra information and completeness.

We will introduce matrix diagonalization and the Jordan normal form of a matrix. The matrices used below will be assumed to be real and square matrices.

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Eigenvalues and Eigenvectors

For any given matrix A one can try to find a number, λ and a vector u such that the equation below is satisfied.

The number λ is called an eigenvalue (or characteristic value) and u is called a corresponding eigenvector (or characteristic vector) of the matrix A. The equation above can also be written as

Non-trivial solutions of this equation exist only if the determinant of the matrix in (A − λ) is zero i.e. det(A − λ) u = 0, so the equation above has a non-zero solution. The polynomial det(A − λ) u = 0 is called the characteristic equation. The characteristic equation is a polynomial equation of degree n in λ and has the following form:

We notice by expansion that one term in this equation is given by and that all other terms contain at most the (n − 2) power. Hence the equation can be written as:

Thus, this equation establishes two relationships between the eigenvalues, the trace and determinant of a matrix. Taking an arbitrary 2×2 matrix A below, we have that:

  1. The trace of matrix A (i.e. the sum of the diagonal entries of matrix A) is equal to the sum of the eigenvalues.
  2. The determinant of A is the product of the eigenvalues.
Proof1 Eg1

Hence in the general case of an n×n matrix, the roots of the characteristic equation give n eigenvalues. In other words, a square matrix of order n always has n eigenvalues. If a solution λ of the characteristic equation is complex then the complex conjugate of λ is a solution also. Having found all the eigenvalues of the matrix A, we now want to find the eigenvectors associated with each of the eigenvalues.

For each given eigenvalue λi, i = 1, 2, ..., n, we can calculate a corresponding eigenvector using the equation

or

Finding Eigenvector
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Similarity

Let M denote the set of all (square) n×n matrices. Let ~ denote a relationship between the matrices in M. This relationship is called an equivalence relation on M if and only if the following three properties are satisfied:

  1. ~ is reflexive: A~A for every A in M,
  2. ~ is symmetric: if A~B then B~A for every A,B in M,
  3. ~ is transitive: if A~B and B~C then A~C for every A,B and C in M.

In addition a similarity relation is an equivalence relation in the sense that:

  1. A is similar to itself
  2. If B is similar to A, then A is similar to B.
  3. If C is similar to B and B is similar to A, then C is similar to A.
Proof2

We aim to find a standard form such that each equivalence class of M contains one and only one matrix. This standard form is called the Jordan normal form. If two matrices have the same standard form they must be in the same equivalence class.

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Algebraic Multiplicity

Recall the characteristic equation from above. In general the characteristic equation can be factored into the following form:

where mj denotes the multiplicity corresponding to the eigenvalue λj for the polynomial det (A − λ I). The multiplicity mj is also called the algebraic multiplicity of λj. In the case where each value of mj = 1, all the eigenvalues are distinct. In general each value of mj is greater than or equal to 1.

Summing all the algebraic multiplicity of the eigenvalues in equation above gives n, the size of the matrix.

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Eigenspaces and Geometric Multiplicity

If λ is an eigenvalue, then ker(A − λI) is known as the eigenspace for λ. The dimension of an eigenspace is the maximum number of linearly independent vectors in the vectorspace, this is called the geometric multiplicity of the eigenvalue λ. So the geometric multiplicity is the dimension of the kernel of the matrix (A − λI).

Theorem: The geometric multiplicity of λ does not exceed the algebraic multiplicity of λ. Proof3

The eigenspace for the eigenvalue λ i will be denoted by Ei. Hence

Theorem: If then Ei and Ej are linearly independent. Proof4

Finding Eigenspace
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Diagonalization

Diagonalization is the process of transforming matrix A to a similar matrix P-1AP which has only non-zero elements along its main diagonal.

We assume that the roots of the characteristic equation of matrix A are real for all eigenvalues and that the geometric multiplicity equals the algebraic multiplicity. Then the matrix A can be transformed to a diagonal form.

Above, it was shown that the eigenvectors of A form a basis. Say that matrix P is a matrix with columns consisting of these n independent eigenvectors of A, i.e.

Since all vectors vi are linearly independent, P is non-singular and

Hence A is diagonalisable.

On the other hand if A is diagonalisable then it has linearly independent eigenvectors. Indeed, D clearly has n independent eigenvectors, the basis vectors e1, e2, ..., en. Previously we have shown this implies that P-1ej is an eigenvector of A. Since P is invertible all the vectors P-1ej are linearly independent.

Note that if all eigenvalues are different then the eigenvectors are linearly independent. Hence in this case it is easy to see that the matrix is diagonal.

Here is an example of a diagonlisable matrix.

Eg4

Here is an example of a non-diagonlisable matrix.

Eg5
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Generalized eigenvectors

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The Jordan Normal Form

When looking at diagonalization of matrices we looked for a basis made up of eigenvectors. But we were faced with the case that such basis does not always exist. In such cases we seek to find a general basis, which consists of eigenvectors and generalised eigenvectors. Then we can show that the matrix is similar to so-called Jordan normal form of a matrix

The Jordan normal form is an upper triangular matrix consisting of repeated (grouped) eigenvalues on the main diagonal. The n entries on the main diagonal of the Jordan normal form matrix J, equal the n eigenvalues (which are repeated according to their algebraic multiplicity. The elements in the superdiagonal (the diagonal one row above and parallel to the main diagonal) of J equal either a 0 or 1. The number of zeros is one less than the number of linearly independent eigenvectors of A.

The Jordan normal form is important when the geometric multiplicity is less than the algebraic multiplicity as there are not enough eigenvectors in the basis.

First we will look at the case of a 3x3 matrix, which has only one eigenvalue hence the algebraic multiplicity is 3.

If the geometric multiplicity is 1

For a more general case the matrix will look like

Have a look at a detailed example of the Jordan Normal Form.

Eg6
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