Frobenius Normal Form - Motivating Example

Motivating Example

For a matrix A with entries in a field F the Frobenius normal form M may be found by conjugation with an invertible matrix P:

The matrices A and M are similar matrices. For example, let F be the field R of real numbers, consider the following matrix A, over R:

A=\begin{pmatrix} -2 & -1 & -2 & -1 & 1 & 0 \\ -2 & -1 & -2 & -1 & 1 & 1 \\ 2 & 1 & 2 & 1 & 0 & 0 \\ 2 & 1 & 0 & 1 & -3 & -1\\ -2 & 0 & -2 & 0 & 0 & 0 \\ 2 & -2 & 0 & 0 & 0 & 0 \end{pmatrix}.

The characteristic polynomial of A is x6 + 6x4 + 12x2 + 8 = (x2 + 2)3 = p(x). The minimal polynomial of this matrix is x2 + 2.

We will then have one block in the Frobenius normal form as

A_1 = \begin{pmatrix} 0 & -2 \\ 1 & 0 \end{pmatrix}.

The characteristic polynomial of A1 is indeed x2 + 2.

Since p factors into solely terms of the form x2 + 2, we expect the other two blocks comprising the normal form of the matrix to be identical to A1. So, we can simply write down the Frobenius normal form:

M = A_1 \oplus A_2 \oplus A_3 = \begin{pmatrix} 0 & -2 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -2 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & -2 \\ 0 & 0 & 0 & 0 & 1 & 0 \end{pmatrix}.

The characteristic and minimal polynomials of M are the same as those of A, which we would expect, since M can be obtained via a similarity transformation P−1AP = M, and determinants are similarity invariant. For this matrix A, P is

P = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & -2 \\ 0 & 0 & 0 & 1 & 0 & -2 \\ -1/2 & 0 & 0 & 0 & 1 & 2 \\ 1 & 1 & 0 & -1 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & -2 \\ 0 & 0 & 1 & 0 & 0 & 0 \end{pmatrix}.

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