LAC Minimal and characteristic polynomials

Section 4.5 Minimal and characteristic polynomials

This section contains somewhat more advanced material; we review a few important facts about minimal and characteristic polynomials.

Subsection 4.5.1 Annihilating polynomials

Let \(A \in M_n(F)\) be a square matrix. One can ask if there is a nonzero polynomial \(f(x) = a_mx^m + \cdots + a_0 \in F[x]\) for which \(f(A) = a_mA^m+ \cdots + a_1 A + a_0 I_n = \0,\) the zero matrix. If we think of trying to find a polynomial, this may seem a challenging task.

However, if we consider that \(M_n(F)\) is a vector space of dimension \(n^2\text{,}\) then Theorem 3.1.5 tells us that the set

\begin{equation*} \{I_n, A, A^2, \dots, A^{n^2}\} \end{equation*}

must be a linearly dependent set, and that means there are scalars \(a_0, a_1, \dots, a_{n^2} \in F\text{,}\) not all zero, for which \(a_{n^2}A^{n^2}+ \cdots + a_1 A + a_0 I_n =\0\text{,}\) so that \(f(x) = a_{n^2}x^{n^2} + \cdots + a_0\) is one nonzero polynomial which annihilates \(A.\)

Subsection 4.5.2 The minimal polynomial

Given a matrix \(A \in M_n(F)\text{,}\) we have seen there is a nonzero polynomial which annihilates it, so we consider the set

\begin{equation*} J = \{ f\in F[x] \mid f(A) = \0\}. \end{equation*}

In the language of abstract algebra, \(J\) is an ideal in the polynomial ring \(F[x],\) and since \(F\) is a field, \(F[x]\) is a PID (principal ideal domain), the ideal \(J\) is principally generated: \(J = \la \mu_A\ra\text{,}\) where \(\mu_A\) is the monic generator of this ideal. In less technical terms, \(\mu_A\) is the monic polynomial of least degree which annihilates \(A,\) and every element of \(J\) is a (polynomial) multiple of \(\mu_A.\) The polynomial \(\mu_A\) is called the minimal polynomial of the matrix \(A.\)

A more constructive version of finding the minimal polynomial comes from the observation that if \(f,g \in J\text{,}\) that if \(f(A)=g(A)=\0,\) then \(h(A) = 0,\) where \(h\) is the greatest common divisor gcd, of \(f\) and \(g.\) In particular, if \(f(A) = \0,\) then \(\mu_A\) must divide \(f\text{,}\) so if we can factor \(f,\) there are only finitely many possibilities for \(\mu_A.\)

Example 4.5.1. \(A^8 = I_n\).

Let’s suppose that \(A\in M_n(\Q)\) and \(A^8=I_n.\) This means that \(f(x) = x^8-1\) is a polynomial which annihilates \(A,\) so \(\mu_A\) must divide it. Over \(\Q,\) we have the following factorization into irreducibles:

\begin{equation*} x^8-1 = \Phi_8\Phi_4\Phi_2\Phi_1= (x^4+1)(x^2+1)(x+1)(x-1), \end{equation*}

where (for those with abstract algebra background) the \(\Phi_d\) are the \(d\)th cyclotomic polynomials defined recursively as an (irreducible) factorization over \(\Q\) by

\begin{equation*} x^n -1 = \prod_{d\mid n} \Phi_d. \end{equation*}

Thus \(\Phi_1 = (x-1)\text{,}\) \(x^2-1 = \Phi_1\Phi_2,\) so \(\Phi_2 = x+1,\) and \(x^8-1\) has the factorization given above.

Subsection 4.5.3 The characteristic polynomial

Given a matrix \(A \in M_n(F)\text{,}\) we have seen that there is a polynomial of degree at most \(n^2\) which annihilates \(A\text{,}\) and given one such nonzero polynomial there is one of minimal degree. But the key to finding a minimal polynomial is obtaining at least one. The idea of trying to find a linear dependence relation among \(I_n, A, A^2, \dots, A^{n^2}\) is far from appealing, but fortunately there is a polynomial we have used before which annhilates \(A.\)

Theorem 4.5.2. Cayley-Hamilton.

Let \(A \in M_n(F)\text{,}\) and \(\chi_A(x) = \det(xI_n - A)\) be its characteristic polynomial. Then \(\chi_A(A) = \0,\) that is \(\chi_A\) is a monic polynomial of degree \(n\) which annihilates \(A.\)

In particular, the minimal polynomial, \(\mu_A\text{,}\) divides the characteristic polynomial, \(\chi_A.\)

Example 4.5.3. Are there any elements of order 8 in \(GL_3(\Q)\text{?}\)

The question asks whether there is an invertible \(3\times 3\) matrix \(A\) so that 8 is the smallest positive integer \(k\) with \(A^k = I_3.\)

Since \(A^8 = I_3\text{,}\) we know \(\det A \ne 0,\) so such a matrix will necessarily be invertible, hence an element of \(GL_3(\Q).\) In the example above, we saw that any matrix which satisfies \(A^8 = I_3\) must have minimal polynomial \(\mu_A\) which divides \(x^8-1= (x^4+1)(x^2+1)(x+1)(x-1).\) But the Cayley-Hamilton theorem tells us that \(\mu_A\) must also divide the characteristic polynomial \(\chi_A\) which must have degree 3, and the only way to create a polynomial of degree 3 with the factors listed above is to have \(\chi_A \mid x^4-1,\) which forces \(A^4=I_3,\) so there are no elements of order 8 in \(GL_3(\Q).\)

On the other hand,

\begin{equation*} A = \begin{bmatrix} 0 \amp0\amp0\amp-1\\ 1\amp0\amp0\amp0\\ 0\amp1\amp0\amp0\\ 0\amp0\amp1\amp0 \end{bmatrix} \end{equation*}

has \(\mu_A=\chi_A = x^4+1,\) so \(A\) is an element of order 8 in \(GL_4(\Q).\) The matrix \(A\) is the companion matrix to the polynomial \(x^4+1\text{.}\) See Example 4.3.6 for more detail.

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