Minimal polynomial (linear algebra)

For the minimal polynomial of an algebraic element of a field, see Minimal polynomial (field theory).

In linear algebra, the minimal polynomial μ_A of an n-by-n matrix A over a field F is the monic polynomial P over F of least degree such that P(A)=0. Any other polynomial Q with Q(A) = 0 is a (polynomial) multiple of μ_A.

The following three statements are equivalent:

1. λ is a root of μ_A,
2. λ is a root of the characteristic polynomial χ_A of A,
3. λ is an eigenvalue of matrix A.

The multiplicity of a root λ of μ_A is the largest power m such that Ker((A − λI_n)^m) strictly contains Ker((A − λI_n)^m−1) (increasing the exponent up to m will give ever larger kernels, but further increasing m will just give the same kernel).

If the field F is not algebraically closed, then the minimal and characteristic polynomials need not factor according to their roots (in F) alone, in other words they may have irreducible polynomial factors of degree greater than 1. For such factors P one has similar equivalences:

1. P divides μ_A,
2. P divides χ_A,
3. the kernel of P(A) has dimension at least 1.

Like the characteristic polynomial, the minimal polynomial does not depend on the base field, in other words considering the matrix as one with coefficients in a larger field does not change the minimal polynomial. The reason is somewhat different than for the characteristic polynomial (where it is immediate from the definition of determinants), namely the fact that the minimal polynomial is determined by the relations of linear dependence between the powers of A: extending the base field will not introduce any new such relations (nor of course will it remove existing ones).

The minimal polynomial is often the same as the characteristic polynomial, but not always. For example, if A is a multiple aI_n of the identity matrix, then its minimal polynomial is X − a since the kernel of aI_n − A = 0 is already the entire space; on the other hand its characteristic polynomial is (X − a)ⁿ (the only eigenvalue is a, and the degree of the characteristic polynomial is always equal to the dimension of the space). The minimal polynomial always divides the characteristic polynomial, which is one way of formulating the Cayley–Hamilton theorem.

Formal definition

Given an endomorphism T on a finite-dimensional vector space V over a field F, let I_T be the set defined as

$\mathit{I}_T = \{ p \in \mathbf{F}[t] \; | \; p(T) = 0 \}$

where F[t] is the space of all polynomials over the field F. It is easy to show that I_T is a proper ideal of F[t].

• The minimal polynomial is the monic polynomial which generates I_T.

Thus it must be the monic polynomial of least degree in I_T.

Applications

An endomorphism φ of a finite dimensional vector space over a field F is diagonalizable if and only if its minimal polynomial factors completely over F into distinct linear factors. The fact that there is only one factor X − λ for every eigenvalue λ means that the generalized eigenspace for λ is the same as the eigenspace for λ: every Jordan block has size 1. More generally, if φ satisfies a polynomial equation P(φ) = 0 where P factors into distinct linear factors over F, then it will be diagonalizable: its minimal polynomial is a divisor of P and therefore also factors into distinct linear factors. In particular one has:

• P = X^k − 1: finite order endomorphisms of complex vector spaces are diagonalizable. For the special case of involutions, this is even true for endomorphisms of vector spaces over any field of characteristic other than 2, since X² − 1 = (X − 1)(X + 1) is a factorization into distinct factors over such a field. This is a part of representation theory of cyclic groups.
• P = X² − X = X(X − 1): endomorphisms satisfying φ² = φ are called projections, and are always diagonalizable (moreover their only eigenvalues are 0 and 1).
• By contrast if μ_φ = X^k with k ≥ 2 then φ (a nilpotent endomorphism) is not diagonalizable, since X^k has a repeated root 0.

These case can also be proved directly, but the minimal polynomial gives a unified perspective and proof.

Computation

Let I_T,v be defined as

$\mathit{I}_{T, v} = \{ p \in \mathbf{F}[t] \; | \; p(T)(v) = 0 \}.$

This definition satisfies the properties of a proper ideal. Let μ_T,v be the monic polynomial which generates it.

Properties

• Since I_T,v contains minimal polynomial μ_T, the latter is divisible by μ_T,v.

• If d is the least latural number such that v, T(v), ... , T^d(v) are linearly dependent, then there exist unique $a_0, a_1, \cdots, a_{d-1} \in \mathbf{F}$ such that

  $a_0 v + a_1 T(v) + \cdots + a_{d-1} T^{d-1} (v) + T^d (v) = 0$

  and for these coefficients one has $\mu_{T,v} (t) = a_0 + a_1 t + \ldots + a_{d-1} t^{d-1} + t^d.$

• Let the subspace V be the image of μ_T,v(T), which is T-stable. Since μ_T,v(T) annihilates at least the vectors v, T(v), ... , T^d-1(v), the codimension of V is at least d.
• The minimal polynomial μ_T is the product of μ_T,v and the minimal polynomial Q of the restriction of T to V. In the (likely) case that V has dimension 0 one has Q = 1 and therefore μ_T = μ_T,v; otherwise a recursive computation of Q suffices to find μ_T.

Example

Define T to be the endomorphism of R³ with matrix, on the canonical basis,

$\begin{bmatrix} 1 & -1 & -1 \\ 1 & -2 & 1 \\ 0 & 1 & -3 \end{bmatrix}.$

Taking the first canonical basis vector e₁ and its repeated images by T one obtains

$e_1 = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \quad T\cdot e_1 = \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}. \quad T^2\cdot e_1 = \begin{bmatrix} 0 \\ -1 \\ 1 \end{bmatrix} \mbox{ and}\quad T^3\cdot e_1=\begin{bmatrix} 0 \\ 3 \\ -4 \end{bmatrix}$

of which the first three are easily seen to be linearly independent, and therefore span all of R³. The last one then necessarily is a linear combination of the first three, in fact $T^3\cdot e_1 = -4T^2\cdot e_1 -T\cdot e_1 + e_1$, so that $\mu_{T,e_1}=X^3+4X^2+X-1$. This is in fact also the minimal polynomial μ_T and the characteristic polynomial χ_T: indeed $\mu_{T,e_1}$ divides μ_T which divides χ_T, and since the first and last are of degree 3 and all are monic, they must all be the same. Another reason is that in general if any polynomial in T annihilates a vector v, then it also annihilates T⋅v (just apply T to the equation that says that it annihilates v), and therefore by iteration it annihilates the entire space generated by the iterated images by T of v; in the current case we have seen that for v = e₁ that space is all of R³, so $\mu_{T,e_1}(T)=0$. Indeed one verifies for the full matrix that T³ + 4T² + T − I₃ is the null matrix:

$\begin{bmatrix} 0 & 1 & -3 \\ 3 & -13 & 23 \\ -4 & 19 & -36 \end{bmatrix} +4\begin{bmatrix} 0 & 0 & 1 \\ -1 & 4 & -6 \\ 1 & -5 & 10 \end{bmatrix} +\begin{bmatrix} 1 & -1 & -1 \\ 1 & -2 & 1 \\ 0 & 1 & -3 \end{bmatrix} +\begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{bmatrix} =\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$