Minimal polynomial (linear algebra)

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

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

The following three statements are equivalent:

λ∈F is a root of p(x),
λ is a root of the characteristic polynomial of A,
λ is an eigenvalue of A.

The multiplicity of a root λ of p(x) is the size of the largest Jordan block corresponding to λ.

The minimal polynomial is not always the same as the characteristic polynomial. Consider the matrix $4 I n$ , which has characteristic polynomial $(x - 4) n$ . However, the minimal polynomial is $x - 4$ , since $4 I - 4 I = 0$ as desired, so they are different for $n\ge 2$ . That the minimal polynomial always divides the characteristic polynomial is a consequence of the Cayley–Hamilton theorem.

1 Formal definition
2 Applications
3 How to compute it
- 3.1 Example

[edit] Formal definition

Given an endomorphism T on a vector space V over a field $\mathbb{K}$ , let I_T be the set defined as

$\mathit{I}_T = \{ p \in \mathbb{K}[t] \; | \; p(T) = O \}$

where $\mathbb{K}[t]$ is the space of all polynomials over the field $\mathbb{K}$ . It is easy to show that I_T is a proper ideal of $\mathbb{K}[t]$ .

The minimal polynomial is the monic polynomial which generates I_T.

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

[edit] Applications

An endomorphism is diagonalizable if and only if every Jordan block has size 1, which is equivalent to each root of the minimal polynomial having multiplicity 1 (and the minimal polynomial factors completely). Thus an endomorphism is diagonalizable over a field K if and only if its minimal polynomial factors completely over K into distinct roots.

$X k - I$ : finite order endomorphisms (including involutions) are diagonalizable over the complex numbers (as the roots of unity are distinct); this is a part of representation theory of cyclic groups.
$X 2 - X$ : projections are diagonalizable, with 0's and 1's on the diagonal.
$X k$ : Nilpotent endomorphisms are not semisimple, as $X k$ has a repeated root.

The above can also be worked out by hand, but the minimal polynomial gives a unified perspective and proof.

[edit] How to compute it

Let I _{T, v} be defined as

$\mathit{I}_{T, v} = \{ p \in \mathbb{K}[t] \; | \; v \in \mbox{Ker} \; p(T) \} = \{ p \in \mathbb{K}[t] \; | \; p(T)(v) = O \}.$

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

Properties

It is easy to see that I_{T, v} divides I_T.
If d is the greatest natural number such that v, T(v), ... , T^d(v) are linearly independent, then

$\alpha_0 v + \alpha_1 T(v) + \cdots + \alpha_n T^d (v) + T^{d+1} (v) = O$ for some alpha in $\mathbb{K}$ and

$\mu_{T,v} (t) = \alpha_0 + \alpha_1 t + \cdots + \alpha_n t^d + t^{d+1}$

Given a basis {v₁,..., v_n} the minimal polynomial is the least common multiple of $\mu_{T,v_i}$ for all i = 1, ... , n.

[edit] Example

Define T on R³ with the matrix

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

Take

$e_1 = \begin{vmatrix} 1 \\ 0 \\ 0 \end{vmatrix}$

We see that e₁, T(e₁), T²(e₁) are linearly independent while by adding T³(e₁) to this list they are no more. Thus it's easy to find $\mu_{T,e_1}$ , repeat it with e₂ and e₃ and calculate the least common multiple of them.