Minimal_polynomial

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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:

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

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

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

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 $\mathbb{K}[t]$ .

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

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 linear factors.

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

Let I _{T, v} be defined as

This definition satisfies the properties of a proper ideal. Let $Î¼ T, v$ be the monic polynomial which generates it.

Properties

Define T on R³ with the matrix

Take

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.

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Last updated on Monday September 24, 2007 at 05:23:47 PDT (GMT -0700)
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