# Minimal polynomial (linear algebra)

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

The following three statements are equivalent:

1. Template:Mvar is a root of μA,
2. Template:Mvar is a root of the characteristic polynomial χA of Template:Mvar,
3. Template:Mvar is an eigenvalue of matrix Template:Mvar.

The multiplicity of a root Template:Mvar of μA is the largest power Template:Mvar such that Ker((AλIn)m) strictly contains Ker((AλIn)m−1). In other words, increasing the exponent up to Template:Mvar will give ever larger kernels, but further increasing Template:Mvar 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 irreducible polynomials Template:Mvar one has similar equivalences:

1. Template:Mvar divides μA,
2. Template:Mvar divides χA,
3. the kernel of P(A) has dimension at least 1.
4. the kernel of P(A) has dimension at least deg(P).

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 from 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 Template:Mvar: 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 Template:Mvar is a multiple aIn of the identity matrix, then its minimal polynomial is Xa since the kernel of aInA = 0 is already the entire space; on the other hand its characteristic polynomial is (Xa)n (the only eigenvalue is Template:Mvar, 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 (for the case of matrices over a field).

## Formal definition

Given an endomorphism Template:Mvar on a finite-dimensional vector space Template:Mvar over a field F, let IT 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 a proper ideal of F[t]. Then the minimal polynomial is the monic polynomial which generates IT. Thus it must be the monic polynomial of least degree in IT.

## Applications

An endomorphism Template:Mvar 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 Xλ for every eigenvalue Template:Mvar means that the generalized eigenspace for Template:Mvar is the same as the eigenspace for Template:Mvar: every Jordan block has size 1. More generally, if Template:Mvar satisfies a polynomial equation P(φ) = 0 where Template:Mvar factors into distinct linear factors over F, then it will be diagonalizable: its minimal polynomial is a divisor of Template:Mvar 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 k = 2 of involutions, this is even true for endomorphisms of vector spaces over any field of characteristic other than 2, since X 2 − 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 = X 2X = 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 μφ = X k with k ≥ 2 then Template:Mvar (a nilpotent endomorphism) is not necessarily 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

For a vector Template:Mvar in Template:Mvar define:

${\mathit {I}}_{T,v}=\{p\in \mathbf {F} [t]\;|\;p(T)(v)=0\}.$ This definition satisfies the properties of a proper ideal. Let μT,v be the monic polynomial which generates it.

### Properties

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### Example

Define Template:Mvar to be the endomorphism of R3 with matrix, on the canonical basis,

${\begin{pmatrix}1&-1&-1\\1&-2&1\\0&1&-3\end{pmatrix}}.$ Taking the first canonical basis vector e1 and its repeated images by Template:Mvar 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 3e1 = −4T 2e1Te1 + e1,

so that:

μT,e1 = X 3 + 4X 2 + X − 1.

This is in fact also the minimal polynomial μT and the characteristic polynomial χT: indeed μT,e1 divides μT which divides χT, 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 Template:Mvar annihilates a vector Template:Mvar, then it also annihilates Tv (just apply Template:Mvar to the equation that says that it annihilates Template:Mvar), and therefore by iteration it annihilates the entire space generated by the iterated images by Template:Mvar of Template:Mvar; in the current case we have seen that for v = e1 that space is all of R3, so μT,e1(T) = 0. Indeed one verifies for the full matrix that T 3 + 4T 2 + TI3 is the null matrix:

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