Minimum Polynomials of Linear Transformations

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Minimum Polynomials of Linear Transformations Spencer De Chenne University of Puget Sound 30 April 2014

Table of Contents Polynomial Basics Endomorphisms Minimum Polynomial Building Linear Transformations Invariant Subspaces via Minimum Polynomial

Polynomials Given a field F, we denote the set of all polynomials with coefficients in F by F[x].

Polynomials Given a field F, we denote the set of all polynomials with coefficients in F by F[x]. Eg. f(x) = x 4 5 9 x3 + 5 Q[x] g(x) = πx 3 ex 2 + i C[x]

Irreducible Polynomials Definition A non-constant polynomial f(x) F[x] is irreducible if there are no g(x), h(x) F[x], where the degrees of g(x) and h(x) are both less than the degree of f(x), such that f(x) = g(x)h(x).

Irreducible Polynomials Definition A non-constant polynomial f(x) F[x] is irreducible if there are no g(x), h(x) F[x], where the degrees of g(x) and h(x) are both less than the degree of f(x), such that f(x) = g(x)h(x). For our purposes, think of irreducible polynomials as equivalent to prime numbers.

Irreducible examples Consider Q[x], R[x], and C[x]: x 2 2: irreducible in Q[x]

Irreducible examples Consider Q[x], R[x], and C[x]: x 2 2: irreducible in Q[x] x 3 15x 2 45x + 21: irreducible in Q[x]

Irreducible examples Consider Q[x], R[x], and C[x]: x 2 2: irreducible in Q[x] x 3 15x 2 45x + 21: irreducible in Q[x] x 2 + 1: irreducible in R[x] and Q[x]

Irreducible examples Consider Q[x], R[x], and C[x]: x 2 2: irreducible in Q[x] x 3 15x 2 45x + 21: irreducible in Q[x] x 2 + 1: irreducible in R[x] and Q[x] Which polynomials are irreducible in C[x]:

Irreducible examples Consider Q[x], R[x], and C[x]: x 2 2: irreducible in Q[x] x 3 15x 2 45x + 21: irreducible in Q[x] x 2 + 1: irreducible in R[x] and Q[x] Which polynomials are irreducible in C[x]: only linear factors.

Irreducible Factors What is important about irreducible polynomials?

Irreducible Factors What is important about irreducible polynomials? Theorem Let f(x) F[x] be a non-constant polynomial. Then f(x) is a unique (up to order) product of irreducible factors.

Irreducible Factors What is important about irreducible polynomials? Theorem Let f(x) F[x] be a non-constant polynomial. Then f(x) is a unique (up to order) product of irreducible factors. Think about this like an integer being a product of prime numbers.

Monic Polynomials Definition A polynomial f(x) F[x] is monic if its leading coefficient is 1.

Monic Polynomials Definition A polynomial f(x) F[x] is monic if its leading coefficient is 1. Eg. f(x) = x 4 + 3x 3 1 is monic g(x) = 2x 7 6x 3 is not monic

Endomorphisms Minimum polynomials are only used for a specific type of linear transformation: endomorphisms. Definition An endomorphism T is a linear transformation mapping from a vector space V onto itself (i.e. T : V V ). For a vector space V, we shall denote the set of all endomorphisms of V as End(V ).

More Endomorphisms Remark Notice that for R, S End(V ), their composition, R S, is also an endomorphism. Also, for α F, αr End(V ).

More Endomorphisms Remark Notice that for R, S End(V ), their composition, R S, is also an endomorphism. Also, for α F, αr End(V ). We denote the n th iterate of T by T n = } T T {{ T}. n times

More Endomorphisms Remark Notice that for R, S End(V ), their composition, R S, is also an endomorphism. Also, for α F, αr End(V ). We denote the n th iterate of T by T n = } T T {{ T}. n times From the previous two remarks, we can see that for T End(V ) and p(x) F[x], then p(t ) End(V ).

Annihilator Polynomial Theorem Let V be a vector space of dimension n, v V a non-zero vector, and T an endomorphism of V. Then there is a unique monic polynomial of minimum degree, m T,v (x), such that m T,v (v) = 0. This polynomial has degree at most n. This polynomial, m T,v (x), is called the T -annihilator polynomial for v.

Proof of Annihilator Polynomial Proof Sketch The set {T n (v), T n 1 (v),..., T (v), v} is a set of n + 1 vectors in an n-dimensional vector space, and must be linearly dependent.

Proof of Annihilator Polynomial Proof Sketch The set {T n (v), T n 1 (v),..., T (v), v} is a set of n + 1 vectors in an n-dimensional vector space, and must be linearly dependent. There exist scalars a n,..., a 1, a 0 such that a n T n (v) + + a 1 T (v) + a 0 v = 0.

Proof of Annihilator Polynomial Proof Sketch The set {T n (v), T n 1 (v),..., T (v), v} is a set of n + 1 vectors in an n-dimensional vector space, and must be linearly dependent. There exist scalars a n,..., a 1, a 0 such that a n T n (v) + + a 1 T (v) + a 0 v = 0. Define f(x) = a n x n + + a 1 x + a 0. We can make this polynomial monic and show it satisfies the other properties of the T -annihilator polynomial of v.

Minimum Polynomial Theorem Let V be an n-dimensional vector space, and T and endomorphism of V. Then there exists a unique monic polynomial of minimum degree, m T (x), such that m T (v) = 0 for every v V. This polynomial has degree at most n. We call this polynomial, m T (x), the minimum polynomial of T.

Characteristic Polynomial Definition For an endomorphism T of V with matrix representation [T ] B relative to basis B, the characteristic polynomial of T, c T (x), is the polynomial c T (x) = det (xi [T ] B ).

Characteristic Polynomial Definition For an endomorphism T of V with matrix representation [T ] B relative to basis B, the characteristic polynomial of T, c T (x), is the polynomial c T (x) = det (xi [T ] B ). Theorem If A and B be similar matrices, then the characteristic polynomials of A and B, c A (x) and c B (x), are equal. We can see that the characteristic polynomial of T is a well-defined polynomial.

Companion Matrix Definition Let f(x) = x n + a n 1 x n 1 + + a 1 x + a 0 be a monic polynomial of degree n 1. Then the companion matrix of f(x), C(f(x)), is the n n matrix a n 1 1 0 0 a n 2 0 1 0 C(f(x)) =...., a 1 0 0 1 a 0 0 0 0 where the 1 s are located on the super-diagonal.

Applications of Companion Matrix Why do we care about the Companion Matrix?

Applications of Companion Matrix Why do we care about the Companion Matrix? Theorem Let f(x) be a polynomial, and A = C(f(x)) its companion matrix. Then c A (x) = det (xi A) = f(x). Further, m A (x) = f(x). We can create linear transformations with eigenvalue properties we want.

Building Endomorphisms Suppose we want a linear transformation, T, with eigenvalues λ = 1, 3, 4, and algebraic multiplicities α( 1) = α(3) = α(4) = 1.

Building Endomorphisms Suppose we want a linear transformation, T, with eigenvalues λ = 1, 3, 4, and algebraic multiplicities α( 1) = α(3) = α(4) = 1. First build c T (x) : c T (x) = (x + 1)(x 3)(x 4) = x 3 6x 2 + 5x + 12

Building Endomorphisms Suppose we want a linear transformation, T, with eigenvalues λ = 1, 3, 4, and algebraic multiplicities α( 1) = α(3) = α(4) = 1. First build c T (x) : c T (x) = (x + 1)(x 3)(x 4) = x 3 6x 2 + 5x + 12 Then build C(c T (x)) : Boom. C(c T (x)) = 6 1 0 5 0 1 12 0 0

Another Example Let s do a larger example: c T (x) = (x 2 + 2) 2 (x 4 + 1).

Another Example Let s do a larger example: c T (x) = (x 2 + 2) 2 (x 4 + 1). Then 0 1 0 0 0 0 0 0 4 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 C(c T (x)) = 5 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0. 4 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 4 0 0 0 0 0 0 0 This linear transformation preserves eigenvalues and algebraic multiplicities of eigenvalues.

Relationship between m T (x) and c T (x) So far, our examples have shown that m T (x) = c T (x). This is not true in general. Consider T = 3 3 3 4 4 4 5 5 5.

Relationship between m T (x) and c T (x) So far, our examples have shown that m T (x) = c T (x). This is not true in general. Consider T = 3 3 3 4 4 4 5 5 5. Then c T (x) = x 2 (x 12). However, we can compute that ker (T (T 12I)) = Q 3.

Relationship between m T (x) and c T (x) So far, our examples have shown that m T (x) = c T (x). This is not true in general. Consider T = 3 3 3 4 4 4 5 5 5. Then c T (x) = x 2 (x 12). However, we can compute that We will see that this implies and m T (x) c T (x). ker (T (T 12I)) = Q 3. m T (x) = x(x 12),

Relationship between m T (x) and c T (x) Theorem Let V be a finite-dimensional vector space, and T and endomorphism of V. Let m T (x) and c T (x) be the minimum and characteristic polynomials of T, respectively. Then m T (x) divides c T (x), and every irreducible factor of c T (x) is also an irreducible factor of m T (x).

Kernels of Polynomials Theorem Let V be an n-dimensional vector space, T and endomorphism of V, and p(x) F[x]. Then, ker (p(t )) = {v V : p(t )(v) = 0} is a T -invariant subspace of V.

Direct Sums via Minimum Polynomials For a special case of an endomorphism, we can use the minimum polynomial to write V as the direct sum of invariant subspaces. Theorem Let V be a vector space, and T an endomorphism of V. Suppose m T (x) factors into pairwise relatively prime polynomials m T (x) = p 1 (x)p 2 (x) p k (x). For each i, let W i = ker (p i (T )). Then each W i is T -invariant, and V = W 1 W 2 W k.

Last Example Recall A = 6 1 0 5 0 1 12 0 0.

Last Example Recall A = 6 1 0 5 0 1 12 0 0 Then, m A (x) = (x + 1)(x 3)(x 4)..

Last Example Recall A = 6 1 0 5 0 1 12 0 0 Then, m A (x) = (x + 1)(x 3)(x 4). We know. Q 3 = ker (T + 1) ker (T 3) ker (T 4) 1 9 16 = 1 3 4. 1 1 1

Bibliography [1] Weintraub, Steven H. A Guide to Advanced Linear Algebra. United States of America: The Mathematical Association of America, 2011. [2] Curtis, Morton L. Abstract Linear Algebra. New York: Springer-Verlag, 1990.

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