Linear Algebra III Lecture 11 - PDF Free Download

Linear Algebra III Lecture 11 Xi Chen 1 1 University of Alberta February 13, 2015

Outline Minimal Polynomial 1 Minimal Polynomial

Minimal Polynomial The minimal polynomial f (x) of a square matrix A is a nonzero polynomial of minimal degree such that f (A) = 0. Minimal polynomial is a similarity invariance. If B = P 1 AP, then f (B) = P 1 f (A)P. So f (A) = 0 if and only if f (B) = 0. By Caley-Hamilton, g(a) = 0 for g(x) = det(xi A). So deg f (x) deg g(x) = n for a minimal polynomial f (x) of an n n matrix A. If A is diagonalizable and det(xi A) = (x λ 1 ) a 1 (x λ 2 ) a 2...(x λ k ) a k with λ 1, λ 2,..., λ k distinct eigenvalues of A, then the minimal polynomial of A is f (x) = (x λ 1 )(x λ 2 )...(x λ k ).

Generalized Eigenspace and Minimal Polynomial Theorem Let A be a square matrix with characteristic polynomial det(xi A) = (x λ 1 ) a 1 (x λ 2 ) a 2...(x λ k ) a k and distinct eigenvalues λ 1, λ 2,..., λ k. Then the minimal polynomial of A is f (x) = (x λ 1 ) m 1 (x λ 2 ) m 2...(x λ k ) m k, where m i a i are the numbers such that dim Nul(A λ i I) < dim Nul(A λ i I) 2 <... < dim Nul(A λ i I) m i = dim Nul(A λ i ) m i +1 =... = a i

Existence of Minimal Polynomial Theorem For a square matrix A M n n (R), there exists a nonzero polynomial f (x) R[x] such that f (A) = 0. So the minimal polynomial of A exists. Proof. Since M n n (R) is a vector space of dimension n 2 over R, there exist c 0, c 1,..., c n 2 R, not all zero, such that c 0 I + c 1 A +... + c n 2A n2 = 0. Then f (A) = 0 for f (x) = c 0 + c 1 x +... + c n 2x n2. Since c 0, c 1,..., c n 2 are not all zero, f (x) 0.

Basics of Polynomials (Long Division) For a pair of polynomials f (x) and g(x) 0, there exist polynomials q(x) and r(x) such that deg r(x) < deg g(x) and f (x) = q(x)g(x) + r(x). (Unique Factorization) Every nonzero polynomial f (x) can be uniquely factored into a product of irreducible polynomials: f (x) = f 1 (x)f 2 (x)...f k (x) where f i (x) are irreducible, i.e., f i (x) g 1 (x)g 2 (x) for any polynomials g 1 (x) and g 2 (x) of deg g 1 (x), deg g 2 (x) < f i (x).

Basics of Polynomials Theorem (Fundamental Theorem of Algebra) Every complex polynomial f (x) C[x] is a product of polynomials of degree 1: f (x) = c(x λ 1 )(x λ 2 )...(x λ n ). Every real polynomial f (x) R[x] is a product of polynomials of degree 1 or 2: f (x) = c(x λ 1 )(x λ 2 )...(x λ k ) (x 2 + a 1 x + b 1 )(x 2 + a 2 x + b 2 )...(x 2 + a l x + b l ).

Basics of Polynomials (Greatest Common Divisor) The gcd of two polynomials f 1 (x) and f 2 (x) is the polynomial g(x) of the highest degree such that g(x) divides both f 1 (x) and f 2 (x), written as gcd(f 1 (x), f 2 (x)) = g(x). We say that f 1 (x) and f 2 (x) are coprime if gcd(f 1 (x), f 2 (x)) = 1. If f 1 (x) = c 1 (x λ 1 ) a 1 (x λ 2 ) a 2...(x λ k ) a k f 2 (x) = c 2 (x λ 1 ) b 1 (x λ 2 ) b 2...(x λ k ) b k for c 1, c 2 0 and λ 1, λ 2,..., λ k distinct, then gcd(f 1 (x), f 2 (x)) = (x λ 1 ) min(a 1,b 1 ) (x λ 2 ) min(a 2,b 2 )...(x λ k ) min(a k,b k )

Basics of Polynomials (Euclidean Algorithm) Given two polynomials f 1 (x) and f 2 (x), we do long divisions f 1 (x) = q 1 (x)f 2 (x) + f 3 (x), deg f 3 (x) < deg f 2 (x) f 2 (x) = q 2 (x)f 3 (x) + f 4 (x), deg f 4 (x) < deg f 3 (x). =. f n 1 (x) = q n 1 (x)f n (x) + f n+1 (x), deg f n+1 (x) < deg f n (x) and stop when f n+1 (x) = 0. Then gcd(f 1 (x), f 2 (x)) = f n (x).

Bezout Identity Minimal Polynomial Theorem For two polynomials f 1 (x) and f 2 (x), there exist polynomials g 1 (x) and g 2 (x) such that g 1 (x)f 1 (x) + g 2 (x)f 2 (x) = gcd(f 1 (x), f 2 (x)). Proof. Apply Euclidean algorithm to f 1 (x) and f 2 (x): f 1 (x) = q 1 (x)f 2 (x) + f 3 (x), deg f 3 (x) < deg f 2 (x) f 2 (x) = q 2 (x)f 3 (x) + f 4 (x), deg f 4 (x) < deg f 3 (x). =. f n 1 (x) = q n 1 (x)f n (x) + f n+1 (x), f n+1 (x) = 0 Then f n (x) = gcd(f 1 (x), f 2 (x)).

Bezout Identity Minimal Polynomial Proof (CONT). a n = 1, b n = 0 a n f n + b n f n+1 = gcd(f 1, f 2 ) a n 1 = b n, b n 1 = a n q n 1 b n a n 1 f n 1 + b n 1 f n = gcd(f 1, f 2 ).. a 1 = b 2, b 1 = a 2 q 1 b 2 a 1 f 1 + b 1 f 2 = gcd(f 1, f 2 )

Uniqueness of Minimal Polynomial Theorem For a square matrix A, if f 1 (A) = f 2 (A) = 0 for two polynomials f 1 (x) and f 2 (x), then g(a) = 0 for g(x) = gcd(f 1 (x), f 2 (x)). So the minimal polynomial of A is unique up to a scalar. Proof. There exist polynomials g 1 (x) and g 2 (x) such that g 1 (x)f 1 (x) + g 2 (x)f 2 (x) = g(x) = gcd(f 1 (x), f 2 (x)). So g(a) = g 1 (A)f 1 (A) + g 2 (A)f 2 (A) = 0. If f 1 (x) and f 2 (x) are two minimal polynomials of A with deg f 1 (x) = deg f 2 (x) = d, then g(a) = 0 for g(x) = gcd(f 1 (x), f 2 (x)). Since deg g d and f 1 (x) and f 2 (x) are minimal, f 1 (x) = cf 2 (x) for some c 0.