Minimal polynomial, transpose, similar

[clg211]

Homework Statement

a) Prove that if a polynomial f(lambda) has f(A)=0, then f(A^T)=0

b) Prove that A and A^T have the same minimal polynomial.

c) If A has a cyclic vector, prove that A^T is similar to A.


2. The attempt at a solution

a) I know that I need to show that f(A^T) = f(A)^T.

b) The main problem is that I don't think I understand completely what the minimal polynomial is. I know how to get the minimal polynomial from the last column of a matrix in rational canonical form, and that's about it.

c) This should be a direct consequence of part (b) since same minimal polynomial --> similar.

 

[mighty2000]

Thank you for your post. I am happy to help you with your questions.

a) To prove that f(AT) = f(A)T, we can use the fact that a polynomial function can be written as a linear combination of its powers. In other words, f(lambda) = a0 + a1*lambda + a2*lambda^2 + ... + an*lambda^n. Therefore, we can write f(AT) = a0I + a1*AT + a2*(AT)^2 + ... + an*(AT)^n. Using the property of matrix multiplication, we can rewrite this as f(AT) = a0I + a1*A*T + a2*A^2*T^2 + ... + an*A^n*T^n. Since f(A) = 0, this means that a0I + a1*A + a2*A^2 + ... + an*A^n = 0. Therefore, we can substitute this in the previous equation to get f(AT) = 0 + 0 + 0 + ... + 0 = 0. Hence, f(AT) = f(A)T.

b) The minimal polynomial of a matrix A is the smallest degree polynomial that satisfies p(A) = 0. This means that the minimal polynomial is the polynomial of smallest degree that has A as a root. To prove that A and AT have the same minimal polynomial, we can use the fact that the minimal polynomial is unique for a given matrix. This means that if two matrices have the same minimal polynomial, then they must be equal. To show this, let p(x) be the minimal polynomial of A. Then, we have p(A) = 0. Now, consider p(AT). Using part (a), we can write p(AT) = p(A)T = 0. This means that p(AT) is also a polynomial of degree smaller than or equal to p(x) that has AT as a root. Since the minimal polynomial is unique, this means that p(AT) = p(x). Hence, A and AT have the same minimal polynomial.

c) As you correctly mentioned, this follows directly from part (b). If A has a cyclic vector, this means that there exists a vector v such that the set {v, Av, A^2v, ..., A^(n-1)v} spans the entire vector space. This means that the matrix