91Ó°ÊÓ

Consider an eigenvalue λ of A with corresponding eigenvector x, i.e., Ax = λx. To show that λ is also an eigenvalue of A^T, we need to find an eigenvector for A^T corresponding to λ. Taking the transpose of the above equation, we get (Ax)^T = (λx)^T x^T A^T = λ x^T. Now, notice that if x is an eigenvector of A, then x^T is a left-eigenvector of A^T. Thus, λ is indeed an eigenvalue of both A and A^T. As a result, A and A^T share the same set of eigenvalues.

The characteristic polynomial p(λ) of a matrix A is given by the determinant of (A - λI), where I is the identity matrix of the same size as A. And the characteristic polynomial of A^T is given by the determinant of (A^T - λI). We need to show that these polynomials are the same. Using the property of determinants, we have: \(\text{det}(A^T - \lambda I) = \text{det}((A-\lambda I)^T)\) Since the determinant of a matrix and its transpose are equal, we get: \(\text{det}(A^T - \lambda I) = \text{det}(A - \lambda I)\). Thus, the characteristic polynomials of A and A^T are the same.

Recall that the minimal polynomial m_A(λ) of a matrix A is the monic polynomial of least degree such that m_A(A) = 0. Since A and A^T have the same characteristic polynomial, they also have the same algebraic multiplicities for their eigenvalues. Hence, for any polynomial P(λ) that annihilates A, P(A) = 0, P(λ) must also annihilate A^T, satisfying P(A^T) = 0. Now, consider the minimal polynomial m_A(λ) of A. Since m_A(λ) divides any polynomial P(λ) that annihilates A, m_A(λ) must also divide any polynomial Q(λ) that annihilates A^T. Therefore, m_A(λ) annihilates A^T as well. Similarly, if m_{A^T}(λ) is the minimal polynomial of A^T, then m_{A^T}(λ) annihilates A too. Since m_A(λ) and m_{A^T}(λ) are monic polynomials, and they annihilate both A and A^T, we can conclude that both minimal polynomials are the same, i.e., m_A(λ) = m_{A^T}(λ). Thus, a matrix A and its transpose A^T have the same minimal polynomial.