If lambda is an eigenvalue of A then lambda is an eigenvalue

If lambda is an eigenvalue of A then lambda\" is an eigenvalue of A\", and A\" has the same eigenvectors as A.

Solution

We know that (aⁿ –bⁿ) = (a –b)(a^n-1 + ba^n-2 + … +b^n-1) . Therefore, A – I is part of the matrix decomposition of (Aⁿ – ⁿ Iⁿ) i.e. (Aⁿ – ⁿ Iⁿ) = (A – I)P where, P is another matrix of suitable dimension.

We also know that det (AB) = det(A)det(B). Thus, if is an eigenvalue of A , then det (A – I) = 0, so that det (Aⁿ – ⁿ Iⁿ ) = or    det (Aⁿ – ⁿ I) = 0 as Iⁿ = I . Therefore ⁿ is an eigenvalue of Aⁿ .

If there is a nontrivial solution x of Ax = x , such an x is called an eigenvector of A corresponding to the eigenvalue . Now, if Ax = x , then (A – I)x = 0. Then, since A – I is part of the decomposition of (Aⁿ – ⁿ Iⁿ), we also have (Aⁿ – ⁿ Iⁿ)x = 0,   or, (Aⁿ – ⁿ I)x so that x is an eigenvector of Aⁿ .