Let n-order matrix a satisfy a ^ 2 + 2A – 3E = 0, prove that a + 4E is invertible, and find their inverse

Let n-order matrix a satisfy a ^ 2 + 2A – 3E = 0, prove that a + 4E is invertible, and find their inverse

(a + 4e) (a-2e) = A & # 178; + 2e-8e, from the known condition, left formula = - 5E, then the inverse of a + 4E is - 1 / 5 (a-2e)

Let a be a square matrix of order n, and the square of a = e, it is proved that: (1) the eigenvalue of a can only be 1 or - 1; (2) 3e-a is invertible

(1) Let λ be the eigenvalue of a, then λ ^ 2-1 is the eigenvalue of a ^ 2-e, and a ^ 2-e = 0, so λ ^ 2-1 = 0, so λ = 1 or - 1. So the eigenvalue of a can only be 1 or - 1. (2) a (a-3e) + 3 (a-3e) = - 8e, so (a + 3e) (3e-a) = 8e, so 3e-a is reversible, and (3e-a) ^ - 1 = (1 / 8) (...)

Let a be a fourth-order square matrix with eigenvalues of 1,3,3, - 2. If a can be diagonalized, then R (3e-a) =?

The eigenvalues of a are 1,3,3, - 2
The eigenvalues of 3e-a are 2,0,0,5
So r (3e-a) = 2