Let the eigenvalues of a square matrix of order 3 be 1,2,0, and its corresponding eigenvectors A1, A2, A3. B = a ^ 3-2a + 3E, and find the eigenvector of B ^ - 1 Why A1, A2, A3? It's not mentioned in the book. What's the basis

Let the eigenvalues of a square matrix of order 3 be 1,2,0, and its corresponding eigenvectors A1, A2, A3. B = a ^ 3-2a + 3E, and find the eigenvector of B ^ - 1 Why A1, A2, A3? It's not mentioned in the book. What's the basis

If α is the eigenvector of a belonging to the eigenvalue λ, then α is the eigenvector of F (a) belonging to the eigenvalue f (λ), so A1, A2, A3 are still the eigenvectors of B = f (a). If α is the eigenvector of a belonging to the eigenvalue λ, and a is reversible, then α is the eigenvector of a ^ - 1 belonging to the eigenvalue 1 / λ

Let a be a square matrix of order 3, x1, X2 and X3 be the three different eigenvalues of a, and the corresponding eigenvectors are A1, A2 and A3 respectively, so that B = a1 + A2 + a3
It is proved that B, AB and a ^ 2B are linearly independent. If a ^ 3B = 3ab-2a ^ 2B, find the eigenvalue of a and calculate the determinant a + E

First, note that A1, A2, A3 are linearly independent, and then (B, AB, a ^ 2b) = (A1, A2, A3) * V, where V = 1 X1 X1 ^ 21 x2 x2 ^ 21 X3 X3 ^ 2 is Vandermonde matrix. Because x1, X2, X3 are different from each other and V is nonsingular, B, AB, a ^ 2B are linearly independent. 0 = a ^ 3B - (3ab-2a ^ 2b) = (x1 ^ 3 + 2x1 ^ 2-3x1) a1 + (x2

Let a 3-order square matrix A have three different eigenvalues N1, N2, N3, and the corresponding eigenvectors are A1, A2, A3. Let B = a1 + A2 + a3,
It is proved that B, AB, a and B are linearly independent

AB = a (a1 + A2 + a3) = Aa1 + aa2 + aa3 = n1a1 + n2a2 + n3a3a ^ 2B = a (AB) = a (n1a1 + n2a2 + n3a3) = N1 ^ 2A1 + N2 ^ 2A2 + N3 ^ 2A3, so (B, AB, a ^ 2b) = (A1, A2, A3) k, where k = 1 N1 N1 N1 ^ 21 N2 ^ 21 N3 N3 N3 ^ 2 because N1, N2, N3 are different, so | K ≠ 0, so K is reversible