7 4 0 :9 points) The matrix A = 8 6 3 has an eigenvalue A = 3. 3 1 3 =ind an eigenvector for this eigenvalue. Try to solve the following problem WITHOUT computing all eigenvalues. 0 0 1 _' 1 The matrix B = 5 5 7 has an eigenvector u = 3 1 1 1 1 =ind the eigenvalue for this eigenvector. A: Llire the following helpful? ? V Example 8.2. ? v Definition 8.6. ? V Theorem 8.18. 5 2 Example 8.2. Let A = and X = Then 6 5 2 2 4 AX = = = 2X 6 6 So, A = 2 is an eigenvalue of A with corresponding eigenvector X.Definition 8.6 (Characteristic polynomial). Let A be a square n x n matrix, and let x denote a variable. The characteristic polynomial of A, denoted by CA (x), is defined by CA() = det (A - x1). The eigenvalues A1, 12, . .., Ak of A are solutions to equation (23) (or, roots of the characteristic polynomial CA (x) ).Theorem 8.18. Let X1, X2, ..., Xk be eigenvectors corresponding to distinct eigenvalues 1, 12, ..., Ak of anxn matrix A. Then the vectors X1, X2, . .., Xk are linearly independent. We can try and find a basis for R" which consists of eigenvectors of a matrix A, by taking bases for distinct eigenspaces of A