Final answer:
To determine if matrix A is diagonalizable, we find its eigenvalues from the characteristic equation, then find the corresponding eigenvectors. If two linearly independent eigenvectors are found, A is diagonalizable, and we can form Q from eigenvectors and D from eigenvalues.
Step-by-step explanation:
The given matrix A in Mnxn(R) is:
[ 1 3 ]
[ 3 1 ]
To test the matrix for diagonalizability, we first need to find the eigenvalues of A. The characteristic equation of A is given by:
|A - λI| = 0
Where λ is an eigenvalue and I is the identity matrix. The determinant of (A - λI) leads to a quadratic equation in λ.
Step 1: Calculate the determinant of A - λI
det(A - λI) = | 1-λ 3 | = (1-λ)(1-λ) - 3·λ = λ^2 - 2λ - 8
| 3 1-λ |
Step 2: Find the eigenvalues by solving the characteristic equation λ^2 - 2λ - 8 = 0. The solutions are λ = 4 and λ = -2.
Step 3: Find the eigenvectors corresponding to each eigenvalue. For λ = 4, we solve (A - 4I)x = 0. For λ = -2, we solve (A + 2I)x = 0.
Step 4: Form the matrix Q using the eigenvectors as columns, and D as the diagonal matrix with the eigenvalues.
If two linearly independent eigenvectors are found, then the matrix A is diagonalizable. If not, then A is not diagonalizable. In this case, since we have two distinct eigenvalues, we expect to find two linearly independent eigenvectors, indicating that A is likely diagonalizable.