Answer: The matrix A is not diagonalizable.
Explanation:
The basis of each eigenspace of the matrix A = [2 0 1; 1 1 1; 1 0 2], we need to find the eigenvectors corresponding to each eigenvalue.
First, let's find the eigenvalues by solving the characteristic equation det(A - λI) = 0, where λ is the eigenvalue and I is the identity matrix.
The characteristic equation is:
|2-λ 0 1| |2-λ 0 1|
|1 1-λ 1| = det|1 1-λ 1|
|1 0 2-λ| |1 0 2-λ|
Expanding the determinant, we get:
(2-λ)((1-λ)(2-λ) - 1) - (0 - (1(2-λ) - 1)) + (1(0) - (1(1-λ))) = 0
Simplifying the equation, we get:
(2-λ)(λ^2 - 3λ + 1) + (2-λ) + (λ-1) = 0
(2-λ)(λ^2 - 3λ + 1 + 1) + (λ-1) = 0
(2-λ)(λ^2 - 3λ + 2) + (λ-1) = 0
(2-λ)(λ-2)(λ-1) + (λ-1) = 0
(λ-1)[(2-λ)(λ-2) + 1] = 0
From this equation, we can see that λ = 1 is an eigenvalue.
Now let's find the eigenvectors corresponding to λ = 1 by solving the equation (A - λI)v = 0, where v is the eigenvector.
Substituting λ = 1 and setting up the equation, we get:
(2-1)v1 + 0v2 + 1v3 = 0
v1 + v3 = 0
We can choose v2 = 1 as a free variable.
Setting v2 = 1, we have v1 = -v3.
Therefore, the eigenvector corresponding to λ = 1 is v = [-1; 1; 1].
Since we found one eigenvector v = [-1; 1; 1], we need to check if there are two more linearly independent eigenvectors. If we find two more, then A is diagonalizable; otherwise, it is not.
Unfortunately, in this case, we do not find two additional linearly independent eigenvectors, so A is not diagonalizable.