For A = [[4,1],[0,2]], which statement is true?

• A. A has eigenvalues 4 and 2; A is not diagonalizable.
• B. A has eigenvalues 4 and 2; A is diagonalizable.
• C. A has eigenvalues 4 and -2; A is diagonalizable.
• D. A has eigenvalues -4 and -2; A is not diagonalizable.

Answer: (B)

Knowing how eigenvalues determine diagonalizability is the key. For a 2×2 matrix, if you have two distinct eigenvalues, you automatically get two linearly independent eigenvectors, which lets you form a basis of the space and transform the matrix into a diagonal form.

Compute the eigenvalues from det(A − λI) = 0. Here det[[4−λ, 1],[0, 2−λ]] = (4−λ)(2−λ), so the eigenvalues are 4 and 2. They’re distinct, so A will be diagonalizable.

Find eigenvectors to see the independence clearly. For λ = 4, solve (A − 4I)x = 0, which gives y = 0 and x free, so an eigenvector is (1, 0). For λ = 2, solve (A − 2I)x = 0, yielding 2x + y = 0, so take (1, −2) as an eigenvector. These two vectors are not multiples, so they are linearly independent and span the plane.

Since A has two independent eigenvectors, it is diagonalizable and similar to diag(4, 2). Hence the statement that the eigenvalues are 4 and 2 and that A is diagonalizable is the true one.