For the matrix A = [ [2, 1], [0, 3] ], which statement about eigenvalues, eigenvectors and diagonalizability is correct?

• A. A has eigenvalues 2 and 3 with eigenvectors (1,0) for λ=2 and (1,1) for λ=3; A is diagonalizable.
• B. A has eigenvalues 2 and 3 with eigenvectors (1,0) for λ=3 and (0,1) for λ=2; A is diagonalizable.
• C. A has eigenvalues 2 and 3 with eigenvectors (1,0) for λ=2 and (0,1) for λ=3; A is not diagonalizable.
• D. A has eigenvalues 0 and 5; A is diagonalizable.

Answer: (A)

When a matrix is upper triangular, its eigenvalues are the diagonal entries, so here they are 2 and 3. To find eigenvectors, solve (A − λI)x = 0. For λ = 2, A − 2I = [[0, 1], [0, 1]], which gives x2 = 0 and x1 free, so eigenvectors are multiples of (1, 0). For λ = 3, A − 3I = [[−1, 1], [0, 0]], giving −x1 + x2 = 0, so eigenvectors are multiples of (1, 1). These two eigenvectors are linearly independent, so A is diagonalizable. The statement that matches this is that the eigenvectors are (1, 0) for λ = 2 and (1, 1) for λ = 3, and A is diagonalizable. The other options misassign the eigenvectors, claim non-diagonalizability, or give eigenvalues that don’t fit the matrix (trace and determinant).