Which statement about eigenpairs of A = [[2,1],[0,3]] is true?

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Multiple Choice

Which statement about eigenpairs of A = [[2,1],[0,3]] is true?

Think in terms of solving (A − λI)v = 0 to find eigenvectors for each eigenvalue λ. For this matrix, the eigenvalues are the diagonal entries 2 and 3 since det(A − λI) = (2 − λ)(3 − λ). For λ = 2, A − 2I = [[0, 1], [0, 1]]. The equations are 0·x + 1·y = 0, which gives y = 0, while x is free. So eigenvectors are nonzero multiples of (1, 0). That means (1, 0) is indeed an eigenvector corresponding to λ = 2. If you check the other options, you’ll see they don’t satisfy (A − λI)v = 0 for their claimed λ: for λ = 3, eigenvectors must satisfy y = x, so (1, 0) or (0, 1) don’t fit; for λ = 2, (0, 1) doesn’t fit since y must be 0. Therefore, the statement about (1, 0) for λ = 2 is correct.

Think in terms of solving (A − λI)v = 0 to find eigenvectors for each eigenvalue λ. For this matrix, the eigenvalues are the diagonal entries 2 and 3 since det(A − λI) = (2 − λ)(3 − λ).

For λ = 2, A − 2I = [[0, 1], [0, 1]]. The equations are 0·x + 1·y = 0, which gives y = 0, while x is free. So eigenvectors are nonzero multiples of (1, 0). That means (1, 0) is indeed an eigenvector corresponding to λ = 2.

If you check the other options, you’ll see they don’t satisfy (A − λI)v = 0 for their claimed λ: for λ = 3, eigenvectors must satisfy y = x, so (1, 0) or (0, 1) don’t fit; for λ = 2, (0, 1) doesn’t fit since y must be 0. Therefore, the statement about (1, 0) for λ = 2 is correct.

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