For A = [[4,1],[0,2]], which vector is an eigenvector corresponding to λ = 4?

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Multiple Choice

For A = [[4,1],[0,2]], which vector is an eigenvector corresponding to λ = 4?

To find eigenvectors for a given eigenvalue, use Av = λv, which means (A − λI)v = 0. Here, with λ = 4, subtract 4 from the diagonal: A − 4I = [[0, 1], [0, −2]]. Solving [[0, 1], [0, −2]] [x; y] = [0; 0] gives y = 0 from the first row, and the second row also enforces y = 0. The variable x is free, so the eigenvectors are all nonzero multiples of (1, 0). The simplest representative is (1, 0), and indeed A*(1, 0) = (4, 0) = 4*(1, 0). That’s why the vector with y = 0 and x ≠ 0 works. The other vectors don’t satisfy Av = 4v: for (0, 1), A*(0, 1) = (1, 2) which is not a multiple of (0, 1); for (1, −1), A*(1, −1) = (3, −2) which is not 4*(1, −1). Note that any nonzero scalar multiple of an eigenvector is also an eigenvector, so (2, 0) would also be valid, illustrating the eigenvector direction along the x-axis.

To find eigenvectors for a given eigenvalue, use Av = λv, which means (A − λI)v = 0. Here, with λ = 4, subtract 4 from the diagonal: A − 4I = [[0, 1], [0, −2]]. Solving [[0, 1], [0, −2]] [x; y] = [0; 0] gives y = 0 from the first row, and the second row also enforces y = 0. The variable x is free, so the eigenvectors are all nonzero multiples of (1, 0). The simplest representative is (1, 0), and indeed A*(1, 0) = (4, 0) = 4*(1, 0).

That’s why the vector with y = 0 and x ≠ 0 works. The other vectors don’t satisfy Av = 4v: for (0, 1), A*(0, 1) = (1, 2) which is not a multiple of (0, 1); for (1, −1), A*(1, −1) = (3, −2) which is not 4*(1, −1). Note that any nonzero scalar multiple of an eigenvector is also an eigenvector, so (2, 0) would also be valid, illustrating the eigenvector direction along the x-axis.

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