12Eigenvector for eigenvalue 3
has eigenvalue . Find an eigenvector for it, scaled so its first entry is .
Solve (A − 3I)x = 0: subtracting 3 from the diagonal gives [[-1,1],[1,-1]], and both rows say y = x. Scaling so the first entry is 1 gives the eigenvector (1, 1). Check: A(1,1)ᵗ = (3,3)ᵗ = 3(1,1)ᵗ ✓.