For , extend the proposed vector by . The sine addition identity gives . Thus row-by-row multiplication yields
These vectors do not depend on . To prove they form a basis even when eigenvalues of coincide, apply the calculation first to the symmetric matrix with , whose eigenvalues are distinct. Its nonzero eigenvectors are orthogonal and form a basis. Their squared lengths are , so multiplying each by gives a common orthonormal eigenbasis for every matrix in the family.

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