Eigenvalues and eigenvectors are typically associated with matrices, not quaternions. But there is a standard injection of quaternions into 2x2 complex matrices: The quaternion Q=A+Bj, for a pair A,B of complex numbers, can be represented by the 2x2 complex matrix: [ A B ] Q = [-B' A'] If A=ar+i*ai and B=br+i*bi, then the matrix for our quaternion Q is: [ ar+i*ai br+i*bi] Q = [-br+i*bi ar-i*ai] Q's eigenvalues are lambda = ar +- i*sqrt(ai^2+br^2+bi^2) = sQ +- i*|VQ|, where sQ=ar, VQ=(ai,br,bi). We seek an invertible matrix X, s.t. [sQ+i*|VQ| ] (1/X)QX = [ 0 sQ-i*|VQ|] = L Here, the matrix X is the matrix of eigenvectors; i.e., the columns of X are eigenvectors for the matrix Q. But *because the eigenvalues are complex conjugates*, the diagonal matrix of eigenvalues L can also represent the quaternion L =(sQ+|VQ|*i)+0j. We have thus converted our *matrix* eigenvalue problem into a *quaternion* problem: We seek a quaternion X, s.t. (1/X)QX = L = (sQ+i*|VQ|) + 0j But we already know the solution to the equation (1/X)QX = L, due to a previous posting! If (1/X)QX = L => QX = XL then the following X is a solution: X = 1-(VQ/|VQ|)(VL/|VL|) = 1-(VQ/|VQ|)(VL/|VQ|) = 1-(VQ/|VQ|)(i*|VQ|/|VQ|) = 1-(VQ/|VQ|)*i = 1-(VQ*i)/|VQ| Now we can clear fractions for X: X = |VQ| - VQ*i QX = XL Q(|VQ| - VQ*i) = (|VQ| - VQ*i) (sQ + |VQ|*i) (sQ+VQ) (|VQ| - VQ*i) = (|VQ| - VQ*i) (sQ+|VQ|*i) sQ*|VQ| - sQ*VQ*i +VQ*|VQ|- VQ*VQ*i = (|VQ| - VQ*i) (sQ+|VQ|*i) sQ*|VQ| - sQ*VQ*i + |VQ|*VQ + |VQ|^2*i = (|VQ| - VQ*i) (sQ+|VQ|*i) = |VQ|*sQ + |VQ||VQ|*i - (VQ*i)*sQ - (VQ i)*|VQ|*i = sQ*|VQ| + |VQ|^2*i - sQ*VQ*i - |VQ|*VQ*i*i = sQ*|VQ| + |VQ|^2*i - sQ*VQ*i + |VQ|*VQ = sQ*|VQ| - sQ*VQ*i + |VQ|*VQ + |VQ|^2*i QED Note that X = |VQ| - VQ*i is a full quaternion whose scalar and vector parts are: sX = |VQ|+(VQ.i) VX = - (VQ x i) = i x VQ where VQ.i is the classical vector dot product, and i x VQ is the classical vector cross product for the 'i' vector and the 'VQ' vector. We thus have an elegant one-step solution for the eigenvalue/eigenvector problem in quaternions: L contains both eigenvalues and X contains both eigenvectors. No polynomials, no determinants, no Gaussian eliminations. We do have one scalar square root needed to calculate |VQ|, but no other square roots are necessary unless we wish to normalize X. A normalized X/|X| provides the special unitary matrix that diagonalizes Q, should that be necessary.