Frobenius covariant

Let be a diagonalizable matrix with eigenvalues λ1, ..., λk. The Frobenius covariant , for i = 1,..., k, is the matrix : F_i (A) \equiv \prod_{j=1 \atop j \ne i}^k \frac{1}{\lambda_i-\lambda_j} (A - \lambda_j I)~. It is essentially the Lagrange polynomial with matrix argument. If the eigenvalue λi is simple, then as an idempotent projection matrix to a one-dimensional subspace, has a unit trace. ==Computing the covariants==

(1849–1917), German mathematician. His main interests were elliptic functions, differential equations, and later group theory. The Frobenius covariants of a matrix can be obtained from any eigendecomposition , where is non-singular and is diagonal with . The matrix is defined up to multiplication on the right by a diagonal matrix. If has no multiple eigenvalues, then let ci be the th right eigenvector of , that is, the th column of ; and let ri be the th left eigenvector of , namely the th row of −1. Then . As a projection matrix, the Frobenius covariant satisfies the relation : F_i (A)F_j (A) = \delta_{ij}F_i (A), which leads to Given that and are the right and left vectors satisfying , the right and left eigenvectors of may be written as and . The orthonormality of the eigenvectors gives one constraint for the normalization coefficients. The remaining freedom is related to the choice of representation for the matrix . If has an eigenvalue λi appearing multiple times, then , where the sum is over all rows and columns associated with the eigenvalue λi. ==Example==

Consider the two-by-two matrix: : A = \begin{bmatrix} 1 & 3 \\ 4 & 2 \end{bmatrix}. This matrix has two eigenvalues, 5 and −2, which can be found by solving the characteristic equation. By virtue of the Cayley–Hamilton theorem, . The corresponding eigen decomposition is : A = \begin{bmatrix} 3 & 1/7 \\ 4 & -1/7 \end{bmatrix} \begin{bmatrix} 5 & 0 \\ 0 & -2 \end{bmatrix} \begin{bmatrix} 3 & 1/7 \\ 4 & -1/7 \end{bmatrix}^{-1} = \begin{bmatrix} 3 & 1/7 \\ 4 & -1/7 \end{bmatrix} \begin{bmatrix} 5 & 0 \\ 0 & -2 \end{bmatrix} \begin{bmatrix} 1/7 & 1/7 \\ 4 & -3 \end{bmatrix}. Hence the Frobenius covariants, manifestly projections, are : \begin{array}{rl} F_1(A) &= c_1 r_1 = \begin{bmatrix} 3 \\ 4 \end{bmatrix} \begin{bmatrix} 1/7 & 1/7 \end{bmatrix} = \begin{bmatrix} 3/7 & 3/7 \\ 4/7 & 4/7 \end{bmatrix} = F_1^2(A)\\ F_2(A) &= c_2 r_2 = \begin{bmatrix} 1/7 \\ -1/7 \end{bmatrix} \begin{bmatrix} 4 & -3 \end{bmatrix} = \begin{bmatrix} 4/7 & -3/7 \\ -4/7 & 3/7 \end{bmatrix}=F_2^2(A) ~, \end{array} with :F_1(A) F_2(A) = 0 , \qquad F_1(A) + F_2(A) = I ~. Note , as required. ==References==