Frobenius covariant

In matrix theory, the Frobenius covariants of a square matrix $A$ are special polynomials of it, namely  projection  matrices Ai associated with the eigenvalues and eigenvectors of  $A$. They are named after the mathematician Ferdinand Frobenius.

Each covariant is a projection on the eigenspace associated with the eigenvalue $λ_{i}$. Frobenius covariants are the coefficients of Sylvester's formula, which expresses a function of a matrix $f(A)$ as a matrix polynomial, namely a linear combination of that function's values on the eigenvalues of $A$.

Formal definition
Let $A$ be a diagonalizable matrix with eigenvalues λ1, ..., λk.

The Frobenius covariant $A_{i}$, for i = 1,..., k, is the matrix
 * $$ A_i \equiv \prod_{j=1 \atop j \ne i}^k \frac{1}{\lambda_i-\lambda_j} (A - \lambda_j I)~. $$

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,  $A_{i}$ has a unit trace.

Computing the covariants


The Frobenius covariants of a matrix $A$ can be obtained from any eigendecomposition $A = SDS^{−1}$, where $S$ is non-singular and $D$ is diagonal with $D_{i,i} = λ_{i}$. If $A$ has no multiple eigenvalues, then let ci be the $i$th right eigenvector of $A$, that is, the $i$th column of $S$; and let ri be the $i$th left eigenvector of $A$, namely the $i$th row of $S$−1. Then $A_{i} = c_{i} r_{i}$.

If $A$ has an eigenvalue λi appearing multiple times, then $A_{i} = Σ_{j} c_{j} r_{j}$, 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; hence $(A − 5)(A + 2) = 0$.

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}

A_1 &= 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} = A_1^2\\ A_2 &= 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}=A_2^2 ~, \end{array} $$ with
 * $$A_1 A_2 = 0, \qquad A_1 + A_2 = I ~.$$

Note $trA_{1} = trA_{2} = 1$, as required.