Frobenius covariant

In matrix theory, the Frobenius covariants of a square matrix A are special polynomials of it, namely projection matrices F_i(A) associated with the eigenvalues and eigenvectors of A.^{[1]: pp.403, 437–8} 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.

Let A be a diagonalizable matrix with eigenvalues λ₁, ..., λ_k.

The Frobenius covariant F_i(A), for i = 1,..., k, is the matrix

${\displaystyle F_{i}(A)\equiv \prod _{j=1 \atop j\neq i}^{k}{\frac {1}{\lambda _{i}-\lambda _{j}}}(A-\lambda _{j}I)~.}$

It is essentially the Lagrange polynomial with matrix argument. If the eigenvalue λ_i is simple, then as an idempotent projection matrix to a one-dimensional subspace, F_i(A) has a unit trace.

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. The matrix S is defined up to multiplication on the right by a diagonal matrix. If A has no multiple eigenvalues, then let c_i be the ith right eigenvector of A, that is, the ith column of S; and let r_i be the ith left eigenvector of A, namely the ith row of S⁻¹. Then F_i(A) = c_i r_i. As a projection matrix, the Frobenius covariant satisfies the relation

${\displaystyle F_{i}(A)F_{j}(A)=\delta _{ij}F_{i}(A),}$

which leads to r_i c_j = δ_ij.

Given that v and w are the right and left vectors satisfying w F_i(A) v ≠ 0, the right and left eigenvectors of A may be written as c_i = N_ci F_i(A) v and r_i = N_ri w F_i(A). 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 S.

If A has an eigenvalue λ_i appearing multiple times, then F_i(A) = Σ_j c_j r_j, where the sum is over all rows and columns associated with the eigenvalue λ_i.^[1]: p.521

Consider the two-by-two matrix:

${\displaystyle A={\begin{bmatrix}1&3\\4&2\end{bmatrix}}.}$

This matrix has two eigenvalues, 5 and −2, which can be found by solving the characteristic equation. By virtue of the Cayley–Hamilton theorem, (A − 5)(A + 2) = 0.

The corresponding eigen decomposition is

${\displaystyle A={\begin{bmatrix}3&1/7\\4&-1/7\end{bmatrix}}{\begin{bmatrix}5&0\\0&-2\end{bmatrix}}{\begin{bmatrix}3&1/7\\4&-1/7\end{bmatrix}}^{-1}={\begin{bmatrix}3&1/7\\4&-1/7\end{bmatrix}}{\begin{bmatrix}5&0\\0&-2\end{bmatrix}}{\begin{bmatrix}1/7&1/7\\4&-3\end{bmatrix}}.}$

Hence the Frobenius covariants, manifestly projections, are

${\displaystyle {\begin{array}{rl}F_{1}(A)&=c_{1}r_{1}={\begin{bmatrix}3\\4\end{bmatrix}}{\begin{bmatrix}1/7&1/7\end{bmatrix}}={\begin{bmatrix}3/7&3/7\\4/7&4/7\end{bmatrix}}=F_{1}^{2}(A)\\F_{2}(A)&=c_{2}r_{2}={\begin{bmatrix}1/7\\-1/7\end{bmatrix}}{\begin{bmatrix}4&-3\end{bmatrix}}={\begin{bmatrix}4/7&-3/7\\-4/7&3/7\end{bmatrix}}=F_{2}^{2}(A)~,\end{array}}}$

with

${\displaystyle F_{1}(A)F_{2}(A)=0,\qquad F_{1}(A)+F_{2}(A)=I~.}$

Note tr F₁(A) = tr F₂ (A)= 1, as required.