Frobenius covariant

In matrix theory, the Frobenius covariants of a square matrix A are special polynomials of it, namely projection matrices A_i 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, ..., λ_k.

The Frobenius covariant A_i, for i = 1,..., k, is the matrix

$A_i\equiv \prod _{\genfrac{}{}{0ex}{}{j=1}{j\ne i}}^k\frac{1}{{\lambda }_i-{\lambda }_j}(A-{\lambda }_{j}I).$

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

Computing the covariants

Ferdinand Georg Frobenius (1849–1917), German mathematician. His main interests were elliptic functions differential equations, and later group theory.

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 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 A_i = c_i r_i.

If A has an eigenvalue λ_i 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 λ_i.

Example

Consider the two-by-two matrix:

$A=\left[\begin{array}{cc}1& 3\\ 4& 2\end{array}\right].$

This matrix has two eigenvalues, 5 and −2; hence (A − 5)(A + 2) = 0.

The corresponding eigen decomposition is

$A=\left[\begin{array}{cc}3& 1/7\\ 4& -1/7\end{array}\right]\left[\begin{array}{cc}5& 0\\ 0& -2\end{array}\right]{\left[\begin{array}{cc}3& 1/7\\ 4& -1/7\end{array}\right]}^{-1}=\left[\begin{array}{cc}3& 1/7\\ 4& -1/7\end{array}\right]\left[\begin{array}{cc}5& 0\\ 0& -2\end{array}\right]\left[\begin{array}{cc}1/7& 1/7\\ 4& -3\end{array}\right].$

Hence the Frobenius covariants, manifestly projections, are

$\begin{array}{rl}{A}_1& =c_1{r}_1=\left[\begin{array}{c}3\\ 4\end{array}\right]\left[\begin{array}{cc}1/7& 1/7\end{array}\right]=\left[\begin{array}{cc}3/7& 3/7\\ 4/7& 4/7\end{array}\right]=A_1^2\\ A_2& =c_2{r}_2=\left[\begin{array}{c}1/7\\ -1/7\end{array}\right]\left[\begin{array}{cc}4& -3\end{array}\right]=\left[\begin{array}{cc}4/7& -3/7\\ -4/7& 3/7\end{array}\right]=A_2^2,\end{array}$

with

$A_1{A}_2=0,A_1+A_2=I.$

Note tr A₁ = tr A₂ = 1, as required.