Frobenius covariant

In matrix theory, the Frobenius covariants of a square matrix Template:Mvar are special polynomials of it, namely projection matrices A_i associated with the eigenvalues and eigenvectors of Template:Mvar.Template:Rp They are named after the mathematician Ferdinand Frobenius.

Each covariant is a projection on the eigenspace associated with the eigenvalue Template:Math. Frobenius covariants are the coefficients of Sylvester's formula, which expresses a function of a matrix Template:Math as a matrix polynomial, namely a linear combination of that function's values on the eigenvalues of Template:Mvar.

Formal definition

Let Template:Mvar be a diagonalizable matrix with eigenvalues λ₁, ..., λ_k.

The Frobenius covariant Template:Math, 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, Template:Math 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 Template:Mvar can be obtained from any eigendecomposition Template:Math, where Template:Mvar is non-singular and Template:Mvar is diagonal with Template:Math. If Template:Mvar has no multiple eigenvalues, then let c_i be the Template:Mvarth right eigenvector of Template:Mvar, that is, the Template:Mvarth column of Template:Mvar; and let r_i be the Template:Mvarth left eigenvector of Template:Mvar, namely the Template:Mvarth row of Template:Mvar⁻¹. Then Template:Math.

If Template:Mvar has an eigenvalue λ_i appearing multiple times, then Template:Math, where the sum is over all rows and columns associated with the eigenvalue λ_i.Template:Rp

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 Template:Math.

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

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

with

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

Note Template:Math, as required.