# 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.:pp.403,437–8 They are named after the mathematician Ferdinand Frobenius.

Each covariant is a projection on the eigenspace associated with the eigenvalue λi. 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 Ai, for i = 1,, k, is the matrix

$A_{i}\equiv \prod _{j=1 \atop j\neq i}^{k}{\frac {1}{\lambda _{i}-\lambda _{j}}}(A-\lambda _{j}I)~.$ It is essentially the Lagrange polynomial with matrix argument. If the eigenvalue λi is simple, then as an idempotent projection matrix to a one-dimensional subspace, Ai 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 Di,i = λi. If A has no multiple eigenvalues, then let ci be the ith right eigenvector of A, that is, the ith column of S; and let ri be the ith left eigenvector of A, namely the ith row of S−1. Then Ai = ci ri.

If A has an eigenvalue λi appear multiple times, then Ai = Σj cj rj, where the sum is over all rows and columns associated with the eigenvalue λi.:p.521

## 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 trA1=trA2=1, as required.

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