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

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.

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 $D i, i = λ i$ . 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 $i$ th right eigenvector of $A$ , that is, the $i$ th column of $S$ ; and let r_i be the $i$ th left eigenvector of $A$ , namely the $i$ th row of $S$ ⁻¹. Then $F i (A) = c i r i$ . As a projection matrix, the Frobenius covariant satisfies the relation

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 $w F i (A) v \neq 0$ , 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

Example

Consider the two-by-two matrix:

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

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}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

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

Note $tr F 1 (A) = tr F 2 (A)= 1$ , as required.

Frobenius covariant

Computing the covariants

Example

References

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