Inlinear algebra, theFrobeniuscompanion matrix of themonic polynomial$${\displaystyle p(x)=c_0+c_{1}x+\cdots +c_{n-1}{x}^{n-1}+x^n}$$is thesquare matrix defined as

Some authors use thetranspose of this matrix,${\displaystyle C(p{)}^T}$, which is more convenient for some purposes such as linearrecurrence relations (see below).

${\displaystyle C(p)}$ is defined from the coefficients of${\displaystyle p(x)}$, while thecharacteristic polynomial as well as theminimal polynomial of${\displaystyle C(p)}$ are equal to${\displaystyle p(x)}$.^[1] In this sense, the matrix${\displaystyle C(p)}$ and the polynomial${\displaystyle p(x)}$ are "companions".

Any matrixA with entries in afieldF has characteristic polynomial${\displaystyle p(x)=det(xI-A)}$, which in turn has companion matrix${\displaystyle C(p)}$. These matrices are related as follows.

The following statements are equivalent:

• A issimilar overF to${\displaystyle C(p)}$, i.e.A can be conjugated to its companion matrix by matrices in GL_n(F);
• the characteristic polynomial${\displaystyle p(x)}$ coincides with the minimal polynomial ofA , i.e. the minimal polynomial has degreen;
• the linear mapping${\displaystyle A:F^n\to F^n}$ makes${\displaystyle F^n}$ acyclic${\displaystyle F[A]}$-module, having a basis of the form${\displaystyle \{v,Av,\dots ,A^{n-1}v\}}$; or equivalently${\displaystyle F^n\cong F[X]/(p(x))}$ as${\displaystyle F[A]}$-modules.

If the above hold, one says thatA isnon-derogatory.

Not every square matrix is similar to a companion matrix, but every square matrix is similar to ablock diagonal matrix made of companion matrices. If we also demand that the polynomial of each diagonal block divides the next one, they are uniquely determined byA, and this gives therational canonical form ofA.

The roots of the characteristic polynomial${\displaystyle p(x)}$ are theeigenvalues of${\displaystyle C(p)}$. If there aren distinct eigenvalues${\displaystyle {\lambda }_1,\dots ,{\lambda }_n}$, then${\displaystyle C(p)}$ isdiagonalizable as${\displaystyle C(p)=V^{-1}DV}$, whereD is the diagonal matrix andV is theVandermonde matrix corresponding to theλ's:Indeed, a reasonably hard computation shows that the transpose${\displaystyle C(p{)}^T}$ has eigenvectors${\displaystyle v_i=(1,{\lambda }_i,\dots ,{\lambda }_i^{n-1})}$ with${\displaystyle C(p{)}^T(v_i)={\lambda }_i{v}_i}$, which follows from${\displaystyle p({\lambda }_i)=c_0+c_1{\lambda }_i+\cdots +c_{n-1}{\lambda }_i^{n-1}+{\lambda }_i^n=0}$. Thus, its diagonalizingchange of basis matrix is${\displaystyle V^T=[v_1^T\dots v_n^T]}$, meaning${\displaystyle C(p{)}^T=V^{T}D(V^T{)}^{-1}}$, and taking the transpose of both sides gives${\displaystyle C(p)=V^{-1}DV}$.

We can read the eigenvectors of${\displaystyle C(p)}$ with${\displaystyle C(p)(w_i)={\lambda }_i{w}_i}$ from the equation${\displaystyle C(p)=V^{-1}DV}$: they are the column vectors of theinverse Vandermonde matrix${\displaystyle V^{-1}=[w_1^T\cdots w_n^T]}$. This matrix is known explicitly, giving the eigenvectors${\displaystyle w_i=(L_{0i},\dots ,L_{(n-1)i})}$, with coordinates equal to the coefficients of theLagrange polynomials$${\displaystyle L_i(x)=L_{0i}+L_{1i}x+\cdots +L_{(n-1)i}{x}^{n-1}=\prod _{j\ne i}\frac{x-{\lambda }_j}{{\lambda }_j-{\lambda }_i}=\frac{p(x)}{(x-{\lambda }_i)p^{\prime }({\lambda }_i)}.}$$Alternatively, the scaled eigenvectors${\displaystyle {\tilde{w}}_i=p^{\prime }({\lambda }_i)w_i}$ have simpler coefficients.

If${\displaystyle p(x)}$ has multiple roots, then${\displaystyle C(p)}$ is not diagonalizable. Rather, theJordan canonical form of${\displaystyle C(p)}$ contains oneJordan block for each distinct root; if the multiplicity of the root ism, then the block is anm ×m matrix with${\displaystyle \lambda }$ on the diagonal and 1 in the entries just above the diagonal. in this case,V becomes aconfluent Vandermonde matrix.^[2]

Alinear recursive sequence defined by${\displaystyle a_{k+n}=-c_0{a}_k-c_1{a}_{k+1}\cdots -c_{n-1}{a}_{k+n-1}}$ for${\displaystyle k\ge 0}$ has the characteristic polynomial${\displaystyle p(x)=c_0+c_{1}x+\cdots +c_{n-1}{x}^{n-1}+x^n}$, whose transpose companion matrix${\displaystyle C(p{)}^T}$ generates the sequence:The vector${\displaystyle v=(1,\lambda ,{\lambda }^2,\dots ,{\lambda }^{n-1})}$ is an eigenvector of this matrix, where the eigenvalue${\displaystyle \lambda }$ is a root of${\displaystyle p(x)}$. Setting the initial values of the sequence equal to this vector produces a geometric sequence${\displaystyle a_k={\lambda }^k}$ which satisfies the recurrence. In the case ofn distinct eigenvalues, an arbitrary solution${\displaystyle a_k}$ can be written as a linear combination of such geometric solutions, and the eigenvalues of largest complex norm give anasymptotic approximation.

Similarly to the above case of linear recursions, consider a homogeneouslinear ODE of ordern for the scalar function${\displaystyle y=y(t)}$:$${\displaystyle y^{(n)}+c_{n-1}{y}^{(n-1)}+\cdots +c_1{y}^{(1)}+c_{0}y=0.}$$This can be equivalently described as a coupled system of homogeneous linear ODE of order 1 for the vector function${\displaystyle z(t)=(y(t),y^{\prime }(t),\dots ,y^{(n-1)}(t))}$:$${\displaystyle z^{\prime }=C(p{)}^{T}z}$$where${\displaystyle C(p{)}^T}$ is the transpose companion matrix for the characteristic polynomial$${\displaystyle p(x)=x^n+c_{n-1}{x}^{n-1}+\cdots +c_{1}x+c_0.}$$Here the coefficients${\displaystyle c_i=c_i(t)}$ may be also functions, not just constants.

If${\displaystyle C(p{)}^T}$ is diagonalizable, then a diagonalizing change of basis will transform this into a decoupled system equivalent to one scalar homogeneous first-order linear ODE in each coordinate.

An inhomogeneous equation$${\displaystyle y^{(n)}+c_{n-1}{y}^{(n-1)}+\cdots +c_1{y}^{(1)}+c_{0}y=f(t)}$$is equivalent to the system:$${\displaystyle z^{\prime }=C(p{)}^{T}z+F(t)}$$with the inhomogeneity term${\displaystyle F(t)=(0,\dots ,0,f(t))}$.

Again, a diagonalizing change of basis will transform this into a decoupled system of scalar inhomogeneous first-order linear ODEs.

In the case of${\displaystyle p(x)=x^n-1}$, when the eigenvalues are the complexroots of unity, the companion matrix and its transpose both reduce to Sylvester's cyclicshift matrix, acirculant matrix.

Consider a polynomial${\displaystyle p(x)=x^n+c_{n-1}{x}^{n-1}+\cdots +c_{1}x+c_0}$ with coefficients in afield${\displaystyle F}$, and suppose${\displaystyle p(x)}$ isirreducible in thepolynomial ring${\displaystyle F[x]}$. Thenadjoining a root${\displaystyle \lambda }$ of${\displaystyle p(x)}$ produces afield extension${\displaystyle K=F(\lambda )\cong F[x]/(p(x))}$, which is also a vector space over${\displaystyle F}$ with standard basis${\displaystyle \{1,\lambda ,{\lambda }^2,\dots ,{\lambda }^{n-1}\}}$. Then the${\displaystyle F}$-linear multiplication mapping

${\displaystyle m_{\lambda }:K\to K}$ defined by${\displaystyle m_{\lambda }(\alpha )=\lambda \alpha }$

has ann ×n matrix${\displaystyle [m_{\lambda }]}$ with respect to the standard basis. Since${\displaystyle m_{\lambda }({\lambda }^i)={\lambda }^{i+1}}$ and${\displaystyle m_{\lambda }({\lambda }^{n-1})={\lambda }^n=-c_0-\cdots -c_{n-1}{\lambda }^{n-1}}$, this is the companion matrix of${\displaystyle p(x)}$:$${\displaystyle [m_{\lambda }]=C(p).}$$Assuming this extension isseparable (for example if${\displaystyle F}$ hascharacteristic zero or is afinite field),${\displaystyle p(x)}$ has distinct roots${\displaystyle {\lambda }_1,\dots ,{\lambda }_n}$ with${\displaystyle {\lambda }_1=\lambda }$, so that$${\displaystyle p(x)=(x-{\lambda }_1)\cdots (x-{\lambda }_n),}$$and it hassplitting field${\displaystyle L=F({\lambda }_1,\dots ,{\lambda }_n)}$. Now${\displaystyle m_{\lambda }}$ is not diagonalizable over${\displaystyle F}$; rather, we mustextend it to an${\displaystyle L}$-linear map on${\displaystyle L^n\cong L{\otimes }_{F}K}$, a vector space over${\displaystyle L}$ with standard basis${\displaystyle \{1\otimes 1,1\otimes \lambda ,1\otimes {\lambda }^2,\dots ,1\otimes {\lambda }^{n-1}\}}$, containing vectors${\displaystyle w=({\beta }_1,\dots ,{\beta }_n)={\beta }_1\otimes 1+\cdots +{\beta }_n\otimes {\lambda }^{n-1}}$. The extended mapping is defined by${\displaystyle m_{\lambda }(\beta \otimes \alpha )=\beta \otimes (\lambda \alpha )}$.

The matrix${\displaystyle [m_{\lambda }]=C(p)}$ is unchanged, but as above, it can be diagonalized by matrices with entries in${\displaystyle L}$:$${\displaystyle [m_{\lambda }]=C(p)=V^{-1}DV,}$$for the diagonal matrix${\displaystyle D=\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_n)}$ and theVandermonde matrixV corresponding to${\displaystyle {\lambda }_1,\dots ,{\lambda }_n\in L}$. The explicit formula for the eigenvectors (the scaled column vectors of theinverse Vandermonde matrix${\displaystyle V^{-1}}$) can be written as:$${\displaystyle {\tilde{w}}_i={\beta }_{0i}\otimes 1+{\beta }_{1i}\otimes \lambda +\cdots +{\beta }_{(n-1)i}\otimes {\lambda }^{n-1}=\prod _{j\ne i}(1\otimes \lambda -{\lambda }_j\otimes 1)}$$where${\displaystyle {\beta }_{ij}\in L}$ are the coefficients of the scaled Lagrange polynomial$${\displaystyle \frac{p(x)}{x-{\lambda }_i}=\prod _{j\ne i}(x-{\lambda }_j)={\beta }_{0i}+{\beta }_{1i}x+\cdots +{\beta }_{(n-1)i}{x}^{n-1}.}$$

• Frobenius endomorphism
• Cayley–Hamilton theorem
• Krylov subspace

• Matrices (mathematics)
• Matrix theory
• Matrix normal forms

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