Companion matrix explained

In linear algebra, the Frobenius companion matrix of the monic polynomialp(x)=c_0 + c_1 x + \cdots + c_x^ + x^nis the square matrix defined as

C(p)=\begin0 & 0 & \dots & 0 & -c_0 \\1 & 0 & \dots & 0 & -c_1 \\0 & 1 & \dots & 0 & -c_2 \\\vdots & \vdots & \ddots & \vdots & \vdots \\0 & 0 & \dots & 1 & -c_\end.

Some authors use the transpose of this matrix,

C(p)T

, which is more convenient for some purposes such as linear recurrence relations (see below).

C(p)

is defined from the coefficients of

p(x)

, while the characteristic polynomial as well as the minimal polynomial of

C(p)

are equal to

p(x)

.[1] In this sense, the matrix

C(p)

and the polynomial

p(x)

are "companions".

Similarity to companion matrix

Any matrix with entries in a field has characteristic polynomial

p(x)=\det(xI-A)

, which in turn has companion matrix

C(p)

. These matrices are related as follows.

The following statements are equivalent:

C(p)

, i.e. A can be conjugated to its companion matrix by matrices in GLn(F);

p(x)

coincides with the minimal polynomial of A, i.e. the minimal polynomial has degree n;

A:Fn\toFn

makes

Fn

a cyclic

F[A]

-module, having a basis of the form

\{v,Av,\ldots,An-1v\}

; or equivalently

Fn\congF[X]/(p(x))

as

F[A]

-modules.

If the above hold, one says that A is non-derogatory.

Not every square matrix is similar to a companion matrix, but every square matrix is similar to a block 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 by A, and this gives the rational canonical form of A.

Diagonalizability

The roots of the characteristic polynomial

p(x)

are the eigenvalues of

C(p)

. If there are n distinct eigenvalues

λ1,\ldots,λn

, then

C(p)

is diagonalizable as

C(p)=V-1DV

, where D is the diagonal matrix and V is the Vandermonde matrix corresponding to the 's: D = \begin\lambda_1 & 0 & \!\!\!\cdots\!\!\! & 0\\0 & \lambda_2 & \!\!\!\cdots\!\!\! & 0\\0 & 0 & \!\!\!\cdots\!\!\! & \lambda_n\end, \qquad V = \begin1 & \lambda_1 & \lambda_1^2 & \!\!\!\cdots\!\!\! & \lambda_1^n\\1 & \lambda_2 & \lambda_2^2 & \!\!\!\cdots\!\!\! & \lambda_2^n\\[-1em]\vdots & \vdots & \vdots & \!\!\!\ddots\!\!\! &\vdots \\1 & \lambda_n & \lambda_n^2 & \!\!\!\cdots\!\!\! & \lambda_n^n\end.Indeed, an easy computation shows that the transpose

C(p)T

has eigenvectors

vi=(1,λi,\ldots,λ

n-1
i

)

with
T(v
C(p)
i)

=λivi

, which follows from

p(λi)=c0+ci+ … +cn-1

n-1
λ
i

in=0

. Thus, its diagonalizing change of basis matrix is

VT=

T
[v
1

\ldots

T]
v
n
, meaning

C(p)T=VTD(VT)-1

, and taking the transpose of both sides gives

C(p)=V-1DV

.

We can read the eigenvectors of

C(p)

with

C(p)(wi)=λiwi

from the equation

C(p)=V-1DV

: they are the column vectors of the inverse Vandermonde matrix

V-1=

T
[w
1

T]
w
n
. This matrix is known explicitly, giving the eignevectors

wi=(L0i,\ldots,L(n-1)i)

, with coordinates equal to the coefficients of the Lagrange polynomialsL_i(x) = L_ + L_x + \cdots + L_x^ = \prod_ \frac = \frac. Alternatively, the scaled eigenvectors

\tildewi=p'(λi)wi

have simpler coefficients.

If

p(x)

has multiple roots, then

C(p)

is not diagonalizable. Rather, the Jordan canonical form of

C(p)

contains one diagonal block for each distinct root, an m × m block with

λ

on the diagonal if the root

λ

has multiplicity m.

Linear recursive sequences

A linear recursive sequence defined by

ak+n=-c0ak-c1ak+1-cn-1ak+n-1

for

k\geq0

has the characteristic polynomial

p(x)=c0+c1x++cn-1xn-1+xn

, whose transpose companion matrix

C(p)T

generates the sequence:\begina_\\a_\\\vdots \\a_\\a_\end=\begin0 & 1 & 0 & \cdots & 0\\0 & 0 & 1 & \cdots & 0\\\vdots & \vdots & \vdots & \ddots & \vdots \\0 & 0 & 0 & \cdots & 1\\-c_0 & -c_1 & -c_2 & \cdots & -c_\end\begina_k\\a_\\\vdots \\a_\\a_\end.The vector

v=(1,λ,λ2,\ldots,λn-1)

is an eigenvector of this matrix, where the eigenvalue

λ

is a root of

p(x)

. Setting the initial values of the sequence equal to this vector produces a geometric sequence

ak=λk

which satisfies the recurrence. In the case of n distinct eigenvalues, an arbitrary solution

ak

can be written as a linear combination of such geometric solutions, and the eigenvalues of largest complex norm give an asymptotic approximation.

From linear ODE to first-order linear ODE system

Similarly to the above case of linear recursions, consider a homogeneous linear ODE of order n for the scalar function

y=y(t)

:y^ + c_y^ + \dots + c_y^ + c_0 y = 0 .This can be equivalently described as a coupled system of homogeneous linear ODE of order 1 for the vector function

z(t)=(y(t),y'(t),\ldots,y(n-1)(t))

:z' = C(p)^T zwhere

C(p)T

is the transpose companion matrix for the characteristic polynomialp(x)=x^n + c_x^ + \cdots + c_1 x + c_0 .Here the coefficients

ci=ci(t)

may be also functions, not just constants.

If

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 equationy^ + c_y^ + \dots + c_y^ + c_0 y = f(t)is equivalent to the system:z' = C(p)^T z + F(t)with the inhomogeneity term

F(t)=(0,\ldots,0,f(t))

.

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

Cyclic shift matrix

In the case of

p(x)=xn-1

, when the eigenvalues are the complex roots of unity, the companion matrix and its transpose both reduce to Sylvester's cyclic shift matrix, a circulant matrix.

Multiplication map on a simple field extension

Consider a polynomial

p(x)=xn+cn-1xn-1++c1x+c0

with coefficients in a field

F

, and suppose

p(x)

is irreducible in the polynomial ring

F[x]

. Then adjoining a root

λ

of

p(x)

produces a field extension

K=F(λ)\congF[x]/(p(x))

, which is also a vector space over

F

with standard basis

\{1,λ,λ2,\ldots,λn-1\}

. Then the

F

-linear multiplication mappinghas an n × n matrix

[mλ]

with respect to the standard basis. Since
i)
m
λ(λ

=λi+1

and
n-1
m
λ(λ

)=λn=-c0- … -cn-1λn-1

, this is the companion matrix of

p(x)

:[m_\lambda] = C(p).Assuming this extension is separable (for example if

F

has characteristic zero or is a finite field),

p(x)

has distinct roots

λ1,\ldots,λn

with

λ1=λ

, so thatp(x)=(x-\lambda_1)\cdots (x-\lambda_n),and it has splitting field

L=F(λ1,\ldots,λn)

. Now

mλ

is not diagonalizable over

F

; rather, we must extend it to an

L

-linear map on

Ln\congLFK

, a vector space over

L

with standard basis

\{1{}1,1{}λ,1{}λ2,\ldots,1{}λn-1\}

, containing vectors

w=(\beta1,\ldots,\betan)=\beta1{}1+ … +\betan{}λn-1

. The extended mapping is defined by

mλ(\beta\alpha)=\beta(λ\alpha)

.

The matrix

[mλ]=C(p)

is unchanged, but as above, it can be diagonalized by matrices with entries in

L

: [m_\lambda]=C(p)= V^\! D V,for the diagonal matrix

D=\operatorname{diag}(λ1,\ldots,λn)

and the Vandermonde matrix V corresponding to

λ1,\ldots,λn\inL

. The explicit formula for the eigenvectors (the scaled column vectors of the inverse Vandermonde matrix

V-1

) can be written as:\tilde w_i = \beta_ 1+\beta_ \lambda +\cdots + \beta_ \lambda^= \prod_ (1\lambda - \lambda_j 1) where

\betaij\inL

are the coefficients of the scaled Lagrange polynomial\frac= \prod_ (x - \lambda_j) = \beta_ + \beta_ x + \cdots + \beta_ x^.

See also

Notes and References

  1. Book: Horn, Roger A. . Matrix Analysis . Charles R. Johnson . Cambridge University Press . 1985 . 0-521-30586-1 . Cambridge, UK . 146–147 . 2010-02-10.