# What is the origin of the companion matrix?

I understand that every polynomial of degree $$k$$ has a companion matrix with a characteristic polynomial of the form $$(-1)^kp(\lambda)$$. I also understand the proof of this fact by induction.

My question is: what was the inspiration behind the idea of building a matrix of this form and study its properties?

$$\begin{vmatrix} -a_{k-1} - \lambda & -a_{k-2} & \dots & -a_{1} & -a_{0} \\ 1 & -\lambda & \dots & 0 & 0 \\ 0 & 1 & \ddots & 0 & 0 \\ 0 & 0 & \dots & -\lambda & 0 \\ 0 & 0 & \dots & 1 & -\lambda \\ \end{vmatrix}$$

• It's designed as a matrix with your favourite polynomial as its minimal polynomial. – Angina Seng Jun 15 '20 at 19:39
• If you transpose and then move the first column to the last column, the case with $\lambda = 0$ becomes the matrix representation of multiplication by $x$ on the vector space $F[x] / \langle p \rangle$ with respect to the basis $(1, x, x^2, \ldots, x^{k-1})$. And the latter linear transformation is somewhat of a natural way to construct a linear transformation which is annihilated by $p$. – Daniel Schepler Jun 15 '20 at 19:43
• The origin is probably in someone's brain, attempting to solve the problem: How do I construct a matrix with a given characteristi polynomial? – Lee Mosher Jun 16 '20 at 0:10

The canonical matrices as representatives of similarity classes of matrices were pretty much understood in the 1870's.

In 1870, Jordan gives his canonical form over $$F_p$$.

In 1878, Frobenius gives his one, over any field $$K$$. It is based on the notion of cyclic subspaces and vectors.

Let $$A\in M_n(K)$$ and $$v\in K^n\setminus\{0\}$$. If $$\{v,Av,\cdots, A^{n-1}v\}$$ is a basis of $$K^n$$, then, in this basis, $$A$$ becomes the matrix $$C_p$$, where $$p(x)=x^n-\sum_{i=1}^{n} C_p[i,n]x^{i-1}$$.

Note that Krylov will take up this idea in 1931.

My favourite polynomial (cf. the @Angina Seng ' post) is $$q(x)=x^5-x-1$$; then, over $$\mathbb{Q},\mathbb{R}$$ or $$\mathbb{C}$$,

$$C_q=\begin{pmatrix}0&0&0&0&1\\1&0&0&0&1\\0&1&0&0&0\\0&0&1&0&0\\0&0&0&1&0\end{pmatrix}$$. Here, $$C_q$$ is a cyclic matrix and $$e_1$$ is a cyclic vector.

Frobenius showed that every matrix has a canonical form as $$diag(C_{p_1},\cdots,C_{p_k})$$ where $$p_i\in K[x]$$ and $$p_1|p_2|\cdots|p_k$$, $$p_k$$ is the minimal polynomial of $$A$$ over $$K$$ and $$p_1\cdots p_k$$ its characteristic polyn.

By construction, the char. pol. of $$C_p$$ is $$p$$. Then (exercise), the set of $$A\in M_n(\mathbb{Q})$$ s.t. $$A^5-A-I=0_n$$ are the matrices in the form $$Q^{-1}diag(C_q,\cdots,C_q)Q$$, where $$Q\in GL_n(\mathbb{Q})$$ and $$n$$ is a multiple of $$5$$.

Frobenius proved also the Cayley Hamilton theorem; indeed, Cayley showed only the cases $$n=2,3$$ and Hamilton the case $$n=4$$ (quaternions when you hold us...).

In fact, C.H. is a direct consequence of the Jordan decomposition; this was not very well understood at the time because Jordan, very busy with the study of Galois theory, remained confined to finite fields.

$$\textbf{Remarks.}$$ For the Frobenius form

Advantage: we obtain the invariant polynomials $$p_i$$ and we stay in the field $$K$$.

Disadvantage: we cannot calculate a closed form for $$A^k$$. Yet, using the remain of the division of $$x^{1000}$$ by $$q(x)$$, (in $$0"015$$)we obtain $$C_q^{1000}=$$

For the Jordan form

Advantage: the calculation of $$A^n$$ is easy (a priori).

Disadvantage (a hilarious one) if $$A\in M_n(\mathbb{Q})$$ with $$n\geq 5$$, we don't know how to calculate its Jordan form -because, in general (for example when $$A=C_q$$), the eigenvalues ​​of A cannot be written using radicals-