Return to Answer

When $F$ is a companion matrix a polynomial $f(x)=x^n+a_1x^{n-1}+\cdots+a_{n-1}x+a_n$, let $$ B = \begin{bmatrix} a_{n-1} & a_{n-2} & \cdots & a_1 & 1 \\ a_{n-2} & a_{n-3} & \cdots & 1 & 0 \\ \vdots & \ \vdots & & \vdots & \vdots \\ a_{1} & 1 & \cdots & 0 & 0 \\ 1 & 0 & \cdots & 0 & 0 \end{bmatrix} \qquad \mathrm{and} \qquad F = \begin{bmatrix} 0 & 0 & \cdots & 0& -a_{n}\\ 1 & 0 & \cdots & 0 & -a_{n-1}\\ \vdots & \ \vdots & & \vdots & \vdots \\ 0 & 0 & \cdots & 0 & -a_{2} \\ 0 & 0 & \cdots & 1 & -a_1 \end{bmatrix}.$$ We have $B^{-1}FB=F^T$. Then a companion matrix $F$ and its transpose $F^T$ are similar.

Here is an answer in a very particular but possibly useful case: if $A$ is the companion matrix of the polynomial $f(x)=x^n+a_1x^{n-1}+\cdots+a_{n-1}x+a_n$, i.e. $$A = \begin{bmatrix} 0 & 0 & \cdots & 0& -a_{n}\\ 1 & 0 & \cdots & 0 & -a_{n-1}\\ \vdots & \ \vdots & & \vdots & \vdots \\ 0 & 0 & \cdots & 0 & -a_{2} \\ 0 & 0 & \cdots & 1 & -a_1 \end{bmatrix}\ ,$$ if one defines $$ P = \begin{bmatrix} a_{n-1} & a_{n-2} & \cdots & a_1 & 1 \\ a_{n-2} & a_{n-3} & \cdots & 1 & 0 \\ \vdots & \ \vdots & & \vdots & \vdots \\ a_{1} & 1 & \cdots & 0 & 0 \\ 1 & 0 & \cdots & 0 & 0 \end{bmatrix}$$ then $P^{-1}AP=A^T$, i.e. the companion matrix $A$ and its transpose $A^T$ are similar.

When $F$ is a companion matrix, let $$ B = \begin{bmatrix} a_{n-1} & a_{n-2} & \cdots & a_1 & 1 \\ a_{n-2} & a_{n-3} & \cdots & 1 & 0 \\ \vdots & \ \vdots & & \vdots & \vdots \\ a_{1} & 1 & \cdots & 0 & 0 \\ 1 & 0 & \cdots & 0 & 0 \end{bmatrix} \qquad \mathrm{and} \qquad F = \begin{bmatrix} 0 & 0 & \cdots & 0& -a_{n}\\ 1 & 0 & \cdots & 0 & -a_{n-1}\\ \vdots & \ \vdots & & \vdots & \vdots \\ 0 & 0 & \cdots & 0 & -a_{2} \\ 0 & 0 & \cdots & 1 & -a_1 \end{bmatrix}.$$ We have $B^{-1}FB=F^T$. Then a companion matrix $F$ and its transpose $F^T$ are similar.

When $F$ is a companion matrix a polynomial $f(x)=x^n+a_1x^{n-1}+\cdots+a_{n-1}x+a_n$, let $$ B = \begin{bmatrix} a_{n-1} & a_{n-2} & \cdots & a_1 & 1 \\ a_{n-2} & a_{n-3} & \cdots & 1 & 0 \\ \vdots & \ \vdots & & \vdots & \vdots \\ a_{1} & 1 & \cdots & 0 & 0 \\ 1 & 0 & \cdots & 0 & 0 \end{bmatrix} \qquad \mathrm{and} \qquad F = \begin{bmatrix} 0 & 0 & \cdots & 0& -a_{n}\\ 1 & 0 & \cdots & 0 & -a_{n-1}\\ \vdots & \ \vdots & & \vdots & \vdots \\ 0 & 0 & \cdots & 0 & -a_{2} \\ 0 & 0 & \cdots & 1 & -a_1 \end{bmatrix}.$$ We have $B^{-1}FB=F^T$. Then a companion matrix $F$ and its transpose $F^T$ are similar.

When $F$ is a companion matrix,let $$ B = \begin{bmatrix} a_{n-1} & a_{n-2} & \cdots & a_1 & 1 \\ a_{n-2} & a_{n-3} & \cdots & 1 & 0 \\ \vdots & \ \vdots & & \vdots & \vdots \\ a_{1} & 1 & \cdots & 0 & 0 \\ 1 & 0 & \cdots & 0 & 0 \end{bmatrix} \qquad \mathrm{and} \qquad F = \begin{bmatrix} 0 & 0 & \cdots & 0& -a_{n}\\ 1 & 0 & \cdots & 0 & -a_{n-1}\\ \vdots & \ \vdots & & \vdots & \vdots \\ 0 & 0 & \cdots & 0 & -a_{2} \\ 0 & 0 & \cdots & 1 & -a_1 \end{bmatrix}.$$ We have $B^{-1}FB=F^T$. Then a companion matrix $F$ and its transpose $F^T$ are similar.

When $F$ is a companion matrix, let $$ B = \begin{bmatrix} a_{n-1} & a_{n-2} & \cdots & a_1 & 1 \\ a_{n-2} & a_{n-3} & \cdots & 1 & 0 \\ \vdots & \ \vdots & & \vdots & \vdots \\ a_{1} & 1 & \cdots & 0 & 0 \\ 1 & 0 & \cdots & 0 & 0 \end{bmatrix} \qquad \mathrm{and} \qquad F = \begin{bmatrix} 0 & 0 & \cdots & 0& -a_{n}\\ 1 & 0 & \cdots & 0 & -a_{n-1}\\ \vdots & \ \vdots & & \vdots & \vdots \\ 0 & 0 & \cdots & 0 & -a_{2} \\ 0 & 0 & \cdots & 1 & -a_1 \end{bmatrix}.$$ We have $B^{-1}FB=F^T$. Then a companion matrix $F$ and its transpose $F^T$ are similar.