Give a direct proof that if \(A\) is the companion matrix of the monic polynomial \(p\), then \(p\) is the characteristic polynomial for \(A\).
Solution
Write \[ p(x)=x^n+c_{n-1}x^{n-1}+\cdots+c_1x+c_0. \] With the companion-matrix convention of this section, \[ A= \begin{pmatrix} 0&0&\cdots&0&-c_0\\ 1&0&\cdots&0&-c_1\\ 0&1&\ddots&0&-c_2\\ \vdots&\ddots&\ddots&\vdots&\vdots\\ 0&\cdots&0&1&-c_{n-1} \end{pmatrix} \] [@HK71]. We prove directly that \(\det(xI-A)=p(x)\).
1. The determinant satisfies the companion recurrence.
One has \[ xI-A= \begin{pmatrix} x&0&\cdots&0&c_0\\ -1&x&\cdots&0&c_1\\ 0&-1&\ddots&0&c_2\\ \vdots&\ddots&\ddots&x&\vdots\\ 0&\cdots&0&-1&x+c_{n-1} \end{pmatrix}. \] Expand along the first row. The cofactor of the entry \(x\) is the same determinant of size \(n-1\) with coefficient list \(c_1,\ldots,c_{n-1}\). The minor of the entry \(c_0\) is upper triangular with all diagonal entries \(-1\), so its determinant is \((-1)^{n-1}\). Its cofactor sign is \((-1)^{1+n}\), and therefore the total sign multiplying \(c_0\) is \[ (-1)^{n+1}(-1)^{n-1}=1. \] Hence, if \(D_n\) denotes this determinant, \[ D_n=xD_{n-1}+c_0. \]
2. Induction gives the polynomial \(p\).
For \(n=1\), \[ D_1=x+c_0. \] Assume the formula for size \(n-1\). By step 1, \[ \begin{aligned} D_n &=x\bigl(x^{n-1}+c_{n-1}x^{n-2}+\cdots+c_2x+c_1\bigr)+c_0\\ &=x^n+c_{n-1}x^{n-1}+\cdots+c_1x+c_0\\ &=p(x). \end{aligned} \] Thus the characteristic polynomial of the companion matrix is exactly \[ \boxed{\chi_A=p}. \]