Then prove that the characteristic polynomial $\det(xI-A)$ of $A$ is the polynomial $p(x)$.
The matrix is called the companion matrix of the polynomial $p(x)$.

Use the mathematical induction. The base case is clear.
For the induction step, use the cofactor expansion and apply the induction hypothesis.

Proof.

We prove that $p(x)=\det(xI-A)$ by induction on $n$.
The base case $n=1$ is clear since $A=[-a_0]$ is a $1 \times 1$ matrix and $\det(xI-A)=\det[x+a_0]=x+a_0$.

Induction step is as follows. Suppose that we have $p(t)=\det(xI-A)$ is true for a degree $n-1$ polynomial $p(t)$ and its companion matrix $A$.

We prove the statement for a degree $n$ polynomial.
Use the cofactor expansion corresponding to the first row, we obtain
\begin{align*}
\det(xI-A)&=\begin{vmatrix}
x & 0 & \dots & 0 &a_0 \\
-1 & x & \dots & 0 & a_1 \\
0 & -1 & \dots & 0 & a_2 \\
\vdots & & \ddots & & \vdots \\
0 & 0 & \dots & -1 & x+a_{n-1}
\end{vmatrix}\\[8pt]
&=x \begin{vmatrix}
x & 0 & \dots & 0 &a_1 \\
-1 & x & \dots & 0 & a_2 \\
\vdots & & \ddots & & \vdots \\
0 & 0 & \dots & 1 & x+a_{n-1}
\end{vmatrix}
+(-1)^{n+1}a_0
\begin{vmatrix}
-1 & x & \dots & 0 \\
0 & -1 & \dots & 0 \\
\vdots & & \ddots & \vdots \\
0 & 0 & \dots & -1
\end{vmatrix}.
\end{align*}

Now by the induction hypothesis, the first determinant is
\[x^{n-1}+a_{n-1}x^{n-2}+\cdots+a_2 x+a_1.\]
The second determinant is $(-1)^{n-1}$ since it is an $(n-1)\times (n-1)$ triangular matrix, determinant is the product of diagonal entries.

Therefore we have
\begin{align*}
\det(xI-A)&=x(x^{n-1}+a_{n-1}x^{n-2}+\cdots+a_2x+a_1)+(-1)^{n+1}a_0(-1)^{n-1}\\
&=x^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0=p(x).
\end{align*}
Thus the statement is true for a degree $n$ polynomial. By induction, we complete the proof.

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