In Linear Algebra Done Right, Axler proves the following theorem:
Suppose $a_0,a_1,\dots,a_m \in F$. If $a_0+a_1 z+⋯+a_mz^m=0$ for every $z \in F$, then $a_0, \dots, a_m = 0$.
by contrapositive.
He starts his proof by letting $z$ (the input to the polynomial) equal
$$
\frac{|a_0| + |a_1| + \cdots +|a_{m-1}|}{|a_m|} + 1.
$$
He then goes on to show that $ |a_0 + a_1 z + \cdots +a_{m-1} z^{m-1}| < |a_m z^m| $ by stating two successive inequalities,
$$
\begin{eqnarray}
|a_0 + a_1 z + \cdots +a_{m-1} z^{m-1}| & \leq & (|a_0| + |a_1| + \cdots +|a_{m-1}|) z^{m-1} \\
& \lt & |a_m z^m|.
\end{eqnarray}
$$
While I understand why this proves the theorem, I don't understand:
- Why he chose $z$ equal to the above fraction as opposed to an
arbitrary element of $F$? Is its value being used by either of the
above two inequalities? It doesn't seem to be on the surface. - Why it's guaranteed that $ (|a_0| + |a_1| + \cdots +|a_{m-1}|)
z^{m-1} \lt |a_m z^m| $ given that some $ a_0, \dots, a_{m-1} $ may
be non-zero.
I know that this previously unanswered question asks similar questions but I think/hope I've provided more specific questions regarding the proof.
Best Answer
The value of $z$ was chosen in order to make both inequalities true. The next-to-last step, $$ |a_0 + a_1 z + \cdots +a_{m-1} z^{m-1}| \leq (|a_0| + |a_1| + \cdots +|a_{m-1}|) z^{m-1}, $$ is true for any value of $z\ge 1$ (it requires $z^{m-1}$ to be at least as large as the lower powers of $z$). However, the next, $$ (|a_0| + |a_1| + \cdots +|a_{m-1}|) z^{m-1} \lt |a_m z^m|, $$ may require $z$ to be larger. Notice that it fails if $z = 1$ and $|a_0| + |a_1| + \cdots +|a_{m-1}| \ge |a_m|$.
How large does $z$ need to be? Assuming $z\gt 0$, the desired inequality can be divided by $|a_m|z^{m-1}$ to get an equivalent statement: $$ \frac{|a_0| + |a_1| + \cdots +|a_{m-1}|}{|a_m|} \lt z. $$ To make sure that this is true, and also that $z\ge 1$ for the previous step, it will work to let $$ z = \frac{|a_0| + |a_1| + \cdots +|a_{m-1}|}{|a_m|} + 1. $$ (By the way, notice the absolute value in the denominator. You omitted that in your question, but it is needed.)