$(x-\lambda)$ and $(x-\overline{\lambda})$ appear the same number of times in the factorization of $p$ Linear Algebra Done Right 3rd Edition Axler

Question

I am reading "Linear Algebra Done Right 3rd Edition" by Sheldon Axler.

4.15 Polynomials with real coefficients have zeros in pairs
Suppose $p\in\mathcal{P}(\mathbb{C})$ is a polynomial with real coefficients. If $\lambda\in\mathbb{C}$ is a zero of $p$, then so is $\overline{\lambda}$.

Suppose $\lambda$ is a nonreal complex number and $(x-\lambda)$ is a term in the factorization of $p$ as an element of $\mathcal{P}(\mathbb{C})$. We are guaranteed by 4.15 that $(x-\overline{\lambda})$ also appears as a term in the factorization, but 4.15 does not state that these two factors appear the same number of times, $\dots$

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Fact A:
$(x-\lambda)$ and $(x-\overline{\lambda})$ appear the same number of times in the factorization of $p$.

Let $p$ be a nonconstant real polynomial and $p(\lambda)=0$ for $\lambda\in\mathbb{C}-\mathbb{R}$.
By 4.15, we can write $p(x)=(x-\lambda)(x-\overline{\lambda})q(x)$, where $q(x)$ is a complex polynomial.
If we can prove $q(x)$ is a real polynomial, then by induction, it is obvious that $(x-\lambda)$ and $(x-\overline{\lambda})$ appear the same number of times in the factorization of $p$.

My proof:
Let $g(x):=(x-\lambda)(x-\overline{\lambda})$.
Then, $g(x)$ is a real polynomial.
By a famous algorithm, we can write $p(x)=q(x)g(x)+ax+b$, where $q(x)$ is a real polynomial and $a,b\in\mathbb{R}$.
$0=p(\lambda)=q(\lambda)g(\lambda)+a\lambda+b=a\lambda+b$.
$0=p(\overline{\lambda})=q(\overline{\lambda})g(\overline{\lambda})+a\overline{\lambda}+b=a\overline{\lambda}+b$.
If $a\neq 0$, then the equation $az+b=0$ has two different solutions.
So, $a=0$.
So, $b=0$.
So, $p(x)=q(x)g(x)$ where $q(x)$ is a real polynomial.

My 2nd Proof:
Let $g(x):=(x-\lambda)(x-\overline{\lambda})$.
Then, $g(x)$ is a real polynomial.
We can write $p(x) = q(x)g(x)$.
Assume that $q(x)=a_nx^n+\dots+a_1x+a_0$ is not a real polynomial.
Let $k:=\min\{i\in\{0,1,\dots\}\mid a_i\in\mathbb{C}-\mathbb{R}\}$.
Let $q(x)g(x)=b_{n+2}x^{n+2}+\dots+b_1x+b_0$.
Then, $b_k\in\mathbb{C}-\mathbb{R}$.
So, $q(x)g(x)$ is not a real polynomial.
Since $p(x)$ is a real polynomial, this is a contradiction.

The author's Proof:
To prove that $q$ has real coefficients, we solve the equation above for $q$, getting $$q(x)=\frac{p(x)}{x^2-2(\operatorname{Re}\lambda)x+|\lambda|^2}$$ for all $x\in\mathbb{R}$. The equation above implies that $q(x)\in\mathbb{R}$ for all $x\in\mathbb{R}$. Writing $$q(x)=a_0+a_1x+\dots+a_{n-2}x^{n-2},$$ where $n=\deg p$ and $a_0,\dots,a_{n-2}\in\mathbb{C}$, we thus have $$0=\operatorname{Im}q(x)=(\operatorname{Im}a_0)+(\operatorname{Im}a_1)x+\dots+(\operatorname{Im}a_{n-2})x^{n-2}$$ for all $x\in\mathbb{R}$. This implies that $\operatorname{Im}a_0,\dots,\operatorname{Im}a_{n-2}$ all equal $0$ (by 4.7). Thus all the coefficients of $q$ are real, as desired.

Intuitively, I expect Fact A holds by symmetry.
Fact A is not obvious?
Do we really need to prove Fact A like the above proofs?
Is there any intuitive explanation which explains Fact A holds obviouly?

Answer 1

I would use the following argument.

If $\lambda$ is a root of $p$ then it exists $q \in \mathbb C[x]$ such that

$$p(x)=(x-\lambda)q(x)$$ Conjugating this equality you get

$$\overline{p}(x) = p(x) = (x- \overline{\lambda})\overline{q}(x)$$ proving that $x- \overline{\lambda}$ also divides $p$. As $x-\lambda$ and $x- \overline{\lambda}$ are coprime in $\mathbb C[x]$ if $\lambda \notin \mathbb R$, you get the desired result that $(x - \lambda)(x- \overline{\lambda})$ divides $p$.

Then proving that $(x - \lambda)$ and $(x- \overline{\lambda})$ appear the same number of time in the factorization of $p$ is immediate by induction on the degree of $p$.

Answer 2

You can express $\lambda $ in polar coordinates, i.e. $\lambda = r e^{i \theta }, \;r, \theta \in \mathbb{R}$. Now if you have some polynomial of the form \begin{align*} p( z) = \sum_{k = 0}^{n} a_{k}z^{k} \end{align*} with $p( \lambda ) = 0$, then \begin{align*} \overline{\lambda } = \overline{r e^{i \theta }} = r e^{-i \theta } \implies p( \overline{\lambda }) = \sum_{k = 0}^{n} a_{k}e^{-i\theta k} = e^{-1} \sum_{k = 0}^{n} a_{k}e^{i\theta k} = e^{-1} p( \lambda ) = 0 .\end{align*}

Answer 3

Let $p(z)=a_0+a_1z+\dots+a_nz^n$.
Let $\overline{p}(z)=\overline{a_0}+\overline{a_1}z+\dots+\overline{a_n}z^n$.
Then,
$p(z)=0.$
$\Leftrightarrow$
$\overline{p}(\overline{z})=0.$
Let $r(z)=p(z)q(z)$.
Then,
$\overline{r}(z)=\overline{p}(z)\overline{q}(z)$.

$p(z)=(z-\lambda)^n q(z)$ and $q(\lambda)\neq 0.$
$\Leftrightarrow$
$\overline{p}(z)=(z-\overline{\lambda})^n\overline{q}(z)$ and $\overline{q}(\overline{\lambda})\neq 0$.

Since $p(z)$ is a real polynomial, $p(z)=\overline{p}(z)$.

So,
$p(z)=(z-\lambda)^n q(z)$ and $q(\lambda)\neq 0.$
$\Leftrightarrow$
$p(z)=(z-\overline{\lambda})^n\overline{q}(z)$ and $\overline{q}(\overline{\lambda})\neq 0$.