Orthogonal Polynomials Let $\{P_n\}_{n=0}^{\infty}$ a family of polynomials in $[a,b]$ where $n$ is the degree of $P_n$. Suppose they are orthogonal with respect to a positive wight function $\rho$. Show that $P_n$ has $n$ zeros in $[a,b]$.
From the data we have that for any $n,m$, it is true that:
$$\displaystyle \int_{a}^{b} P_n(x)P_m(x)\rho (x) = 0$$
How should I proceed for the desired result?
 A: Presumably $P_0$ is a nonzero constant and $a<b$. You may first prove that $P_n(x)$ is orthogonal to all arbitrary polynomials with lower degrees. Now, as $\int_a^b P_n(x)1\rho(x)=0$, $P_n$ must have a zero $c\in[a,b]$. Let $P_n(x)=(x-c)q_{n-1}(x)$. Since $P_n$ is orthogonal to all polynomials with lower degree, we have, in particular, $\int_a^b P_n(x)(x-c)\rho(x)=0$, i.e. $\int_a^b q_{n-1}(x)(x-c)^2\rho(x)=0$. So $q_{n-1}$ must have a zero in $[a,b]$ and you may proceed recursively.
A: First of all, wlog. we can assume that the constant polynomial is $P_0(x)=1$.
Intuitively it can go like this:
We need $\langle P_1,P_0\rangle=0$, so $P_1$ cannot be only positive (or only negative) all over $[a,b]$.
For $P_2$, consider both $\langle P_2,P_0\rangle=0$ and $\langle P_2,P_1\rangle=0$. The first one tells you (just as before) that $P_2$ must also change sign at least once, and if you spell out the other one, regarding possible signs, and let's say $P_1$'s root is $x_1$, it will lead that $P_2$ must change sign in both $(a,x_1)$ and $(x_1,b)$. 
I hope it gets the picture..
