# Inner Product Spaces and Minimum Polynomials

Problem

Let $V$ be the vector space of real polynomials $\mathbb{R}[x]$ endowed with the inner product

$$(f,g)=\int_{-\infty}^{\infty}e^{-|x|}f(x)g(x)dx$$

By considering the series of subspaces $\{V_n\}$ where $V_n=\{f(x) \in \mathbb{R}[X]: \deg(f) \leq n\}$ or otherwise, show that there exist unique monic polynomials $\varphi_n(x)$ for $n \geq 0$ such that

$$\int_{-\infty}^{\infty}e^{-|x|}\varphi_n(x)g(x)dx=0$$ whenever deg $g<n$, and find $\varphi_n(x)$ for $n=0,1,2$.

Progress

We look to identify $(V_n)^{\perp}$ for each $n$, i.e.

$$(V_n)^{\perp}=\{f(x) \in \mathbb{R}[X]: \langle f,g \rangle=0 \text{ for all } g \in V_n\}$$

If we can show this has dimension $1$, then we can show the existence of a unique monic polynomial. Any thoughts on how this is done (or an alternative method for the problem) would be greatly appreciated. Regards.

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I think there's a condition missing for unicity, namely $\deg \varphi_n = n$. Stated otherwise $V_n^{\perp}$ is not one-dimensional (it's infinite-dimensional), but $V_n^{\perp} \cap V_{n+1}$ is. – Joel Cohen Dec 23 '11 at 15:40
Offsetting indices does not change the fact that $V_n^{\perp}$ is not finite-dimensional (let alone one-dimensional). In fact, if $V$ is any finite-dimensional subspace of $\mathbb{R}[X]$, then $V^{\perp}$ is infinite dimensional. For example, you can check that $V_0^{\perp}$ contains $X^{2n+1}$ for all $n \le 0$. – Joel Cohen Dec 23 '11 at 15:59
For existence I'd use Gram-Schmidt on $\{1,x,x^2,x^3,\ldots\}$ first, then normalize appropriately to ensure that the polynomial becomes monic. Uniqueness will become clear afterwards. – t.b. Dec 23 '11 at 16:03
@TJO : I think you made progress in the right direction : your interpretation in terms of orthogonal complement is the right one, and the fact that you have to prove some space is one-dimensional to prove unicity is also spot on. The only problem is that you're missing a condition to conclude (because $V_n^{\perp}$ is actually infinite dimensional). If you require $\deg \varphi_n = n$, then you'll be able to proceed with your idea. – Joel Cohen Dec 23 '11 at 16:13
@TJO : Yes, the Gram-Schmidt process yields the solution which satisfies $\deg \varphi_n = n$ (because from the Gram-Schmidt process, we have $\textrm{Span}(\varphi_0, \ldots, \varphi_n) = \textrm{Span}(X^0, \ldots, X^{n}) = V_{n}$ so $\deg \varphi_n \le n$, but from the orthagonality condition, $\deg \varphi_n \ge n$). – Joel Cohen Dec 23 '11 at 18:42

This problem should remind you of Gram-Schmidt, but if you are interested in showing only the existence (and uniqueness) of the $\varphi_n$, and not interested in the final formula, then one can stop short of all the calculations. That said, I recommend that you are not as lazy as I am; the calculations might end up illuminating. :)

All one needs to understand is the abstract operation of orthogonal projection. Consider a vector space $V$ and a finite dimensional subspace $E \subseteq V$. Then the orthogonal projection of $v$ onto $E$ is the unique vector $w \in E$ such that $v - w \perp E$. It is a standard fact that such a projection always exists and is unique; let us denote this $\operatorname{Proj}(v, E)$.

If you know this fact (or take it for granted), then this problem is easy: $$\varphi_n(x) = x^n - \operatorname{Proj}(x^n, V_{n-1})$$ is the unique monic polynomial of degree $n$ such that $(\varphi_n, g) = 0$ for all $g \in V_{n-1}$. (Note that the projection, being in $V_{n-1}$, is a polynomial of degree $n-1$.) Now, it is an altogether a different matter how to evaluate the projection term. It does not come out to be a nice expression, and moreover, it's not at all clear to me that it is intended here.

To complete the picture, let us also see why the projection as defined always exists and is unique.

Uniqueness. Suppose there exist $w_1, w_2 \in E$ such that $v - w_1 \perp E$ and $v - w_2 \perp E$. Then $w_1 - w_2 = (v - w_2) - (v- w_1)$ is also orthogonal to $E$. In particular, since $w_1 - w_2$ itself is in $E$, it is orthogonal to itself; that is, $w_1 - w_2 = 0$. Thus the projection -- when it exists -- is unique.

Existence. Since $E$ is a finite dimensional real* inner product vector space, it possesses an orthonormal basis. Let $u_1, u_2, \ldots, u_r$ be such an orthonormal basis, where $r = \dim E$. Then the projection of a given vector can simply be written down as: $$w = \operatorname{Proj}(v, E) := \sum_{i=1}^r (v, u_i) u_i.$$ It is easy to see that $v-w$ is orthogonal to each basis vector $u_i$, and hence is orthogonal to the whole subspace $E$.

Of course, if, in a concrete setting, we want to find the projection, we should compute an orthonormal basis for $E$. This is usually done using the Gram-Schmidt process, as was alluded to in the comments.

*I assume real base field; the complex case is not too different.

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