3.4: Inner Product Spaces
We have discussed the inner product (or dot product) between two column matrices. Recall that the inner product between, say, two three-by-one column matrices
\[\text{u}=\left(\begin{array}{c}u_1\\u_2\\u_3\end{array}\right),\quad\text{v}=\left(\begin{array}{c}v_1\\v_2\\v_3\end{array}\right)\nonumber \]
is given by
\[\text{u}^{\text{T}}\text{v}=u_1v_1+u_2v_2+u_3v_3.\nonumber \]
We now generalize the inner product so that it is applicable to any vector space, including those containing functions.
We will denote the inner product between any two vectors \(\text{u}\) and \(\text{v}\) as \((\text{u}, \text{v})\), and require the inner product to satisfy the same arithmetic rules that are satisfied by the dot product. With \(\text{u}\), \(\text{v}\), \(\text{w}\) vectors and \(c\) a scalar, these rules can be written as
\[(\text{u},\text{v})=(\text{v},\text{u}),\quad (\text{u}+\text{v},\text{w})=(\text{u},\text{w})+(\text{v},\text{w}),\quad (\text{cu},\text{v})=\text{c}(\text{u},\text{v})=(\text{u},\text{cv});\nonumber \]
and \((\text{u},\text{u})\geq 0\), where the equality holds if and only if \(\text{u}=0\).
Generalizing our definitions for column matrices, the norm of a vector \(\text{u}\) is defined as
\[||\text{u}||=(\text{u},\text{u})^{1/2}.\nonumber \]
A unit vector is a vector whose norm is one. Unit vectors are said to be normalized to unity , though sometimes we just say that they are normalized . We say two vectors are orthogonal if their inner product is zero. We also say that a basis is orthonormal (as in an orthonormal basis) if all the vectors are mutually orthogonal and are normalized to unity. For an orthonormal basis consisting of the vectors \(v_1,\: v_2,\cdots , v_n\), we write
\[(v_i,v_j)=\delta_{ij},\nonumber \]
where \(\delta_{ij}\) is called the Kronecker delta, defined as
\[\delta_{ij}=\left\{\begin{array}{ll}1,&\text{if }i=j; \\ 0,&\text{if }i\neq j.\end{array}\right.\nonumber \]
Oftentimes, basis vectors are used that are orthogonal but are normalized to other values besides unity.
Define an inner product for \(\mathbb{P}_n\).
Solution
Let \(p(x)\) and \(q(x)\) be two polynomials in \(\mathbb{P}_n\). One possible definition of an inner product is given by
\[(p,q)=\int_{-1}^1 p(x)q(x)dx.\nonumber \]
You can check that all the conditions of an inner product are satisfied.
Show that the first four Legendre polynomials form an orthogonal basis for \(\mathbb{P}_3\) using the inner product defined above.
Solution
The first four Legendre polynomials are given by
\[P_0(x)=1,\quad P_1(x)=x,\quad P_2(x)=\frac{1}{2}(3x^2-1),\quad P_3(x)=\frac{1}{2}(5x^3-3x),\nonumber \]
and these four polynomials form a basis for \(\mathbb{P}_3\). With an inner product defined on \(\mathbb{P}_n\) as
\[(p,q)=\int_{-1}^1p(x)q(x)dx,\nonumber \]
it can be shown by explicit integration that
\[(P_m,P_n)=\frac{2}{2n+1}\delta_{m,n},\nonumber \]
so that the first four Legendre polynomials are mutually orthogonal. They are normalized so that \(P_n(1) = 1\).
Define an inner product on \(\mathbb{P}_n\) such that the Hermite polynomials are orthogonal.
Solution
For instance, the first four Hermite polynomials are given by
\[H_0(x)=1,\quad H_1(x)=2x,\quad H_2(x)=4x^2-2,\quad H_3(x)=8x^3-12x,\nonumber \]
which also form a basis for \(\mathbb{P}_3\). Here, define an inner product on \(\mathbb{P}_n\) as
\[(p,q)=\int_{-\infty}^\infty p(x)q(x)e^{-x^2}dx.\nonumber \]
It can be shown that
\[(H_m,H_n)=2^n\pi^{1/2}n!\delta_{m,n},\nonumber \]
so that the Hermite polynomials are orthogonal with this definition of the inner product. These Hermite polynomials are normalized so that the leading coefficient of \(H_n\) is given by \(2^n\).