The notion of orthonormal system

Inner Product Spaces: Orthonormal systems

The notion of orthonormal system

Of particular importance to the study of inner product spaces are vectors that are mutually perpendicular.

Orthogonal and orthonormal systems

Let $\vec{v}_1,\ldots ,\vec{v}_n$ be a set of vectors of an inner product space $V$ .

The system is called orthogonal if for $1\leq i, j\leq n$ with $i\neq j$ we have
$\dotprod{\vec{v}_i}{\vec{v}_j}=0$
The system is called orthonormal if for $1\leq i, j\leq n$ we have $\dotprod{\vec{v}_i}{\vec{v}_j}=\left\{\,\begin{array}{l} 0\ \text{if}\ i\neq j\\ 1\ \text{if}\ i=j\ \end{array}\right.$

If, in addition, the system $\vec{v}_1,\ldots ,\vec{v}_n$ is a basis for $V$ , we speak of an orthonormal basis of $V$ .

The system
$\frac{1}{\sqrt{2}}\rv{1,-1,0}, \, \frac{1}{\sqrt{2}}\rv{1,1,0},\,\rv{0,0,1}$ in $\mathbb{R}^3$ is an orthonormal system with respect to the standard inproduct. It is not difficult to verify that the system is independent. Consequently, the three vectors form an orthonormal basis of $\mathbb{R}^3$ .

An example is the standard basis $\vec{e}_1,\ldots ,\vec{e}_n$ for $\mathbb{R}^n$ . Using such bases makes projections, coordinates, and distances easy to calculate.

An orthonormal system is an orthogonal system in which all vectors have length $1$ .

A function that takes two integers $i$ and $j$ as arguments and assigns to this pair $0$ if $i\neq j$ and $1$ if $i=j$ is called a Kronecker delta function. It is denoted by $\delta_{ij}$ . The inner product of vectors from the orthonormal system $\basis{\vec{v}_1,\ldots,\vec{v}_n}$ can thus be written as

$\dotprod{\vec{v}_i}{\vec{v}_j}=\delta_{ij}$

Consider the inner product space of polynomial functions of degree at most $1$ (that is, linear functions) on the interval $\ivcc{0}{6}$ with inner product of functions given by $\dotprod{f}{g} = \int_{0}^{6} f(x)\cdot {g(x)}\,\dd x$ .

Determine an orthogonal basis for the space consisting of two functions $f$ and $g$ of degree $1$ , that is, of the form $f(x)=a+b\cdot x\quad\text{ and }\quad g(x)=c+d\cdot x$
for suitable numbers $a$ , $b$ , $c$ and $d$ with $b\ne0$ and $d\ne 0$ .

Give your answer in the form of a list.

$\rv{f(x),g(x)}=$ $\rv{1+x,1-{{4}\over{15}} x}$
Other answers are possible.

To achieve orthogonality of $f$ and $g$ we need $\dotprod{f}{g}=0$ . Writing out this inner product gives $\begin{array}{rcl} \dotprod{f}{g}&=&\displaystyle \int_{0}^{6}f(x)\cdot g(x) \, \dd x \\ &&\phantom{xx}\color{blue}{\text{definition of inner product of functions}}\\ &=&\displaystyle \int_{0}^{6} (a+bx)\cdot(c+dx)\, \dd x \\ &&\phantom{xx}\color{blue}{\text{functions rules of }f(x)\text{ and }g(x)\text{ substituted}}\\ &=&\displaystyle \int_{0}^{6}\left( ac + (ad+bc)x+bdx^2\right)\, \dd x \\ &&\phantom{xx}\color{blue}{\text{expanded}}\\ &=&\displaystyle \left[acx+\frac{ad+bc}{2}x^2+\frac{bd}{3}x^3 \right]_{0}^{6}\\ &&\phantom{xx}\color{blue}{\text{antiderivative taken}}\\ &=&\displaystyle 6 a c+18 \left(a d+b c\right)+72 b d\\ &&\phantom{xx}\color{blue}{\text{boundary values used}} \end{array}$ If we equate this to $0$ , we get an equation with $4$ unknowns. For convenience we take $a=1$ , $b=1$ , $c=1$ , and substitute these values into the equation. We then infer what the value of $d$ should be for these values of $a$ , $b$ , $c$ .
$\begin{array}{rcl} \displaystyle 6 a c+18 \left(a d+b c\right)+72 b d&=&\displaystyle 0\\ &&\phantom{xx}\color{blue}{\text{equation set up}}\\ \displaystyle6+18\cdot \left(d+1\right)+72\cdot d&=&\displaystyle 0\\ &&\phantom{xx}\color{blue}{\text{values for }a,b\text{, and }c\text{ substituted}}\\ \displaystyle 90\cdot d+24 &=&\displaystyle 0\\ &&\phantom{xx}\color{blue}{\text{left-hand side simplified}}\\ d&=&\displaystyle-{{4}\over{15}}\\ &&\phantom{xx}\color{blue}{\text{solved}} \end{array}$ Substituting the values found for $a$ , $b$ , $c$ , $d$ , we find the polynomials $f(x) = 1+x$ and $g(x) =1-{{4}\over{15}} x$ .

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