### 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#.

Consider the inner product space of polynomial functions of degree at most #1# (that is, linear functions) on the interval #\ivcc{0}{2}# with

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.

*inner product of functions*given by \(\dotprod{f}{g} = \int_{0}^{2} f(x)\cdot {g(x)}\,\dd x \).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.

Solution #\rv{f(x),g(x)}=# #\rv{1+x,1-{{6}\over{7}} 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}^{2}f(x)\cdot g(x) \, \dd x \\

&&\phantom{xx}\color{blue}{\text{definition of inner product of functions}}\\

&=&\displaystyle \int_{0}^{2} (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}^{2}\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}^{2}\\

&&\phantom{xx}\color{blue}{\text{antiderivative taken}}\\

&=&\displaystyle 2 a c+2 \left(a d+b c\right)+{{8 b d}\over{3}}\\

&&\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 2 a c+2 \left(a d+b c\right)+{{8 b d}\over{3}}&=&\displaystyle 0\\

&&\phantom{xx}\color{blue}{\text{equation set up}}\\

\displaystyle2+2\cdot \left(d+1\right)+{{8\cdot d}\over{3}}&=&\displaystyle 0\\

&&\phantom{xx}\color{blue}{\text{values for }a,b\text{, and }c\text{ substituted}}\\

\displaystyle {{14\cdot d}\over{3}}+4 &=&\displaystyle 0\\

&&\phantom{xx}\color{blue}{\text{left-hand side simplified}}\\

d&=&\displaystyle-{{6}\over{7}}\\

&&\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-{{6}\over{7}} 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}^{2}f(x)\cdot g(x) \, \dd x \\

&&\phantom{xx}\color{blue}{\text{definition of inner product of functions}}\\

&=&\displaystyle \int_{0}^{2} (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}^{2}\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}^{2}\\

&&\phantom{xx}\color{blue}{\text{antiderivative taken}}\\

&=&\displaystyle 2 a c+2 \left(a d+b c\right)+{{8 b d}\over{3}}\\

&&\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 2 a c+2 \left(a d+b c\right)+{{8 b d}\over{3}}&=&\displaystyle 0\\

&&\phantom{xx}\color{blue}{\text{equation set up}}\\

\displaystyle2+2\cdot \left(d+1\right)+{{8\cdot d}\over{3}}&=&\displaystyle 0\\

&&\phantom{xx}\color{blue}{\text{values for }a,b\text{, and }c\text{ substituted}}\\

\displaystyle {{14\cdot d}\over{3}}+4 &=&\displaystyle 0\\

&&\phantom{xx}\color{blue}{\text{left-hand side simplified}}\\

d&=&\displaystyle-{{6}\over{7}}\\

&&\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-{{6}\over{7}} x#.

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