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}{7}# with inner product of functions given by \(\dotprod{f}{g} = \int_{0}^{7} f(x)\cdot {g(x)}\,\dd x \).
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.
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-{{27}\over{119}} x}#
Other answers are possible.
To achieve orthogonality of #f# and #g# we need #\dotprod{f}{g}=0#. Writing out this inner product gives \[\begin{array}{rcl}
\dotprod{f}{g}&=&\displaystyle \int_{0}^{7}f(x)\cdot g(x) \, \dd x \\
&&\phantom{xx}\color{blue}{\text{definition of inner product of functions}}\\
&=&\displaystyle \int_{0}^{7} (a+bx)\cdot(c+dx)\, \dd x \\
&&\phantom{xx}\color{blue}{\text{functions rules of }f(x)\text{ and }g(x)\text{ substituted}}\\
&=&\displaystyle \int_{0}^{7}\left( ac + (ad+bc)x+bdx^2\right)\, \dd x \\
&&\phantom{xx}\color{blue}{\text{expanded}}\\
&=&\displaystyle \left[acx+\frac{ad+bc}{2}x^2+\frac{bd}{3}x^3 \right]_{0}^{7}\\
&&\phantom{xx}\color{blue}{\text{antiderivative taken}}\\
&=&\displaystyle 7 a c+{{49 \left(a d+b c\right)}\over{2}}+{{343 b d}\over{3}}\\
&&\phantom{xx}\color{blue}{\text{boundary values used}}
\end{array}\] 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#.
\[\begin{array}{rcl}
\displaystyle 7 a c+{{49 \left(a d+b c\right)}\over{2}}+{{343 b d}\over{3}}&=&\displaystyle 0\\
&&\phantom{xx}\color{blue}{\text{equation set up}}\\
\displaystyle7+{{49\cdot \left(d+1\right)}\over{2}}+{{343\cdot d}\over{3}}&=&\displaystyle 0\\
&&\phantom{xx}\color{blue}{\text{values for }a,b\text{, and }c\text{ substituted}}\\
\displaystyle {{833\cdot d}\over{6}}+{{63}\over{2}} &=&\displaystyle 0\\
&&\phantom{xx}\color{blue}{\text{left-hand side simplified}}\\
d&=&\displaystyle-{{27}\over{119}}\\
&&\phantom{xx}\color{blue}{\text{solved}}
\end{array}\] Substituting the values found for #a#, #b#, #c#, #d#, we find the polynomials #f(x) = 1+x# and #g(x) =1-{{27}\over{119}} x#.
Other answers are possible.
To achieve orthogonality of #f# and #g# we need #\dotprod{f}{g}=0#. Writing out this inner product gives \[\begin{array}{rcl}
\dotprod{f}{g}&=&\displaystyle \int_{0}^{7}f(x)\cdot g(x) \, \dd x \\
&&\phantom{xx}\color{blue}{\text{definition of inner product of functions}}\\
&=&\displaystyle \int_{0}^{7} (a+bx)\cdot(c+dx)\, \dd x \\
&&\phantom{xx}\color{blue}{\text{functions rules of }f(x)\text{ and }g(x)\text{ substituted}}\\
&=&\displaystyle \int_{0}^{7}\left( ac + (ad+bc)x+bdx^2\right)\, \dd x \\
&&\phantom{xx}\color{blue}{\text{expanded}}\\
&=&\displaystyle \left[acx+\frac{ad+bc}{2}x^2+\frac{bd}{3}x^3 \right]_{0}^{7}\\
&&\phantom{xx}\color{blue}{\text{antiderivative taken}}\\
&=&\displaystyle 7 a c+{{49 \left(a d+b c\right)}\over{2}}+{{343 b d}\over{3}}\\
&&\phantom{xx}\color{blue}{\text{boundary values used}}
\end{array}\] 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#.
\[\begin{array}{rcl}
\displaystyle 7 a c+{{49 \left(a d+b c\right)}\over{2}}+{{343 b d}\over{3}}&=&\displaystyle 0\\
&&\phantom{xx}\color{blue}{\text{equation set up}}\\
\displaystyle7+{{49\cdot \left(d+1\right)}\over{2}}+{{343\cdot d}\over{3}}&=&\displaystyle 0\\
&&\phantom{xx}\color{blue}{\text{values for }a,b\text{, and }c\text{ substituted}}\\
\displaystyle {{833\cdot d}\over{6}}+{{63}\over{2}} &=&\displaystyle 0\\
&&\phantom{xx}\color{blue}{\text{left-hand side simplified}}\\
d&=&\displaystyle-{{27}\over{119}}\\
&&\phantom{xx}\color{blue}{\text{solved}}
\end{array}\] Substituting the values found for #a#, #b#, #c#, #d#, we find the polynomials #f(x) = 1+x# and #g(x) =1-{{27}\over{119}} x#.
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