91Ó°ÊÓ

To find the orthogonal projection matrix consider the solution.

According to the Exercise 67, the required matrix will be $A{\left(A^{T}A\right)}^{-1}{A}^T$.

Where the matrix $A=\left[{\stackrel{→}{v}}_1\right]$.

Consider the equations below to find the projection matrix.

$A^{T}A={||\stackrel{→}{v}||}^2=1+a^2+a^4+...+a^{2n-2}$

$$\begin{gathered} A{\left(A^{T}A\right)}^{-1}{A}^T=\stackrel{→}{v}\frac{1}{{||\stackrel{→}{v}||}^2}{\stackrel{→}{v}}^T \\ =\frac{1}{{||\stackrel{→}{v}||}^2}\stackrel{→}{v}{\stackrel{→}{v}}^T \\ =\frac{1}{{||\stackrel{→}{v}||}^2}\left[\begin{array}{c}1\\ a\\ a^2\\ .\\ .\\ a^{n-1}\end{array}\right]\left[\begin{array}{ccc}1& a& a^2...a^{n-1}\end{array}\right] \\ =\frac{1}{{||\stackrel{→}{v}||}^2}\left[\begin{array}{ccc}1& a& a^2...a^{n-1}\\ a& a^2& a^3...a^n\\ a^2& a^3& ........a^{n+1}\\ .& .& .........\\ .& .& ........\\ a^{n-1}& a^n& .....a^{2n-2}\end{array}\right] \end{gathered}$$

$p=\frac{1}{1+a^2+a^4+....+a^{2n-2}}\left[\begin{array}{ccc}1& a& a^2...a^{n-1}\\ a& a^2& a^3...a^n\\ a^2& a^3& ........a^{n+1}\\ .& .& .........\\ .& .& ........\\ a^{n-1}& a^n& .....a^{2n-2}\end{array}\right]$

Further solve the above expression,

role="math" localid="1660133190975" $$\begin{gathered} A{\left(A^{T}A\right)}^{-1}{A}^T=\stackrel{→}{v}\frac{1}{{||\stackrel{→}{v}||}^2}{\stackrel{→}{v}}^T \\ =\frac{1}{{||\stackrel{→}{v}||}^2}\stackrel{→}{v}{\stackrel{→}{v}}^T \\ =\frac{1}{{||\stackrel{→}{v}||}^2}\left[\begin{array}{c}1\\ a\\ a^2\\ .\\ .\\ a^{n-1}\end{array}\right]\left[\begin{array}{ccc}1& a& a^2...a^{n-1}\end{array}\right] \\ =\frac{1}{{||\stackrel{→}{v}||}^2}\left[\begin{array}{ccc}1& a& a^2...a^{n-1}\\ a& a^2& a^3...a^n\\ a^2& a^3& ........a^{n+1}\\ .& .& .........\\ .& .& ........\\ a^{n-1}& a^n& .....a^{2n-2}\end{array}\right] \end{gathered}$$

Hence, the matrix will be $P=\frac{1}{1+a^2+a^4+...+a^{2n-2}}\left[\begin{array}{ccc}1& a& a^2...a^{n-1}\\ a& a^2& a^3...a^n\\ a^2& a^3& ........a^{n+1}\\ .& .& .........\\ .& .& ........\\ a^{n-1}& a^n& .....a^{2n-2}\end{array}\right]$.

In order to show that the above matrix is Hankel matrix determine the equation.

$$\begin{gathered} {\left(A{A}^T\right)}_{ij}={\left(\stackrel{→}{v}\stackrel{→}{v}T\right)}_{ij}=v_{i1}{v}_{1j} \\ =a^{i-1}{a}^{j-1} \\ =a^{i-1+1}{a}^{j-1-1} \\ =a^i{a}^{j-2} \\ ={\left(A{A}^T\right)}_{i+1,j+1} \end{gathered}$$

Thus, the matrix is Hankel matrix.

Hence, the matrix P for the vector $\stackrel{→}{v}=\left[\begin{array}{c}1\\ 2\\ 4\end{array}\right]\mathrm{is}P=\left[\begin{array}{ccc}1& 2& 4\\ 2& 4& 8\\ 4& 8& 16\end{array}\right]$.