Question

Let n be an even integer.In both parts of this problem,let Vbe the subspace of all vector$\overrightarrow{\mathbf{x}}$in${\mathit{R}}^{\mathbf{n}}$
such that ${\mathit{x}}_{\mathbf{j}\mathbf{+}\mathbf{2}}\mathbf{=}{\mathit{x}}_{\mathbf{j}}\mathbf{+}{\mathit{x}}_{\mathbf{j}\mathbf{+}\mathbf{1}}\mathbf{\forall }\mathit{j}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{,}\mathit{n}\mathbf{-}\mathbf{2}\mathbf{.}$.Consider the basis $\overrightarrow{\mathbf{v}}\mathbf{,}\overrightarrow{\mathbf{w}}$of V with

$\overrightarrow{\mathbf{a}}\mathbf{=}\left[\begin{array}{l}1\\ a\\ a^2\\ .\\ .\\ .\\ a^{n-1}\end{array}\right]\mathbf{,}\overrightarrow{\mathbf{b}}\mathbf{=}\left[\begin{array}{l}1\\ b\\ b^2\\ .\\ .\\ .\\ b^{n-1}\end{array}\right]$

where $\mathit{a}\mathbf{=}\frac{\mathbf{1}\mathbf{+}\sqrt{\mathbf{5}}}{\mathbf{2}}$and$\mathit{b}\mathbf{=}\frac{\mathbf{1}\mathbf{-}\sqrt{\mathbf{5}}}{\mathbf{2}}$

a.Show that$\overrightarrow{\mathbf{a}}$ is orthogonal to$\overrightarrow{\mathbf{b}}$

b.Explain why the matrix P of the orthogonal projection onto V is a Hankel matrix.

Short answer

a.The solution is orthogonal.

b.It suffices to show that M and N are Hankel matrices.Indeed $m_{ij}=a^{i+j-2}=m_{i+j,j-1}$ and $n_{ij}=b^{i+j-2}=n_{i+1,j-1}$ for all $i=1,n-1andj=2,3,\mathrm{...},n$

Step-by-step solution

(a)Step1:Definition for orthogonal

Two vectors $\overrightarrow{v}$ and $\overrightarrow{w}$ in$R^n$ are called orthogonal if $\overrightarrow{v}.\overrightarrow{w}=0$

Verification of orthogonality

Assume n be an even integer.

let V be the subspace of all vector$\overrightarrow{x}$in $R^n$such that $x_{j+2}=x_j+x_{j+1}\forall j=1,2,\mathrm{..},n-2.$ .Consider the basis $\overrightarrow{v},\overrightarrow{w}$ of V with:

$\overrightarrow{a}=\left[\begin{array}{l}1\\ a\\ a^2\\ .\\ .\\ .\\ a^{n-1}\end{array}\right],\overrightarrow{b}=\left[\begin{array}{l}1\\ b\\ b^2\\ .\\ .\\ .\\ b^{n-1}\end{array}\right]$

Where$a=\frac{1+\sqrt{5}}{2}$and$b=\frac{1-\sqrt{5}}{2}$

Note that ab=-1

Now we have to prove that$\overrightarrow{a}.\overrightarrow{b}=0$for the verification of orthogonality.

$\begin{array}{c}\overrightarrow{a}=\sum _{k=0}^{n-1}{a}^k\\ \overrightarrow{b}=\sum _{k=0}^{n-1}{b}^k\\ \overrightarrow{a}.\overrightarrow{b}=\sum _{k=0}^{n-1}{a}^k\sum _{k=0}^{n-1}{b}^k\\ \overrightarrow{a}.\overrightarrow{b}=\sum _{k=0}^{n-1}{a}^k{b}^k\end{array}$

Further calculation can be as follows:

$\begin{array}{c}\overrightarrow{a}.\overrightarrow{b}=\sum _{k=0}^{n-1}{\left(ab\right)}^k\\ =\frac{1-{\left(ab\right)}^n}{1-ab}\\ =\frac{1-{(-1)}^n}{1-(-1)}\\ =0\forall n\end{array}$

Thus$\overrightarrow{a}.\overrightarrow{b}=0$for all even integer n.

Hence$\overrightarrow{a}$and$\overrightarrow{b}$are orthogonal.

Therefore, the solution.

(b)Step 3:Explanation of orthogonal projection onto Hankel matrix

By a theorem we know that a subspace V of$R^n$with orthonormal basis${\overrightarrow{u}}_1,{\overrightarrow{u}}_2,\mathrm{...},{\overrightarrow{u}}_m$.The matrix P of the orthogonal projection onto V is P=QQ^T.

Where$Q=\left[\begin{array}{l}...\\ {\overrightarrow{u}}_1{\overrightarrow{u}}_2{\overrightarrow{u}}_m\\ ...\end{array}\right]$

Note that the matrix P is symmetric, since:

$\begin{array}{c}{P}^T={\left(Q{Q}^T\right)}^T\\ =Q^T{\left(Q^T\right)}^T\\ =Q^{T}Q\\ =P\end{array}$

Thus, the preceding paragraph P is a linear combination of the matrices$M=\overrightarrow{a}{\overrightarrow{a}}^T$ and $N=\overrightarrow{b}{\overrightarrow{b}}^T$.

It suffices to show that M and N are Hankel matrices.

Indeed$m_{ij}=a^{i+j-2}=m_{i+j,j-1}$and$n_{ij}=b^{i+j-2}=n_{i+1,j-1}$for all$i=1,n-1andj=2,3,\mathrm{...},n$

Hence the explanation.