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Found in: Page 236

### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

# Let n be an even integer.In both parts of this problem,let Vbe the subspace of all vector $\stackrel{\mathbf{\to }}{\mathbf{x}}$ in ${{\mathbit{R}}}^{{\mathbf{n}}}$ such that ${{\mathbit{x}}}_{\mathbf{j}\mathbf{+}\mathbf{2}}{\mathbf{=}}{{\mathbit{x}}}_{{\mathbf{j}}}{\mathbf{+}}{{\mathbit{x}}}_{\mathbf{j}\mathbf{+}\mathbf{1}}{\mathbf{\forall }}{\mathbit{j}}{\mathbf{=}}{\mathbf{1}}{\mathbf{,}}{\mathbf{2}}{\mathbf{,}}{\mathbf{.}}{\mathbf{.}}{\mathbf{,}}{\mathbit{n}}{\mathbf{-}}{\mathbf{2}}{\mathbf{.}}$ .Consider the basis $\stackrel{\mathbf{\to }}{\mathbf{v}}{\mathbf{,}}\stackrel{\mathbf{\to }}{\mathbf{w}}$ of V with $\stackrel{\mathbf{\to }}{\mathbf{a}}{\mathbf{=}}\left[\begin{array}{l}1\\ a\\ {a}^{2}\\ .\\ .\\ .\\ {a}^{n-1}\end{array}\right]{\mathbf{,}}\stackrel{\mathbf{\to }}{\mathbf{b}}{\mathbf{=}}\left[\begin{array}{l}1\\ b\\ {b}^{2}\\ .\\ .\\ .\\ {b}^{n-1}\end{array}\right]$where ${\mathbit{a}}{\mathbf{=}}\frac{\mathbf{1}\mathbf{+}\sqrt{\mathbf{5}}}{\mathbf{2}}$ and ${\mathbit{b}}{\mathbf{=}}\frac{\mathbf{1}\mathbf{-}\sqrt{\mathbf{5}}}{\mathbf{2}}$a.Show that $\stackrel{\mathbf{\to }}{\mathbf{a}}{\mathbf{\text{\hspace{0.17em}}}}$ is orthogonal to $\stackrel{\mathbf{\to }}{\mathbf{b}}$b.Explain why the matrix P of the orthogonal projection onto V is a Hankel matrix.

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 $i=1,n-1\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}and\text{\hspace{0.17em}\hspace{0.17em}}j=2,3,...,n$

See the step by step solution

## (a)Step1:Definition for orthogonal

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

## Step2: Verification of orthogonality

Assume n be an even integer.

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

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

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

Note that ab=-1

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

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

Further calculation can be as follows:

$\begin{array}{c}\stackrel{\to }{a}.\stackrel{\to }{b}=\sum _{k=0}^{n-1}{\left(ab\right)}^{k}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1-{\left(ab\right)}^{n}}{1-ab}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1-{\left(-1\right)}^{n}}{1-\left(-1\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=0\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\forall \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}n\end{array}$

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

Hence $\stackrel{\to }{a}$ and $\stackrel{\to }{b}$ 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 ${\stackrel{\to }{u}}_{1},{\stackrel{\to }{u}}_{2},...,{\stackrel{\to }{u}}_{m}$ .The matrix P of the orthogonal projection onto V is P=QQT.

Where $Q=\left[\begin{array}{l}.\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}.\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}.\\ {\stackrel{\to }{u}}_{1}\text{\hspace{0.17em}\hspace{0.17em}}{\stackrel{\to }{u}}_{2}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\stackrel{\to }{u}}_{m}\\ \text{\hspace{0.17em}\hspace{0.17em}}.\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}.\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}.\text{\hspace{0.17em}}\end{array}\right]$

Note that the matrix P is symmetric, since:

$\begin{array}{c}{P}^{T}={\left(Q{Q}^{T}\right)}^{T}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={Q}^{T}{\left({Q}^{T}\right)}^{T}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={Q}^{T}Q\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=P\end{array}$

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

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 $i=1,n-1\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}and\text{\hspace{0.17em}\hspace{0.17em}}j=2,3,...,n$

Hence the explanation.