Book edition 5th

Author(s) Otto Bretscher

Pages 442 pages

ISBN 9780321796974

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Short Answer

Let n be an even integer.In both parts of this problem,let Vbe the subspace of all vector $\vec{x}$ in $R^{n}$ such that $x_{j + 2} = x_{j} + x_{j + 1} \forall j = 1, 2, . ., n - 2 .$ .Consider the basis $\vec{v}, \vec{w}$ of V with
$\vec{a} = [\begin{array}{l} 1 \\ a \\ a^{2} \\ . \\ . \\ . \\ a^{n - 1} \end{array}], \vec{b} = [\begin{array}{l} 1 \\ b \\ b^{2} \\ . \\ . \\ . \\ b^{n - 1} \end{array}]$
where $a = \frac{1 + \sqrt{5}}{2}$ and $b = \frac{1 - \sqrt{5}}{2}$
a.Show that $\vec{a}$ is orthogonal to $\vec{b}$
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 $m_{i j} = a^{i + j - 2} = m_{i + j, j - 1}$ and $n_{i j} = b^{i + j - 2} = n_{i + 1, j - 1}$ for all $i = 1, n - 1 a n d j = 2, 3, ..., n$

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TABLE OF CONTENTS :

TABLE OF CONTENTS

(a)Step1:Definition for orthogonal

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

Step2: Verification of orthogonality

Assume n be an even integer.

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

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

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

Note that ab=-1

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

$\begin{matrix} \vec{a} = \sum_{k = 0}^{n - 1} a^{k} \\ \vec{b} = \sum_{k = 0}^{n - 1} b^{k} \\ \vec{a} . \vec{b} = \sum_{k = 0}^{n - 1} a^{k} \sum_{k = 0}^{n - 1} b^{k} \\ \vec{a} . \vec{b} = \sum_{k = 0}^{n - 1} a^{k} b^{k} \end{matrix}$

Further calculation can be as follows:

$\begin{matrix} \vec{a} . \vec{b} = \sum_{k = 0}^{n - 1} {(a b)}^{k} \\ = \frac{1 - {(a b)}^{n}}{1 - a b} \\ = \frac{1 - {(- 1)}^{n}}{1 - (- 1)} \\ = 0 \forall n \end{matrix}$

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

Hence $\vec{a}$ and $\vec{b}$ 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 ${\vec{u}}_{1}, {\vec{u}}_{2}, ..., {\vec{u}}_{m}$ .The matrix P of the orthogonal projection onto V is P=QQ^T.

Where $Q = [\begin{array}{l} . . . \\ {\vec{u}}_{1} {\vec{u}}_{2} {\vec{u}}_{m} \\ . . . \end{array}]$

Note that the matrix P is symmetric, since:

$\begin{matrix} P^{T} = {(Q Q^{T})}^{T} \\ = Q^{T} {(Q^{T})}^{T} \\ = Q^{T} Q \\ = P \end{matrix}$

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

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

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

Hence the explanation.

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Linear Algebra With Applications

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Short Answer

Step by Step Solution

(a)Step1:Definition for orthogonal

Step2: Verification of orthogonality

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

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Linear Algebra With Applications

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Step by Step Solution

(a)Step1:Definition for orthogonal

Step2: Verification of orthogonality

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

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