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Q38E

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Linear Algebra With Applications

Found in: Page 401

Book edition 5th

Author(s) Otto Bretscher

Pages 442 pages

ISBN 9780321796974

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Short Answer

For which values of the constants p and q is the matrix
$B = [\begin{matrix} p & q & L & q \\ q & p & L & q \\ M & M & 0 & M \\ q & q & L & p \end{matrix}]$
positive definite? (B has p’s on the diagonal and q’s elsewhere.) Hint: Exercise 8.1.17 is helpful.

$p > q > - \frac{p}{n - 1} where p, q \neq 0$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given Information:

$B = [\begin{matrix} p & q & \dots & q \\ q & p & \dots & q \\ ⋮ & ⋮ & ⋱ & ⋮ \\ q & q & \dots & p \end{matrix}]$

Step 2: Finding the positive definite values:

Consider the following $n \times n$ matrix with the constants p and q:

$B = [\begin{matrix} p & q & \dots & q \\ q & p & \dots & q \\ ⋮ & ⋮ & ⋱ & ⋮ \\ q & q & \dots & p \end{matrix}]$

The eigen values of the following matrix B are p-q (multiplicity n-1) and nq+p-q, respectively (multiplicity 1). Because all eigen values must be positive for matrix B to be positive definite, we have:

$\{\begin{cases} p - q > 0 \\ nq + p - q > 0 \end{cases}$

localid="1659688090121" $\Rightarrow \{\begin{cases} p > q \\ (n - 1) q + p > 0 \end{cases}$

$\Rightarrow \{\begin{cases} p > q \\ (n - 1) q > - p \end{cases}$

$\Rightarrow \{\begin{cases} p > q \\ q > - \frac{p}{(n - 1)} \end{cases}$

As a result, we receive

$p > q > - \frac{p}{n - 1} (where p, q \neq 0)$

As a result, the matrix B is positive definite for these values of the constants p and q.

Step 3: Determining the Result:

$p > q > - \frac{p}{n - 1} where p, q \neq 0$

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