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Q25E

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Found in: Page 400

### Linear Algebra With Applications

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

# Consider a quadratic form ${\mathbf{q}}\left(\stackrel{⇀}{x}\right){\mathbf{=}}\stackrel{\mathbf{⇀}}{\mathbf{x}}{\mathbf{.}}{\mathbf{A}}\stackrel{\mathbf{⇀}}{\mathbf{x}}$where A is a symmetric nxn matrix. Let $\stackrel{\mathbf{⇀}}{\mathbf{v}}$ be a unit eigenvector of A, with associated eigenvalue ${\mathbit{\lambda }}$. Find ${\mathbit{q}}\left(\stackrel{⇀}{v}\right)$.

$\mathrm{q}\left(\stackrel{⇀}{\mathrm{v}}\right)=\mathrm{\lambda }$

See the step by step solution

## Step 1: Given Information:

$\mathrm{q}\left(\stackrel{⇀}{\mathrm{x}}\right)=\stackrel{⇀}{\mathrm{x}}.\mathrm{A}\stackrel{⇀}{\mathrm{x}}$

## Step 2: Determining q(v⇀):

Take a look at the quadratic form:

$\mathrm{q}\left(\stackrel{⇀}{\mathrm{x}}\right)=\stackrel{⇀}{\mathrm{x}}.\mathrm{A}\stackrel{⇀}{\mathrm{x}}$

A is a nxn symmetric matrix. Let $\stackrel{⇀}{v}$be the unit eigenvector of A, with lambda as the eigenvalue. Now, use the following formula to access $\mathrm{q}\left(\stackrel{⇀}{\mathrm{v}}\right)$:

role="math" localid="1659615170927" $\mathrm{q}\left(\stackrel{⇀}{\mathrm{v}}\right)=\stackrel{⇀}{\mathrm{v}}.\mathrm{A}\stackrel{⇀}{\mathrm{v}}\phantom{\rule{0ex}{0ex}}=\stackrel{⇀}{\mathrm{v}}.\mathrm{\lambda }\stackrel{⇀}{\mathrm{v}}\phantom{\rule{0ex}{0ex}}=\mathrm{\lambda }\left(\stackrel{⇀}{\mathrm{v}}.\stackrel{⇀}{\mathrm{v}}\right)\phantom{\rule{0ex}{0ex}}=\mathrm{\lambda }\left(1\right)$

(According to the eigen value definition)

$⇒q\left(\stackrel{⇀}{v}\right)=\lambda$

## Step 3: Determining the Result:

$q\left(\stackrel{⇀}{v}\right)=\lambda$