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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. Find ${\mathbit{q}}\left({\stackrel{⇀}{e}}_{1}\right)$. Give your answer in terms of the entries of the matrix A.

$\mathrm{q}\left({\stackrel{⇀}{\mathrm{e}}}_{1}\right)={\mathrm{a}}_{11}$

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 :

Take a look at the quadratic form:

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

Where A is a nxn symmetric matrix Take into account this:

$\mathrm{A}=\left({\mathrm{a}}_{\mathrm{ij}}\right)$ $\left(\mathrm{where}1\le \mathrm{i},\mathrm{j}\le \mathrm{n}\right)$

Now, use the following formula to access $\mathrm{q}\left({\stackrel{⇀}{\mathrm{e}}}_{1}\right)$:

localid="1659613659189" $\mathrm{q}\left({\stackrel{⇀}{\mathrm{e}}}_{1}\right)={\stackrel{⇀}{\mathrm{e}}}_{1}.\mathrm{A}{\stackrel{⇀}{\mathrm{e}}}_{1}\phantom{\rule{0ex}{0ex}}={\stackrel{⇀}{\mathrm{e}}}_{1}\left(\sum _{\mathrm{j}-1}^{\mathrm{n}}{\mathrm{a}}_{\mathrm{j}1}{\stackrel{⇀}{\mathrm{e}}}_{\mathrm{j}}\right)\phantom{\rule{0ex}{0ex}}=\sum _{\mathrm{j}-1}^{\mathrm{n}}{\mathrm{a}}_{\mathrm{j}1}\left({\stackrel{⇀}{\mathrm{e}}}_{\mathrm{i}}.{\stackrel{⇀}{\mathrm{e}}}_{\mathrm{j}}\right)\phantom{\rule{0ex}{0ex}}⇒\mathrm{q}\left({\stackrel{⇀}{\mathrm{e}}}_{1}\right)={\mathrm{a}}_{11}\left(\mathrm{Since}{\stackrel{⇀}{\mathrm{e}}}_{1}.{\stackrel{⇀}{\mathrm{e}}}_{\mathrm{j}}={\mathrm{\delta }}_{\mathrm{ij}}\right)$

Step 3: Determining the Result:

$\mathrm{q}\left({\stackrel{⇀}{\mathrm{e}}}_{1}\right)={\mathrm{a}}_{11}$