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Q25E

Expert-verifiedFound in: Page 400

Book edition
5th

Author(s)
Otto Bretscher

Pages
442 pages

ISBN
9780321796974

**Consider a quadratic form **

${\mathbf{q}}{\left(\stackrel{\rightharpoonup}{x}\right)}{\mathbf{=}}\stackrel{\mathbf{\rightharpoonup}}{\mathbf{x}}{\mathbf{.}}{\mathbf{A}}\stackrel{\mathbf{\rightharpoonup}}{\mathbf{x}}$

**where A is a symmetric**** nxn**** matrix. Let $\stackrel{\mathbf{\rightharpoonup}}{\mathbf{v}}$ be a unit eigenvector of A, with associated eigenvalue ${\mathit{\lambda}}$. Find ${\mathit{q}}{\left(\stackrel{\rightharpoonup}{v}\right)}$.**

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

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

Take a look at the quadratic form:

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

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

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

(According to the eigen value definition)

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

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

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