91Ó°ÊÓ

$\mathrm{q}\left(\stackrel{â‡\P Ä}{\mathrm{x}}\right)=\stackrel{â‡\P Ä}{\mathrm{x}}.\mathrm{A}\stackrel{â‡\P Ä}{\mathrm{x}}$

Take a look at the quadratic form:

$\mathrm{q}\left(\stackrel{â‡\P Ä}{\mathrm{x}}\right)=\stackrel{â‡\P Ä}{\mathrm{x}}.\mathrm{A}\stackrel{â‡\P Ä}{\mathrm{x}}$

where A is a symmetric nxn matrix with${\mathrm{λ}}_1â‰\yen {\mathrm{λ}}_2â‰\yen ..........â‰\yen {\mathrm{λ}}_{\mathrm{n}}$eigenvalues. Assume that${\mathrm{S}}^{\mathrm{n}-1}$is the set of all unit vectors in ${\mathrm{R}}^{\mathrm{n}}$. We get an orthonomal basis$\left\{{\mathrm{v}}_1,.....{\mathrm{v}}_{\mathrm{n}}\right\}$of ${\mathrm{R}}^{\mathrm{n}}$using the spectral theorem.

${\mathrm{Av}}_{\mathrm{i}}={\mathrm{λ}}_{\mathrm{i}}{\mathrm{v}}_{\mathrm{i}}$

Consider the fact that, for each $\stackrel{â‡\P Ä}{\mathrm{x}}∈{\mathrm{S}}^{\mathrm{n}-1}$,

Consider the following for $\stackrel{â‡\P Ä}{\mathrm{x}}.A\stackrel{â‡\P Ä}{\mathrm{x}}$:

As a result, it indicates

role="math" localid="1660206280724" $$\begin{gathered} ⇒{\mathrm{λ}}_1â‰\yen \stackrel{â‡\P Ä}{\mathrm{x}}.\mathrm{A}\stackrel{â‡\P Ä}{\mathrm{x}}â‰\yen {\mathrm{λ}}_{\mathrm{n}}\left(\mathrm{Since}{||\mathrm{x}||}^2\right) \\ \mathrm{As}\mathrm{a}\mathrm{result},\mathrm{we}\mathrm{may}\mathrm{deduce}\mathrm{that} \\ \stackrel{â‡\P Ä}{\mathrm{x}}.\mathrm{A}\stackrel{â‡\P Ä}{\mathrm{x}}∈\left[{\mathrm{λ}}_{\mathrm{n}},{\mathrm{λ}}_1\right] \\ ⇒\mathrm{q}\left({\mathrm{S}}^{\mathrm{n}-1}\right)⊆\left[{\mathrm{λ}}_{\mathrm{n}},{\mathrm{λ}}_1\right] \\ \mathrm{Consider}\mathrm{the}\mathrm{fact}\mathrm{that},\mathrm{for}\mathrm{each}\mathrm{Î\pm }∈\left[{\mathrm{λ}}_{\mathrm{n}},{\mathrm{λ}}_1\right], \\ {\mathrm{λ}}_1â‰\yen \mathrm{Î\pm }â‰\yen {\mathrm{λ}}_{\mathrm{n}} \\ ⇒\mathrm{Î\pm }={\mathrm{³Ùλ}}_1+\left(1+\mathrm{t}\right){\mathrm{λ}}_{\mathrm{n}}\left(\mathrm{for}\mathrm{some}\mathrm{t}\right) \end{gathered}$$

Take a look at $\stackrel{â‡\P Ä}{\mathrm{x}}=\sqrt{\left(1-\mathrm{t}\right)}{\mathrm{u}}_{\mathrm{n}}$now. Take note of this:

${||\stackrel{â‡\P Ä}{\mathrm{x}}||}^2={\left(\sqrt{\mathrm{t}}\right)}^2+{\left(\sqrt{1-\mathrm{t}}\right)}^2=1$

As a result, $\stackrel{â‡\P Ä}{\mathrm{x}}∈{\mathrm{S}}^{\mathrm{n}-1}$. Now, analyze$\stackrel{â‡\P Ä}{\mathrm{x}}.\mathrm{A}\stackrel{â‡\P Ä}{\mathrm{x}}$ in the following manner:

$\mathrm{q}\left({\mathrm{S}}^{\mathrm{n}-1}\right)=\left[{\mathrm{λ}}_{\mathrm{n}},\mathrm{λ}\right]$