91影视

Theorem 6states that consider \(A\) as a symmetric matrixand \(m = \min \left\{ {{{\bf{x}}^T}A{\bf{x}}:\left\| {\bf{x}} \right\| = 1} \right\},M = \max \left\{ {{{\bf{x}}^T}A{\bf{x}}:\left\| {\bf{x}} \right\| = 1} \right\}\). Then the greatest eigenvalue \({\lambda _1}\) of A is \(M\) and the least eigenvalue of \(A\) is \(m\). \(M\) is the value of \({{\bf{x}}^T}A{\bf{x}}\) if the unit eigenvalue of \({{\bf{u}}_1}\) is \({\bf{x}}\) that corresponds to \(M\). \(m\) is the value of \({{\bf{x}}^T}A{\bf{x}}\) if the unit eigenvalue of \({{\bf{u}}_1}\) is \({\bf{x}}\) that corresponds to \(m\).

The Pythagoras theoremstates that the vectors \({\bf{u}}\) and \({\bf{v}}\) are said to beorthogonal such that if \({\left\| {{\bf{u}} + {\bf{v}}} \right\|^2} = {\left\| {\bf{u}} \right\|^2} + {\left\| {\bf{v}} \right\|^2}\).

When \(m = M\) then consider that \(t = \left( {1 - 0} \right)m + 0M = m\) and \({\bf{x}} = {{\bf{u}}_n}\). Theorem 6 demonstrates that \({{\bf{u}}^T}A{{\bf{u}}_n} = m\).

Assume that \(m < M,\) and consider that \(t\) is between \(m\) and \(M\). Then \(0 \le t - m \le M - m\) and \(0 \le \left( {t - m} \right)\left( {M - m} \right) \le 1\).

Consider that \(\alpha = \frac{{\left( {t - m} \right)}}{{\left( {M - m} \right)}}\) and \({\bf{x}} = \sqrt {1 - \alpha } {{\bf{u}}_n} + \sqrt \alpha {{\bf{u}}_1}\). The vectors \(\sqrt {1 - \alpha } {{\bf{u}}_n}\) and \(\sqrt \alpha {{\bf{u}}_1}\) are orthogonal since the eigenvectors have different eigenvalues (or one of them is 0). According to the Pythagoras theorem

\(\begin{array}{c}{{\bf{x}}^T}{\bf{x}} = {\left\| {\bf{x}} \right\|^2}\\ = {\left\| {\sqrt {1 - \alpha } {{\bf{u}}_n}} \right\|^2} + {\left\| {\sqrt \alpha {{\bf{u}}_1}} \right\|^2}\\ = \left| {1 - \alpha } \right|{\left\| {{{\bf{u}}_n}} \right\|^2} + \left| \alpha \right|{\left\| {{{\bf{u}}_1}} \right\|^2}\\ = \left( {1 - \alpha } \right) + \alpha \\ = 1\end{array}\)

Because \({{\bf{u}}_n}\) and \({{\bf{u}}_1}\) are unit vectors and \(0 \le \alpha \le 1\). Moreover, \({{\bf{u}}_n}\) and \({{\bf{u}}_1}\) are orthogonal, then

\(\begin{array}{c}{{\bf{x}}^T}A{\bf{x}} = {\left( {\sqrt {1 - \alpha } {{\bf{u}}_n} + \sqrt \alpha {{\bf{u}}_1}} \right)^T}A\left( {\sqrt {1 - \alpha } {{\bf{u}}_n} + \sqrt \alpha {{\bf{u}}_1}} \right)\\ = {\left( {\sqrt {1 - \alpha } {{\bf{u}}_n} + \sqrt \alpha {{\bf{u}}_1}} \right)^T}\left( {m\sqrt {1 - \alpha } {{\bf{u}}_n} + M\sqrt \alpha {{\bf{u}}_1}} \right)\\ = \left| {1 - \alpha } \right|m{\bf{u}}_n^T{{\bf{u}}_n} + \left| \alpha \right|M{\bf{u}}_1^T{{\bf{u}}_1}\\ = \left( {1 - \alpha } \right)m + \alpha M\\ = t\end{array}\)

Therefore, the quadratic form \({{\bf{x}}^T}A{\bf{x}}\) takes all values between \(m\) and \(M\) for a relevant unit vector \(x\).

Thus, it is proved that \({{\bf{x}}^T}{\bf{x}} = t\) and \({{\bf{x}}^T}A{\bf{x}} = t\).