91影视

When any Symmetric Matrix\(A\)is diagonalized orthogonallyas \(PD{P^{ - 1}}\)we have:

\(\begin{array}{l}{{\rm{x}}^T}A{\rm{x}} = {{\rm{y}}^T}D{\rm{y}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ {{\rm{as }}x = Py} \right\}\\{\rm{and}}\\\left\| {\rm{x}} \right\| = \left\| {P{\rm{y}}} \right\| = \left\| {\rm{y}} \right\|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ {\forall y \in \mathbb{R}} \right\}\end{array}\)

As per the question, we have:

The matrix\(A\)has eigenvalue\(\lambda \)with eigenvector\({\rm{x}}\), then:

\(\begin{array}{c}{{\rm{x}}^T}A{\rm{x}} = {{\rm{x}}^T}\lambda {\rm{x}}\\ = \lambda \left( {{{\rm{x}}^T}{\rm{x}}} \right)\\ = \lambda {\left\| {\rm{x}} \right\|^2}\\ = \lambda \end{array}\)

Thus,

\({{\rm{x}}^T}A{\rm{x}} = \lambda ,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ {\forall \lambda \in \left[ {m,M} \right]} \right\}\)

The statement\({{\rm{x}}^T}A{\rm{x}} = \lambda \)is justified when\(m \le \lambda \le M\).