= Hermitian Rayleigh quotient range
{c}
{title2=$[\lambda_{\min},\lambda_{\max}]$}
For a finite-dimensional <Hermitian matrix> $A$, the <Rayleigh quotient> has precisely the range $[\lambda_{\min},\lambda_{\max}]$, where the endpoints are its extreme <eigenvalues>. The <finite-dimensional spectral theorem> writes a nonzero <vector> as $x=\sum_jc_je_j$ in an <orthonormal eigenbasis>, giving
$$
R_A(x)=\sum_j\lambda_jw_j,\qquad
w_j=\frac{|c_j|^2}{\sum_k|c_k|^2},\qquad
w_j\geq0,\quad\sum_jw_j=1.
$$
Thus the quotient is a <convex combination> of <eigenvalues>. The bounds follow immediately, and equality at either endpoint means that $x$ lies in the corresponding extreme <eigenspace>, including any repeated <eigenvalue>. Every intermediate value occurs: for orthonormal extreme <eigenvectors>, $x=\sqrt{1-t}\,e_{\min}+\sqrt t\,e_{\max}$ gives $R_A(x)=(1-t)\lambda_{\min}+t\lambda_{\max}$ for $0\leq t\leq1$. If all <eigenvalues> are equal, the quotient is constant.
The Hermitian hypothesis matters. The real <matrix> $B=\begin{pmatrix}1&4\\0&1\end{pmatrix}$ has only the <eigenvalue> $1$, but $(1,1)B(1,1)^T/2=3$. Its quotient is governed by the <symmetric part of a matrix> $(B+B^T)/2$, not by a real <orthonormal eigenbasis> of $B$. A nonsymmetric <matrix> therefore need not obey the displayed eigenvalue bounds.
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