[Math] Largest and smallest eigenvalues of a hermitian matrix

eigenvalues-eigenvectorslinear algebramatricesnumerical linear algebra

How to show that the largest and smallest eigenvalues of a hermitian matrix $A \in \mathbb{C}^{n \times n} $ can be found as:
$\displaystyle \lambda_{max} = \underset{x\neq0}{\max{\frac{x^*Ax}{x^*x}}}$ and $\displaystyle \lambda_{min} = \underset{x\neq0}{\min{\frac{x^*Ax}{x^*x}}}$

Best Answer

Hints:

  • Convince yourself that it is ok to remove the norm term $x^Hx$ from denominator and you have to consider $x^Hx=1$ instead of $x\neq 0$.
  • There exists a decomposition $A=U^HDU$ where $D$ is a diagonal matrix containing the eignevalues and $U$ is the orthonormal matrix with eigenvectors.
  • Define $y=Ux$, then $x^HAx=y^HDy=\sum_{i=1}^{N}|y_i|^2\lambda_i$
  • Let $\lambda_1<\lambda_2<\lambda_3<\dots<\lambda_N$ be $N$ real numbers. Let $\theta_1,\theta_2,\dots,\theta_N$ be $N$ non-negative numbers such that $\sum_{i}^{N}\theta_i=1$, then $\max_{\theta_i}\sum_{i}^{N}\theta_i\lambda_i=\lambda_N$ and $\min_{\theta_i}\sum_{i}^{N}\theta_i\lambda_i=\lambda_1$
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