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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 $\displaystyle \lambda_{min} = \underset{x\neq0}{\min{\frac{x^*Ax}{x^*x}}}$

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Min-max theorem. The proof is straightforward once you have taken an orthonormal basis. But note this gives a neat characterization of every $\lambda_k$, not only the first one and the last one. When they are ordered, I mean. –  1015 May 3 '13 at 1:17
I meant an orthonormal diagonalization basis for $A$. –  1015 May 3 '13 at 1:37

1 Answer 1


  • $A=U^HDU$
  • Define $y=Ux$, then $x^HAx=y^HDy$
  • 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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May be you can also help at this related question, mathoverflow.net/questions/187850/… –  Anirbit Nov 23 '14 at 0:27

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