# Alternative characterization max eigenvalue

I want to see if my line of thinking is correct on the following problem (given that $A$ is Hermitian, the eigenvalues are in non decreasing order $\lambda_\min$ to $\lambda_\max$, and $A$ has at least one positive eigenvalue):

Show that $$\lambda_\max = \max\left\{{1 \over x^*x} \mid \:x^*Ax=1, x \in \mathbb{C}^n\right\}.$$ My thought was to use the Rayleigh Quotient. We know $\lambda_\max = \max\{{x^*Ax\over x^*x} \mid \:x \in \mathbb{C^n\}}$. Since we know $\lambda_\max$ is positive then the first set contains the eigenvector $y\over \sqrt{\lambda_\max}$ corresponding to $\lambda_\max$, since ${y^*Ay \over y^*y}=1$ . Then since the first set is a subset of $\left\{{x^*Ax\over x^*x} \mid \:x \in \mathbb{C}^n\right\}$ and contains the maximum of that set then its maximum is $\lambda_\max$ as well.

Your thought is of course valid. The first observation is that we do not need to maximise all the Rayleigh quotients but only the positive ones, then it's just a simple manipulation (substitution): $$\lambda_{\max} =\max_{y\neq 0}\frac{y^*Ay}{y^*y}=\max_{y^*Ay>0}\frac{y^*Ay}{y^*y} =\max_{y^*Ay>0}\left(\frac{y^*}{\sqrt{y^*Ay}}\frac{y}{\sqrt{y^*Ay}}\right)^{-1} =\max_{x^*Ax=1}\frac{1}{x^*x}.$$