Prove that $$\sqrt{\frac{a^TA^2a}{a^TAa}}\le \sqrt{\|A\|}$$ where $A$ is a $n \times n$ symmetric positive definite matrix and $a \in \mathbb{R}^n \setminus \{0_n\}$.
The norm of a vector here is the Euclidean norm and norm of a matrix $A$ is $$\|A\|=\max\{\|Ax\|:\|x\|=1\}$$ and if $A$ is symmetric then $$\|A\|=\max\{|x^TAx|:\|x\|=1\}$$ and also equal to the largest eigenvalue of $A$.
We have from this min-max theorem that
$$\lambda_{\min}\|x\|^2\leq x^THx \le \lambda_{\max}\|x\|^2$$
where $\lambda_{\min},\lambda_{\max}$ are the smallest and largest eigenvalues of the Hermitian matrix $H$. So we have
$$\sqrt{\frac{a^TA^2a}{a^TAa}}\le \sqrt{\frac{\lambda^2_{\max}}{\lambda_{\min}}}$$
In order to prove the inequality in the question, we need to show that $\frac{\lambda^2_{max}}{\lambda_{\min}}\le \lambda_{\max}$ which implies $\frac{\lambda_{\max}}{\lambda_{\min}}\le 1$. We have $\|A\|=\lambda_{\max}$ and $\|A^{-1}\|=\frac{1}{\lambda_{\min}}$ but on the contrary it is a fact that $\|A\|\|A^{-1}\|\ge 1$. Where have I gone wrong and how do we prove this inequality?
Best Answer
Since $\rm A$ is symmetric and positive definite, it has a symmetric and positive definite square root $\mathrm A^\frac 12$ and $\lambda_{\max} \left( \mathrm A \right) = \| \mathrm A\|_2$, i.e., the spectral radius is equal to the spectral norm. Let $\rm v := \mathrm A^\frac 12 u$. Hence,
$$\dfrac{\mathrm u^\top \mathrm A^2 \mathrm u}{\mathrm u^\top \mathrm A \,\, \mathrm u} = \dfrac{\mathrm u^\top \mathrm A^\frac 12 \mathrm A \, \mathrm A^\frac 12 \mathrm u}{\mathrm u^\top \mathrm A^\frac 12 \mathrm A^\frac 12 \mathrm u} = \dfrac{\mathrm v^\top \mathrm A \, \mathrm v}{\mathrm v^\top \mathrm v} \leq \lambda_{\max} \left( \mathrm A \right) = \| \mathrm A\|_2$$