Minimum and Maximum eigenvalue inequality from a positive definite matrix.

I got a positive definite matrix $B,$ that is, $V(x) = x^TBx > 0$ for any vector $ x \neq 0.$ I want to show that $ \lambda_\min \|x\|_2^2 \leq V(x) \leq \lambda_\max \|x\|_2^2$ for any $x \neq 0,$ where $\lambda_\min$ and $ \lambda_\max$ are defined by $$ \lambda_\min = \min \lbrace | \lambda|: \lambda \text{ is an eigenvalue of } B \rbrace$$ and $$ \lambda_\max = \max \lbrace | \lambda|: \lambda \text{ is an eigenvalue of } B \rbrace$$ Any hint please?

Solution 1:

Define $B=Q^TDQ$ as the eigendecomposition of $B$. Note that diagonal entries of diagonal matrix $D$ will be positive (since $B$ is positive definite) and will be the eigenvalues of $B$ (denoted as $\lambda_i$ hereafter). In the following, I use the substituitions , $u=\frac{x}{\sqrt{x^Tx}}$ and $z=Qu$. Note that $u$ and $z$ are unit norm vectors. \begin{align} \frac{V(x)}{||x||^2}=\frac{x^TBx}{x^{T}x}=u^{T}Bu=u^{T}Q^{T}DQu=z^TDz=\sum_{i=1}^{N}\lambda_i|z_i|^2 \end{align} where $z_i$'s are entries of the vector $z$. Now what is the minimum value of the above expression. Also what is its maximum value?. In essence, what is the minimum and maximum of a sum of weighted combination of positive numbers ($\lambda_i$'s) such that their (non-negative) weights ($|z_i|^2$) sum to one. (Answer will be the minimum of those numbers and the maximum of them respectively). Infact the above argument works for hermitian matrices also where some of the eigenvalues can be negative. In general, what you have stated is courant fischer minmax theorem for highest and smallest eigenvalues.

Solution 2:

Since $B$ is positive definite, you can forget the absolute value signs in the definitions of $\lambda_{max}$ and $\lambda_{min}$. Note that $B - \lambda_{min} I$ and $\lambda_{max} I - B$ are positive semidefinite.