For symmetric, non-singular, positive definite matrix $B$ and some unit vector $u$ show $u^T Bu \ge \frac{1}{ {\lVert B^{-1} \rVert} }$

Question

For symmetric non-singular positive definite matrix $B$, and any unit vector ${\lVert u \rVert} = 1$, show that:

\begin{gather*} u^T Bu \ge \frac{1}{ {\lVert B^{-1} \rVert} } \end{gather*}

Since $B$ is positive definite, we know that:

\begin{gather*} u^T Bu > 0 \\ \end{gather*}

That and the Cauchy Schwarz Inequality give us:

\begin{gather*} 0 < u^T Bu \le {\lVert Bu \rVert} \le {\lVert B \rVert} \\ \end{gather*}

For any invertible matrix:

\begin{gather*} {\lVert B^{-1} \rVert} {\lVert B \rVert} \ge 1, \quad \frac{1}{ {\lVert B^{-1} \rVert} } \le {\lVert B \rVert} \\ \end{gather*}

I'm stuck on what to try from here.

Answer 1

Using Cauchy Schwarz we have, \begin{align*} 1 = u^{\top} u = \langle B^{1/2}u, B^{-1/2}u\rangle \leq (u^{\top} B u) \cdot (u^{\top} B^{-1} u) \leq (u^{\top} B u) \cdot \|B^{-1}\|. \end{align*}