Matrix with positive and negative eigenvalues (proof/saddle point)

Question

Let $f:\mathbb{R}^n \to \mathbb{R}$ with $f(\textbf{x})=\frac{1}{2}\textbf{x}^T\textbf{B}\textbf{x}$ where $\textbf{B} \in \mathbb{R}^{2\times 2}$ has one positive and one negative eigenvalue for the existence of a saddle point.

My question is: Why does $\textbf{B}$ has to have one positive and one negative eigenvalue for the existence of a saddle point?

My idea was:

It follows from the Hessian matrix $\textbf{H}f(\textbf{x})$, since $\textbf{H}f(\textbf{x})$ is indefinite if it has positive and negative eigenvalues. So $\textbf{x}$ is a saddle point. Is this correct?

I know the proof for showing that if $\textbf{H}f(\textbf{x})$ is indefinite then $\textbf{x}$ is a saddle point.

But does it follow from this that if $\textbf{B}$ is indefinite then $\textbf{H}f(\textbf{x})$ is indefinite or how can it be shown correctly?

Answer 1

We have $\textbf{B}= \textbf{H}f(x)$ for all $x$.