A real-valued symmetric square matrix is called positive definite if $(x,Ax)>0$ for all $x\neq0,$ where $(.,.)$ represents the scalar product. For a positive definite matrix determine
$$\max\left\{ \frac{(x,Ax)}{||x||^2};x\neq0\right\}$$
and
$$\min\left\{ \frac{(x,Ax)}{||x||^2};x\neq0\right\}.$$
Also for an infinite-dimensional vector space determine an analogue of this at which
$$\min\left\{ \frac{(x,Ax)}{||x||^2};(x,Ax)>0\right\}=0.$$
I hope this make sense and so far I have looked into Min-Max theorem, Positive definite matrices, and a few other topics but have no idea as to how I should proceed. Any help would be greatly appreciated.
If $A\in\mathbb R^{n\times n}$ is symmetric, then the Spectral Theorem ensures that all its eigenvalues $\lambda_i$ are real and that there exists an orthonormal basis of $\mathbb R^n$ made of eigenvectors of $A$. Then $$ \|A\| = \sup\left\{\|Ax\|\,:~\|x\|=1\right\} =\max_i\left|\lambda_i\right| $$
To see this, let $\{v_1\ldots v_n\}$ be such orthonormal basis, where $v_i$ is the eigenvector with eigenvalue $\lambda_i$ (there might be $i\neq j$ s.t. $\lambda_i=\lambda_j$). That is, $$ Av_i = \lambda_iv_i \quad\text{and}\quad (v_i,v_j) = \begin{cases} 1 &\text{if } i=j\\0&\text{if }i\neq j\end{cases} $$ Any $x$ in $\mathbb R^n$ can be written as a linear combination of the basis $\{v_1\ldots v_n\}$: $x=\sum_{i=1}^nc_iv_i$. Then, by orthonormality, $$ \|x\|^2 = \left\|\sum_{i=1}^nc_iv_i\right\|^2 = \sum_{i=1}^nc_i^2 $$ and $$ \left\|Ax\right\|^2 = \left\|A\sum_{i=1}^nc_iv_i\right\|^2 = \left\|\sum_{i=1}^nc_iAv_i\right\|^2 = \left\|\sum_{i=1}^nc_i\lambda_iv_i\right\|^2 = \sum_{i=1}^n\|c_i\lambda_iv_i\|^2 = \sum_{i=1}^n c_i^2\lambda_i^2 $$ You can easily see that among all the $x\in\mathbb R^n$ with $\|x\|=1$, one that maximizes $\|Ax\|$ is $v_j$ where $|\lambda_j|$ is the largest eigenvalue in absolute value.
If $A$ is positive definite then all the eigenvalues are positive, as you can easily verify by computing $(v_i,Av_i)$ for all $i$.
As to the infinite dimensional case, take $\{e_n\}$ to be an orthonormal basis of the space $\mathcal H$, and define $$ T:\mathcal H\rightarrow\mathcal H \quad\text{s.t.}\quad Te_n = \tfrac1ne_n ~\text{ on the basis} $$ $T$ is positive definite with eigenvalues $\{\frac1n\}_{n\in\mathbb N^+}$, but the infimum of such set is $0$.