Let A denote a symmetric matrix in $\mathbb R^{N\times N}$, and let $f$ denote the function defined for all $x\in \mathbb R^{N}$\{0} by
$$f(x) = \frac{x^TAx}{x^Tx}$$
Let $\lambda_1,....,\lambda_N$ denote the eigenvalues of A. Prove that
$$\min_x f(x) = \min_i \lambda_i$$ and $$\max_x f(x) = \max_i \lambda_i$$
------My Solution (Incomplete)------
I have computed the gradient of $f$, which is $\frac{2[Ax(x^Tx)-x(x^TAx)]}{(x^Tx)^2}$. Therefore when the gradient is 0, from $2[Ax(x^Tx)-x(x^TAx)]$, I get $Ax = \frac{x^TAx}{x^Tx}x$.
This seems to correspond to the eigenvalue formula $Ax = \lambda x$. But from here I am stuck and nowhere to find the solution. Woule anybody please help me?
$A$ is symmetric therefore diagonalisable in an orthonormal basis of $\Bbb{R}^n$. Let $(e_i)$ be that basis and write
$$x=\sum_ix_ie_i,A e_i=\lambda_ie_i$$
One then has
$$f(x)={\sum_i\lambda_ix_i^2\over \sum_ix_i^2}$$
Let’s give the details of the computation
$$\begin{align}x^TAx&=x^TA\sum_ix_ie_i\\ &=x^T\sum_ix_i\lambda_ie_i\text{ By linearity and definition of eigenvector}\\ &=\sum_i\lambda_ix_i^2\text{ Because the basis is orthonormal}\end{align}$$
Now denote $\max\lambda_i=\lambda_M$ and $\min\lambda_i=\lambda_m$ One has $\lambda_m\leq\lambda_i\leq\lambda_M$ and therefore
$$\lambda_m\leq f(x)\leq\lambda_M$$
Besides $f(e_m)=\lambda_m$ and $f(e_M)=\lambda_M$ and we’re done because we’ve proven that $f$ is bounded by $\min\lambda_i$ and $\max\lambda_i$ and reaches those bounds.