Why is the maximum Rayleigh quotient equal to the maximum eigenvalue?

First, note that R does not depend on the length of v, so we might as well impose the constraint $\mid v\mid^2=1$.

We maximize $R$ subject to this constraint by using a Lagrange multiplier: $v^tMv+\lambda (\mid v\mid^2-1)$, and differentiating with respect to the components of v, we obtain the equation $Mv+\lambda v=0$, so the extrema are precisely the eigenvectors of $M$. If $v$ is an eigenvector, then it follows immediately that the value of $R$ is the corresponding eigenvalue.

The matrix $M$ describes a linear map $M:\>{\mathbb R}^n=V\to V$ of the euclidean vector space $V$ in terms of the standard basis of $V$. The Rayleigh quotient $$R(M,v):={\langle v, Mv\rangle \over |v|^2}$$ is defined independently of the chosen basis, and for orthonormal bases is given by the formula you quote. Now for a symmetric matrix $M$ there is an orthonormal basis that diagonalizes $M$. With respect to such a basis we have $$R(M,v)={\sum_{i=1}^n \lambda_i v_i^2\over\sum_{i=1}^nv_i^2}\ ,$$ and this is maximal when $v$ is an element of the eigenspace $E_\lambda$ corresponding to the eigenvalue $\lambda:=\max{\rm spec}(M)$.

I really enjoy the answers above, and they help me gain some geometrical intuition. I cannot comment under the answer, so I share them here. Hopefully, someone could help me to correct or modify my answer. Many thanks.

1) Pre-multiplying a symmetric matrix $M$ to a vector $v$ is the same as rotating and stretching this vector in the original space. The result should be $ku$, $u$ is a vector with some direction in the same space and $k$ is a scalar.

2) The eigenvector of $M$ gives a direction in which a vector will remain the same after pre-multiplying by $M$.

3) The Rayleigh quotient can be viewed as the cosine value of the angle between the original vector $v$ and $Mv$, multiplied by a scalar $k$.

Then the answer can be implied by those facts.