How to prove that the maximal eigenvalue of a matrix is $ \max\left\{ x^{T}Ax\thinspace|\thinspace x\in\mathbb{R}^{n},x^{T}x=1\right\} $ Let $A$ be a symmetric matrix. How can I prove that $$ \max\left\{ x^{T}Ax\thinspace|\thinspace x\in\mathbb{R}^{n},x^{T}x=1\right\}  $$
Is the biggest eigenvalue of $ A $?
After reading the hints from the comments:
Say $ v_{1},...,v_{n} $ is an orthogonal basis composed of eigenvectors of $A$ and each $v_i$ correspond to an eigenvalue $\lambda_i$. Then for any eigenvector $x $ with eigenvalue $\lambda$ we have
$ Ax=\lambda x $
So that $$ A\left(\sum_{k=1}^{n}\frac{<x,v_{k}>}{||v_{k}||^{2}}v_{k}\right)=\sum_{k=1}^{n}\frac{<x,v_{k}>}{||v_{k}||^{2}}Av_{k}=\sum_{k=1}^{n}\frac{<x,v_{k}>}{||v_{k}||^{2}}\lambda_{k}v_{k}=\lambda x=\underset{k=1}{\overset{n}{\sum}}\lambda\frac{<x,v_{k}>}{||v_{k}||^{2}}v_{k} $$
So that $ \sum_{k=1}^{n}\left(\frac{<x,v_{k}>}{||v_{k}||^{2}}\lambda_{k}-\lambda\frac{<x,v_{k}>}{||v_{k}||^{2}}\right)v_{k}=0 $
How do I proceed?
 A: $x^TAx=\langle x, Ax\rangle$ (the standard dot product)$\le |x||Ax|=|Ax|$. So the maximum of $x^TAx$ does not exceed the operator norm of $A$ which is the maximal absolute value of an eigenvalue of $A$.
A: If you take an orthonormal eigenbasis, which exists by the spectral theorem, the problem will boil down to the following: if $(a_1, \ldots, a_n), (\lambda_1, \ldots, \lambda_n)\in \mathbb R^n$ with $\sum a_i^2=1$ and $\lambda_{\max}=\max{\lambda_i}$, then $\sum \lambda_i a_i^2 \leq \sum \lambda_{\max} a_i^2 \leq \lambda_{\max}$, which holds because $a_i^2\geq 0$ for all $i$.
A: Let $a$ be a real number and $v$ be a unit vector such that $a=v^TAv=\max_{x^Tx=1}x^TAx$. Then $P=aI-A$ is positive semidefinite (because it is real symmetric and $x^TPx\ge0$ for every vector $x$). However, from $v^TPv=0$, one can infer that $Pv=0$. Therefore $Av=av$, i.e., $a$ is an eigenvalue of $A$. It must be the largest among all eigenvalues of $A$, because for every unit eigenvector $x$ corresponding to an eigenvalue $\lambda$ of $A$, we have $\lambda=x^TAx\le a$.
