maximum that a function attains on unit ball

Question

Let $A$ be a symmetric matrix of size $n\times n$, and assume that it's eigenvalues, $\lambda_1 < \ldots < \lambda_{n}$ are all different, and that at least one of them is positive.
Also, let B be the unit ball in $\mathbb{R}^{n}$, namely $B = \left\{ x\in \mathbb{R} ^{n}| \left\| x\right\| _{2}\leq 1\right\}$.
we define a function $f:B \rightarrow \mathbb{R}$, by $f(x) = x^{t}Ax$ for every $x \in B$.
find the maximum of $f$ on $B$, and the point $x_0 \in B$ on which $f$ is maximized. write the answer in terms of the eigenvalues and/or eigenvectors of A.
My attempt:
Because $A$ is symmetric, from the spectral theorem it's have an orthonormal basis of eigenvectors, let's denote it by $U = \left( u_{1}, \ldots ,u_{n} \right)$, where each $u_{i}$ is an eigenvector of the eigenvalue $\lambda_{i}$.
So we can write every $x \in \mathbb{R}^{n}$ as: $$x = a_{1}u_{1} + \ldots + a_{n}u_{n}$$ So we get: $$f\left(x\right) = x^{t}Ax = \left(\sum ^{n}_{i=1}a_{i}u_{i}\right)^{t}A\left(\sum ^{n}_{j=1}a_{j}u_{j}\right) = \left(\sum ^{n}_{i=1}a_{i}u_{i}\right)^{t}\left(\sum ^{n}_{j=1}a_{j}\lambda _{j}u_{j}\right)=$$ $$\sum ^{n}_{i=1}\left( a_{i}\right) ^{2}\lambda _{i}\left\| u_{i}\right\| = \sum ^{n}_{i=1}\left( a_{i}\right) ^{2}\lambda _{i}$$
In addition, because $x \in B$, it must follows that: $$\left\| x\right\| =\left\| \sum ^{n}_{i=1}a_{i}u_{i}\right\| \leq 1$$ meaning we must have: $$\left\| \sum ^{n}_{i=1}a_{i}u_{i}\right\| ^{2}= \langle \sum ^{n}_{i=1}a_{i}u_{i}| \sum ^{n}_{j=1}a_{j}u_{j}\rangle = \sum ^{n}_{i=1}\left( a_{i}\right) ^{2}\leq 1$$ but from here on I couldn't see how I can find the maximum point and value of $f$. Also, I was guided that I am not allowed to use lagrange multipliers, so I'm pretty stuck on it for a while, any help would be appreciated.
I thought about taking each $a_{i}$ to be $a_{i} = \dfrac{1}{\sqrt{n}}$, but couldn't prove that it is the maximum

Answer 1

Assume $\lambda_1>0$ and is the largest eigenvalue. Observe that

1. The max is achieved on the boundary of the ball. This is because the max is positive (just take $a_1=1, a_j=0$ for $j\neq 1$) and if it were achieved at some point inside,$a=(a_j)$ with $\sum a_j^2<1$, we could replace $a$ by $ka$ with some $k>1$ so that the new point $ka$ is on the boundary and the value of $f$ increases by $k^2$ times.

2. $\sum (a_j)^2\lambda_j\le \lambda_1\sum (a_j)^2=\lambda_1$

3. $f(1,0,0,\dots)=\lambda_1$

Conclusion: The max of $f$ is $\lambda_1$, achieved at $(1,0,0,\dots)$, which is the corresponding (normalized) eigenvector.