Berkeley Problems in Mathematics 7.5.22 This is the problem: Let $A$ be a real symmetric $n \times n$ matrix with non negative entries. Prove that $A$ has an eigenvector with non-negative entries
I looked at the answer key and don't quite understand it. In the expression containing max, why should it correspond to the eigenvalue $\lambda_0$? I thought that this may be because if Ax is parallel to x, then the dot product between $Ax$ and $x$ is maximised, but is it not possible that it still attains a large value if $A$ transforms $x$ in a way that scales x by so much that Ax is large enough to make $\langle Ax,x\rangle$ large even though they may not be parallel?
Solution(as in answer key):
Let $\lambda_0$ be the largest eigenvalue of $A$. We have
$$\lambda_0 = \max{\{\langle Ax, x\rangle\mid x\in\mathbb{R}^n,\|x\| = 1\}}$$
and the maximum it attains precisely when $x$ is an eigenvector of $A$ with 
eigenvalue $\lambda_0$. Suppose $v$ is a unit vector for which the maximum is attained, and let $u$ be the vector whose coordinates are the absolute values of the coordinates of $v$. Since the entries of $A$ are nonnegative, we have
$$\langle Au,u \rangle \ge \langle Ax,x\rangle =\lambda_0$$ implying that $\langle Au,u\rangle = \lambda_0$, so that $u$ is an eigenvector of $A$ for the eigenvalue $\lambda_0$.

By the spectral theorem, a real symmetric matrix can be diagonalized by an orthogonal matrix.  So by (rigid) change of basis, $A$ is diagonal.  (By rigid, we mean that the initial orthonormal basis may be rotated to the basis that makes $A$ diagonal.)  Consequently, embiggening you describe (which is a valid concern in general) cannot occur.
