Which unit vector $\vec x$ maximises $\langle A\vec x|\vec x\rangle$?

Question

Let $A$ be a linear operator on a real finite-dimensional vector space $V$, equipped with the usual Euclidean inner product. Which unit vector $\vec x\in V$ will maximise the dot/inner product $\langle A\vec x|\vec x\rangle$?

Answer 1

You can assume $A$ is Hermitian without loss of generality. Indeed,

$$ A = \frac{A + A^T}{2} + \frac{A - A^T}{2}. $$

The quadratic form over the antisymmetric part evaluates to

$$ \langle (A - A^T) x \mid x \rangle = \sum_{i = 1}^n \sum_{j = 1}^n x_i x_j (A_{ij} - A_{ji}) = \sum_{i = 1}^n \sum_{j = 1}^n x_i x_j A_{ij} - \sum_{i = 1}^n \sum_{j = 1}^n x_i x_j A_{ji}. $$

By relabeling indices in the second sum and switching the order of summation, you can verify that

$$ \langle (A - A^T)x \mid x \rangle = 0. $$

Therefore, the maximum is achieved by the eigenvector corresponding to the algebraically largest eigenvalue of $(A + A^T) / 2$.

Answer 2

Assume $A$ is a symmetric matrix. Then, to maximize with a constraint, you add in the constraint with a Lagrange multiplier, like this

$$\left<Ax|x\right> + λ(1 - \left<x|x\right>),$$

and maximize this by differentiating both sides with respect to $x$ to get

$$Ax + A^T x - λ(x + x) = 0$$

where $A^T$ denotes the transpose of $A$; i.e. $Ax = λx$, since we're assuming $A^T = A$. So, $x$ is an eigenvector of $A$ with eigenvalue $λ$, and - thus

$$\left<Ax|x\right> = \left<λx|x\right> = λ\left<x|x\right> = λ.$$

The problem reduces to that of finding the largest eigenvalue $λ$ and, with it, the eigenvector(s) $x$ that go with it.