A linear function involving the global optimum of the Rayleigh quotient

Question

Given $\mathbf{A}$ is a real symmetric matrix. The leading eigenvalue $\lambda_1$ of $\mathbf{A}$ is equal to: $$\lambda_1 = max_{||\mathbf{x}||=1} \mathbf{x}^T\mathbf{A}\mathbf{x},$$ where the eigenvector $\mathbf{v}_1$ associated with $\lambda_1$ is the optimum of the above optimization: $$\mathbf{v}_1 = argmax_{||\mathbf{x}||=1} \mathbf{x}^T\mathbf{A}\mathbf{x}.$$ Define $t:=\mathbf{v}_1^T \mathbf{c}$, where $\mathbf{c}$ is a real vector with constants. I know $\lambda_1$ is a convex function of $\mathbf{A}$ by the Courant-Fisher Theorem. I am wondering if $t$ is a convex/concave function of $\mathbf{A}$? Is there a principled way to analyze the relations between $t$ and $\mathbf{A}$?

Answer 1

I think this function is not a convex or concave function of $\mathbf{A}$ since the eigenvector will change a lot even with a small perturbation.

Consider $\mathbf{A} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ and $ \mathbf{B} = \begin{pmatrix} 1 & \delta \\ \delta & 1 \end{pmatrix} $ for small $\delta > 0$. Then, the first eigenvector of $\mathbf{A}$ is $(1,0)$, while the first eigenvector of $\mathbf{B}$ is $(1/\sqrt{2},1/\sqrt{2})$. Taking $c = (0,1)$, you can find a counterexample.