For $d$ and $e$ is this convex function convex?

Question

Let the function $f \colon \mathbb{R}^2 \longrightarrow \mathbb{R}$ be defined by $$ f(\mathbf{v}) = \frac{1}{2}\mathbf{v}^{T}\begin{bmatrix} d & e\\ e & d \end{bmatrix} \mathbf{v} $$

Answer 1

For a quadratic function $\mathbf{v}^TQ\mathbf{v}$ to be convex, the matrix $Q$ must be positive semidefinite (PSD). For strict convexity, $Q$ must be positive definite. [https://web.stanford.edu/~boyd/cvxbook/]

PSD and PD conditions can be verified using eigenvalues of the matrix. Set of two eigenvalues of given matrix is $\lambda = d \pm e$. So

If $\lambda = d \pm e \geq 0 \implies $ matrix $Q$ is PSD $\implies f(\mathbf{v})$ is convex

If $\lambda = d \pm e > 0 \implies $ matrix $Q$ is PD $\implies f(\mathbf{v})$ is strictly convex.