Let f(x,y)=x^2-y^2+2xy. Determine whether f is convex on ℝ^2 and justify using the Hessian.

It is not convex, because the Hessian [2 2; 2 -2] has a negative eigenvalue

It is convex, because all Hessian entries are nonnegative

It is not convex, because the Hessian is the zero matrix

It is convex, because the Hessian has trace zero

Exercise · ZW-CONVEXITY-AND-THE-HESSIAN-010 · ★★★ Advanced

Let \(f(x,y)=x^2-y^2+2xy\). Determine whether \(f\) is convex on \(\mathbb{R}^2\) and justify using the Hessian.

Choose one option.

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C. It is not convex, because the Hessian \(\begin{pmatrix}2&2\\2&-2\end{pmatrix}\) has a negative eigenvalue

Explanation

The Hessian has determinant \(-4-4=-8<0\), so its eigenvalues have opposite signs and one is negative. Zero trace does not imply positive semidefiniteness, and the entry \(-2\) is negative.

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Review the concept: Convexity and the Hessian

Criterion characterizing convexity of a twice differentiable function through positive semidefiniteness of its Hessian matrix \(\nabla^2 f(x)\) at every point of the domain.

\[f \text{ convexa} \iff \nabla^2 f(x) \succeq 0 \ \forall x \in \operatorname{dom} f\]

Let \(f(x,y)=x^2-y^2+2xy\). Determine whether \(f\) is convex on \(\mathbb{R}^2\) and justify using the Hessian.

Explanation

Review the concept: Convexity and the Hessian

More practice problems: convexity and the hessian