Eigenvalues of hessian are (2,-1,-1). Does this mean that the function is not convex?

Question

I have to check the convexity of following function: $$ x_1 x_2+x_1 x_3+x_2 x_3$$ I found the hessian to be $$ \begin{bmatrix}{0} && {1} && {1} \\ {1} && {0} && {1} \\ {1} && {1} && {0}\end{bmatrix} $$ The corresponding eigenvalues for $ |H-\lambda I| = 0 $ are (2,-1,-1). Does this mean that the function is not convex.

Answer 1

Let $f(x_1,x_2,x_3)=x_1 x_2+x_1 x_3+x_2 x_3$. Then $f \in C^2( \mathbb R^3)$.

$f$ is convex $ \iff$ the hessian of $f$ is positive semi- definit.

The matrix $\begin{bmatrix}{0} && {1} && {1} \\ {1} && {0} && {1} \\ {1} && {1} && {0}\end{bmatrix}$ is indefinit. Thus $f$ is not convex.