Let $n\in\mathbb N$ with $n\ge 1$, let $U\subset\mathbb R^n$ be open, let $f:U\to\mathbb R$ belong to $C^2(U)$, and let $x_0\in U$ be a critical point of $f$, meaning $\nabla f(x_0)=0$. Let $Hf_{x_0}\in\mathbb R^{n\times n}$ denote the Hessian matrix of $f$ at $x_0$, with entries $(Hf_{x_0})_{ij}=\partial_{x_i}\partial_{x_j}f(x_0)$. If $v^\top Hf_{x_0}v>0$ for every nonzero $v\in\mathbb R^n$, then $x_0$ is a strict local minimum of $f$, meaning that there exists $\delta>0$ such that $B(x_0,\delta)\subset U$ and $f(x)>f(x_0)$ for every $x\in B(x_0,\delta)$ with $x\ne x_0$. If $v^\top Hf_{x_0}v<0$ for every nonzero $v\in\mathbb R^n$, then $x_0$ is a strict local maximum of $f$, meaning that there exists $\delta>0$ such that $B(x_0,\delta)\subset U$ and $f(x)<f(x_0)$ for every $x\in B(x_0,\delta)$ with $x\ne x_0$. If $Hf_{x_0}$ is indefinite, meaning that there exist nonzero vectors $h,k\in\mathbb R^n$ such that $h^\top Hf_{x_0}h>0$ and $k^\top Hf_{x_0}k<0$, then $x_0$ is not a local extremum of $f$, meaning that $x_0$ is neither a local minimum nor a local maximum of $f$.