Hessian matrix defines bilinear form that outputs second-order directional derivatives

Statement

Suppose $f$ is a function of $n$ variables $x_{1},x_{2},\dots ,x_{n}$ , which we think of as a vector variable ${\overline {x}}$ . Suppose ${\overline {u}},{\overline {v}}$ are unit vectors in $n$ -space. Then, we have the following:

$D_{\overline {v}}(D_{\overline {u}}(f))={\overline {u}}^{T}H(f){\overline {v}}$

where ${\overline {u}},{\overline {v}}$ are treated as column vectors, so ${\overline {u}}^{T}$ is ${\overline {u}}$ as a row vector, and ${\overline {v}}$ is ${\overline {v}}$ as a column vector. The multiplication on the right side is matrix multiplication. Note that this tells us that the bilinear form corresponding to the Hessian matrix outputs second-order directional derivatives.

Note further that if the second-order mixed partials are continuous, this forces the Hessian matrix to be symmetric by Clairaut's theorem on equality of mixed partials, which means that the bilinear form we obtain is symmetric, and hence, we will get:

$D_{\overline {v}}(D_{\overline {u}}(f))=D_{\overline {u}}(D_{\overline {v}}(f))$