# Dot product of the gradient of a function

In matrix calculus, I keep on seeing things like $$\langle \nabla f(x), v\rangle$$, which is the dot product of the gradient of a function with a vector.

I was wondering if there is any intuitive understanding of what this means.

For example, we have the Mean Value Theorem:

Let $$\mathcal{O}$$ be an open subset of $$\mathbb{R}^{n}$$ and suppose the mapping $$F : \mathcal{O} \rightarrow \mathbb{R}^{m}$$ is continuously differentiable. Suppose that the points $$x$$ and $$x + h$$ are in $$\mathcal{O}$$ and that the segment joining these points are also in $$\mathcal{O}$$. Then, there exist numbers $$\theta_1, \theta_2, \ldots, \theta_m$$ in the open interval $$(0, 1)$$ such that $$F_{i}(x + h) - F_{i}(x) = \langle \nabla F_{i}(x + \theta_{i}h), h\rangle$$

I was wondering if there is any good way to interpret $$\langle \nabla F_{i}(x + \theta_{i}h), h\rangle$$ in this context.

Thanks

• It can be shown that $\langle \nabla f(x), v \rangle$ is equal to the directional derivative of $f$ at $x$ in the direction $v$. – littleO Mar 23 at 19:36
• product of magnitudes times the cosine of the angle between v and f(x) "uphill" – phdmba7of12 Mar 23 at 19:41

Note that the derivative of $$f\colon\mathbb R^n\to\mathbb R$$ is not a vector, but a linear form instead.
In presence of an inner product $$\langle.,.\rangle$$ the gradient $$\nabla^{\langle .,.\rangle}f$$ in respect to the inner product $$\langle .,.\rangle$$ is the unique vector which represents this linear form in presence of the specified inner product.