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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

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

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