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Kaplan's Advanced Calculus defines the gradient of a function $f:\mathbb{R^n} \rightarrow \mathbb{R}$ as the $1 \times n$ row vector whose entries respectively contain the $n$ partial derivatives of $f$. By this definition then, the gradient is just the Jacobian matrix of the transformation.

We also know that using the Riesz representation theorem, assuming $f$ is differentiable at the point $x$, we can define the gradient as the unique vector $\nabla f$ such that

$ df(x)(h) = \langle h, \nabla f(x) \rangle, \; h \in \mathbb{R}^n $

Assuming we ignore the distinction between row vectors and column vectors, the former definition follows easily from the latter. But, row vectors and column vectors are not the same things. So, I have the following questions:

  1. Is the distinction here between row/column vectors important?
  2. If (1) is true, then how can we know from the second defintion that the vector in question is a row vector and not a column vector?
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    @ItsNotObvious I'm a little confused about your equation $df(x)(h)=\left\langle{h},\nabla{f}(x)\right\rangle$. I intuit gradients as row vectors (cotangent covectors) that are multiplied *on-the-right* by differential tangent (column) vectors, which would be your $h$'s. I thus expected to see $\left\langle\nabla{f}(x),h\right\rangle$. Maybe my intuition conflates differential forms and gradients (see answer below), and I'm just accustomed always to think about differential forms.2018-03-03

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Yes, the distinction between row vectors and column vectors is important. On an arbitrary smooth manifold $M$, the derivative of a function $f : M \to \mathbb{R}$ at a point $p$ is a linear transformation $df_p : T_p(M) \to \mathbb{R}$; in other words, it's a cotangent vector. In general the tangent space $T_p(M)$ does not come equipped with an inner product (this is an extra structure: see Riemannian manifold), so in general we cannot identify tangent vectors and cotangent vectors.

So on a general manifold one must distinguish between vector fields (families of tangent vectors) and differential $1$-forms (families of cotangent vectors). While $df$ is a differential form and exists for all $M$, $\nabla f$ can't be sensibly defined unless $M$ has a Riemannian metric, and then it's a vector field (and the identification between differential forms and vector fields now depends on the metric).

If one thinks of tangent vectors as column vectors, then $\nabla f$ ought to be a column vector, but the linear functional $\langle -, \nabla f \rangle$ ought to be a row vector. A major problem with working entirely in bases is that distinctions like these are frequently glossed over, and then when they become important students are very confused.


Some remarks about non-canonicity. The tangent space $T_p(V)$ to a vector space at any point can be canonically identified with $V$, so for vector spaces we don't run into quite the same problems. If $V$ is an inner product space, then in the same way it automatically inherits the structure of a Riemannian manifold by the above identification. Finally, when people write $V = \mathbb{R}^n$ they frequently intend $\mathbb{R}^n$ to have the standard inner product with respect to the standard basis, and this equips $V$ with the structure of a Riemannian manifold.

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    Looking at the Jacobian matrix makes it really look like row vectors dont it? i.e thinking of it as a linear map on the space on which it's function act.2018-02-04