Differentiating a vector valued function giving a row vector?

Question

If $f:\mathbb R^n \to \mathbb R$, why is $f'(u)$ a $1 \times n$ row vector? (for any $u \in \mathbb R^n$). Many thanks!

Answer 1

Notice that $f'$ by definition is a linear map $\mathbb{R}^n \to \mathbb{R}$. Therefore it has to be a $1 \times n$-Matrix which is the same as a row vector.

Answer 2

The derivative of $f$ at point $x$ is a linear functional $L$ on $\Bbb R^n$ such that

$$f(x+\Delta x) = f(x) + L(\Delta x) + o(\|\Delta x\|).$$

The linear functionals in $\Bbb R^n$ are represented by a scalar multiplication by some fixed vector (Riesz representation theorem), i.e. $\exists v\in \Bbb R^n$ such that $L(x) = (v,x)$.

Finally, the scalar product of two column vectors $(v,x)$ is equivalent to the product $v^Tx$, that is why we can say that the gradient $\nabla_x f$ is a row vector.