My former multivariable calculus professor denoted the gradient as $$\mathrm{grad}\,\mathit{f}\, (x;y)= \begin{pmatrix}x\\y\end{pmatrix}.$$ Completely forgot that. Since then I've always been writing the gradient as a row vector. Is there a difference? Googled it, but I'm not satisfied. Some even make a difference between the vector derivative and gradient. I'm a bit confused now. What is right? Is there an actual difference?

The reason for my question is a matrix derivate problem. In one of my older question I got a hint to a problem:

$$\frac {\partial(a^Tx)} {\partial x}=\frac {\partial(x^Ta)} {\partial x}=a$$

I tried to learn this rule. My thoughts are as follow. Both numerators are the same because they result in the same scalar. The $x$ in the denominator is a column vector. Therefore the result (here $a$) should be a column vector too when taking the derivate.

On the other hand, taking a vector derivative (or the gradient?) of a scalar (or a function) should result in a row vector $a^T$.

Additionally, if the numerator would be a vector instead of a scalar, then the result would be a Jacobean right? And Jacobeans have on one row all the derivatives respective to a $x_m$, and the $numerators_m$ in the columns. Keeping that form the output of the hint should be $a^T$, shouldn't it?

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    $\begingroup$ The gradient should be a row vector. One reason is that it defines the directional derivative $\nabla_v(f) = v\cdot\nabla f$, and the dot product is natural between row & column vectors, but not between two column vectors; the transpose requires a metric. $\endgroup$
    – mr_e_man
    Aug 7, 2018 at 8:47
  • $\begingroup$ @mr_e_man thank you! does this imply that the result of the hint should also be $a^T$? $\endgroup$
    – Mr.Sh4nnon
    Aug 7, 2018 at 8:48
  • $\begingroup$ Yes; $a\cdot x$ is a scalar, so its gradient should be a row vector. But you should check the definition of $\partial/\partial x$, which might be the "vector derivative". "Vector" usually means a column, while "covector" means a row. $\endgroup$
    – mr_e_man
    Aug 7, 2018 at 8:50
  • $\begingroup$ physics.stackexchange.com/q/126740/27215 $\endgroup$
    – mvw
    Aug 7, 2018 at 8:54
  • $\begingroup$ @mr_e_man so this answer would than be false? or does he just use a different definition? math.stackexchange.com/questions/1887688/… $\endgroup$
    – Mr.Sh4nnon
    Aug 7, 2018 at 9:11


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