Can someone show the equivalence of the following relationship $\dfrac{\partial{v}^Tv}{\partial v}=2v^T$

Question

Let $v$ be an n dimensional vector

Then how can one show that $$\dfrac{\partial{v}^Tv}{\partial v}=2v^T$$

Specifically, I am not quite understanding the difference between taking the derivative against a quantity $v$ versus its transposed quantity $v^T$?

Answer 1

It depends on the convention used to define derivative with respect to a vector.

If you define the derivative with respect to vector $\nu = [\nu_1 \; \nu_2 \; \cdots \;\nu_n]^T$ as: \begin{equation} \frac{\partial }{\partial \nu} = \left[\begin{array}{c}\frac{\partial }{\partial \nu_1} \\ \frac{\partial }{\partial \nu_2} \\ \vdots \\\frac{\partial }{\partial \nu_n} \end{array}\right] \end{equation}

Then the derivative of $\nu^T \nu = (\nu_1^2 + \nu_2^2 + \cdots + \nu_n^2)$ with respect to $\nu$ is given by:

\begin{equation} \frac{\partial \nu^T \nu }{\partial \nu} = \left[\begin{array}{c}\frac{\partial }{\partial \nu_1} (\nu_1^2 + \nu_2^2 + \cdots + \nu_n^2) \\ \frac{\partial }{\partial \nu_2} (\nu_1^2 + \nu_2^2 + \cdots + \nu_n^2) \\ \vdots \\\frac{\partial }{\partial \nu_n} (\nu_1^2 + \nu_2^2 + \cdots + \nu_n^2)\end{array}\right] = \left[\begin{array}{c}2\nu_1 \\ 2\nu_2 \\ \vdots \\ 2\nu_n \end{array}\right] = 2\nu \end{equation}

If you define the derivative with respect to vector $\nu$ as: \begin{equation} \frac{\partial }{\partial \nu} = \left[\begin{array}{c}\frac{\partial }{\partial \nu_1} \frac{\partial }{\partial \nu_2} \cdots \frac{\partial }{\partial \nu_n} \end{array}\right] \end{equation}

Then the derivative of $\nu^T \nu$ with respect to $\nu$ is given by:

\begin{equation} \frac{\partial \nu^T \nu }{\partial \nu} = \left[\begin{array}{c}\frac{\partial }{\partial \nu_1}( \nu_1^2 + \nu_2^2 + \cdots + \nu_n^2 ) \quad \cdots \quad \frac{\partial }{\partial \nu_n} (\nu_1^2 + \nu_2^2 + \cdots + \nu_n^2)\end{array}\right] = \left[\begin{array}{c}2\nu_1 \quad \cdots \quad 2\nu_n \end{array}\right] = 2\nu^T \end{equation}

Answer 2

Actually, it's a matter of convention. Note that $v^Tv$ is a scalar and the derivative $\partial{(v^Tv)}/\partial v$ is a vector with elements $\partial{(v^Tv)}/\partial v_i$. It can be either a row vector ($2v^T$) or a column vector ($2v$).