# Inverse of partial derivative

Good afternoon everyone,

I'm posting a similar but different question with respect to this post.

I have two vectors $a \in \mathbb{R}^M$ and $b \in \mathbb{R}^N$. Assume that I know the partial derivative $\frac{\partial a}{\partial b} \in \mathbb{R}^{M\times N}$ and, from that, I would like to know its inverse $\frac{\partial b}{\partial a} \in \mathbb{R}^{N \times M}$. Also assume that the functions are continuous and differentiable.

I would like to know if there is a relationship between the two maps because I know that, if both vectors are of the same dimensions, it is valid that $\frac{\partial b}{\partial a} = \left(\frac{\partial a}{\partial b}\right)^{-1}$. In my case, do I have to use the Moore-Penrose pseudoinverse? Or is there any more elegant solution?

Moreover, I personally think it is wrong to adjust the previous computation for vectors.

$\frac{\partial b_j}{\partial a} = \left(\frac{\partial a}{\partial b_j}\right)^{-1}$

for any component of $b$. Here I obtain a vector $\in \mathbb{R}^{1 \times M}$ that I should obtain with the pseudoinverse. I think that stacking these vectors ($j \in 1,\dots,N$) is different from the first solution I proposed at the beginning. I think this is wrong. Am I correct?

I also think that computing $\frac{\partial b_j}{\partial a_i} = \left(\frac{\partial a_i}{\partial b_j}\right)^{-1}, \forall i \in \{1,\dots,M\}, j \in \{1,\dots,M\}$ and composing properly the derivative is wrong. In other words: $\left(\begin{matrix} \frac{\partial b_1}{\partial a_1} & \dots & \frac{\partial b_1}{\partial a_M} \\ \dots & \dots & \dots \\ \frac{\partial b_N}{\partial a_1} & \dots & \frac{\partial b_N}{\partial a_M} \end{matrix}\right)$. Am I right that this is wrong?

P.S. just for knowledge, the reason for which the two vectors are of different dimensionality is that $a$ is a unit quaternion in $\mathbb{R}^4$ (even if it has a 1-dim constraint) while $b$ is a standard vector in $\mathbb{R}^3$.

First, I think you should be a little more precise: If you have two plain old vectors $a$ and $b$ of some dimension, then $\partial a/\partial b$ is either undefined or can be considered the zero matrix, because differentiation always needs a function that is being differentiated.
But I assume you meant something like: You have a differentiable function $b\colon \mathbb R^m \to \mathbb R^n$, so that you can consider the differential $d b(a)/da$. (I just write differential '$d$' instead of '$\partial$', because there are no other arguments to $b$ in my considerations here. But the same holds, if $b$ has more arguments).
You asked about '$\partial a/\partial b$' which I assume means something like: There is an inverse function $b^{-1}\colon U\subset\mathbb R^n \to \mathbb R^m$ to the function $b$ and you want to know something about $db^{-1}/dc$ with $c\in U$. But we know that in order for $db^{-1}/dc$ to exist, $b$ must be injective and $b^{-1}$ has to be differentiable. This means that $b$ (or its restriction to some subset of $\mathbb R^m$) must be a diffeomorphism. But this means that $\dim \mathbb R^m = \dim b(\mathbb R^m)$. If $n<m$ this is impossible. If $n=m$, you know what to do. If $n>m$ then you know that $\dim b(\mathbb R^m) = m$, so $b$ maps to an $m$-dimensional space. This must be a manifold. So you can apply the inverse function theorem for manifolds.
• First of all, thanks for your reply. Ok, you are right I was not so specific. Let me give you some additional details. In my case, $a$ is quaternion function, that lies in $\mathbb{R}^4$ while $b$ is a 3-dimensional vector. So, that's why I asked that question and why I have $M>N$. I know the $\frac{\partial a}{\partial b}$ from calculations I'm not describing here for lack of space. – Neostek Oct 30 '17 at 14:56
• I would suggest trying to write down the inverse function $a^{-1}$. This will probably give you insight about the image of $a$. If you have $a^{-1}$ you can simply derive it. – Wauzl Nov 1 '17 at 9:11