I am new to Jacobians and still trying to understand how they work. My understanding so far is as follows:
Suppose I have a function $f$ expressed as:
$$f(b_1(a_1,a_2),b_2(a_1,a_2),b_3(a_1,a_2))$$
And I take the partial differential of it with respect to $[b_1,b_2,b_3]'$ to get:
$$\frac{\partial{f}}{\partial[b_1,b_2,b_3]'}=\begin{bmatrix}\frac{\partial f}{\partial b_1} \\ \frac{\partial f}{\partial b_2} \\ \frac{\partial f}{\partial b_3}\end{bmatrix}$$
Then I can calculate the Jacobian as:
$$J = \begin{bmatrix}\frac{\partial b_1}{\partial a_1} & \frac{\partial b_2}{\partial a_1} & \frac{\partial b_3}{\partial a_1} \\ \frac{\partial b_1}{\partial a_2} & \frac{\partial b_2}{\partial a_2} & \frac{\partial b_3}{\partial a_2} \end{bmatrix}$$
And I can get the partial derivative with respect to $[a_1,a_2]'$ as follows:
$$\frac{\partial{f}}{\partial[a_1,a_2]'}=J\frac{\partial{f}}{\partial[b_1,b_2,b_3]'}$$
$$=\begin{bmatrix}\frac{\partial b_1}{\partial a_1} & \frac{\partial b_2}{\partial a_1} & \frac{\partial b_3}{\partial a_1} \\ \frac{\partial b_1}{\partial a_2} & \frac{\partial b_2}{\partial a_2} & \frac{\partial b_3}{\partial a_2} \end{bmatrix}\begin{bmatrix}\frac{\partial f}{\partial b_1} \\ \frac{\partial f}{\partial b_2} \\ \frac{\partial f}{\partial b_3}\end{bmatrix}$$
$$=\begin{bmatrix}\frac{\partial f}{\partial a_1} \\ \frac{\partial f}{\partial a_2} \end{bmatrix}$$
Before proceeding, I need to ask; is my understanding correct? (i.e. is the above all correct?) If so, and this is my real question, I have some confusion regarding the following example:
Let $f,b_1,b_2$ and $b_3$ be defined as follows:
$$f(b_1(a_1,a_2),b_2(a_1,a_2),b_3(a_1,a_2))=b_1(a_1,a_2)+b_2(a_1,a_2)+b_3(a_1,a_2)$$
$$b_1(a_1,a_2)=2a_1+a_2$$
$$b_2(a_1,a_2)=3a_1+a_2$$
$$b_3(a_1,a_2)=a_2$$
By rearranging the above, it follows that you could define $a_1$ and $a_2$ as functions of $b_1,b_2$ and $b_3$ as follows:
$$a_1(b_1,b_2,b_3)=b_2-b_1$$
$$a_2(b_1,b_2,b_3)=b_3$$
Using the above functions you could now define $\tilde{J}$ to be the following matrix:
$$\tilde{J} = \begin{bmatrix}\frac{\partial a_1}{\partial b_1} & \frac{\partial a_2}{\partial b_1} \\ \frac{\partial a_1}{\partial b_2} & \frac{\partial a_2}{\partial b_2} \\ \frac{\partial a_1}{\partial b_3} & \frac{\partial a_2}{\partial b_3} \end{bmatrix}$$
By the same logic as previously, does it now hold that:
$$\frac{\partial{f}}{\partial[b_1,b_2,b_3]'}=\tilde{J}\frac{\partial{f}}{\partial[a_1,a_2]'}$$
$$=\tilde{J}J\frac{\partial{f}}{\partial[b_1,b_2,b_3]'}$$
This last line confuses me. The reason for my confusion is that the matrix rank of $J$ and $\tilde{J}$ in this example are $2$, yet the above seems to imply that $\tilde{J}J$ is equal to the rank $3$ identity matrix, which isn't possible.
I am very confused by this. Could anyone explain where the mistake in my reasoning/understanding is?
Best Answer
You have the right idea and the full Jacobian (with respect to $[a_1,a_2]$) is correct.
While your equation for for $a_1$ is implied by the definitions of $(b_i)_{1\leq i\leq 3}$, it is not equivalent, i.e. you can not recover all three of the definitions of $(b_i)_{1\leq i\leq 3}$ from your definitions describing $(a_i)_{1\leq i\leq 2}$.
The first system in your example involves three equation constraints and two free variables. The second system involves two equation constraints and three free variables. Naturally, this changes the structure of the resulting Jacobian.