matrix multiplication can be thought of as taking partial derivatives?

Question

I'm struggling with a sentence in the following text: ross.

It says that the action $y=Ax$ is

like performing many inner products, one for each row of $A$ -

which I agree with. But then it says:

Generally, the matrix elements can be thought of as the partial derivatives

$A_{ij} = \frac{dy_i}{dx_j}$

Why is that true?

Answer 1

Consider a linear function $y:\mathbb{R}^n\longrightarrow \mathbb{R}^n$. $$y_i = \sum_{l=1}^n A_{il}x_l$$then $$\frac{\partial y_i}{\partial x_j} = \frac{\partial }{\partial x_j}\left(\sum_{l=1}^n A_{il}x_l\right) = \sum_{l=1}^n A_{il}\frac{\partial x_l}{\partial x_j} = \sum_{l=1}^n A_{il}\delta_{lj} = A_{ij}$$

Answer 2

Let's take an example: $$\mathbf{y} = A\mathbf{x},$$ where $$A=\begin{bmatrix}1&2\\3&4\end{bmatrix}.$$

So, for instance, if $\mathbf{x}=(1,1)$, we have $$\mathbf{y} = \begin{bmatrix}1&2\\3&4\end{bmatrix} \begin{bmatrix}1\\1\end{bmatrix}=\begin{bmatrix}3\\7\end{bmatrix}$$

The partial derivative $\frac{\partial y_i}{\partial x_j}$ describes what happens to $y_i$ if you change $x_j$, leaving the rest constant. Let's see what happens if you increase $x_1$ by $1$, that is $$\mathbf{y} = \begin{bmatrix}1 & 2\\ 3 & 4 \end{bmatrix} \begin{bmatrix}2\\ 1\end{bmatrix}=\begin{bmatrix}4\\10\end{bmatrix}.$$ As you can see, the $y_1$ coordinate increases by $1$, and the $y_2$ coordinate increases by $3$. This directly corresponds to the elements $A_{11}$ and $A_{21}$ in the matrix, and that is why we can consider them as partial derivatives. You can check for yourself that if you increase $x_2$ by $1$, then $y_1$ will increase by $2$ (corresponding to $A_{12}$), and $y_2$ will increase by $4$ (corresponding to $A_{22}$).