Is it right in saying that the Jacobian Matrix of a Matrix output of an elementwise operation to the same input is a diagonal matrix ?

Context below.

From ref 1 it is clear that when you have an elementwise operation on a vector; the Jacobian matrix of the function wrto its input vector is a diagonal matrix

For an input vector $\textbf{x} = \{x_1, x_2, \dots, x_n\}$ on which an element wise function is applied; say the activation function sigmoid $\sigma$; and it give the output vector $\textbf{a} = \{a_1, a_2, \dots, a_n\}$ $a_i= f(x_i); \text{ what is } \frac { \partial a}{ \partial x} $

In scalar case this becomes $\frac { \partial f(x)}{ \partial x} = f'(x)$

In Vector case, that is when we take there derivative of a vector with respect to another vector we get the following (square) Jacobian matrix

Example from ref 2

$$ \begin{aligned} \\ \\ \text{The Jacobain, J } = \frac {\partial a}{\partial x} = \begin{bmatrix} \frac{\partial a_{1}}{\partial x_{1}} & \frac{\partial a_{2}}{\partial x_{1}} & \dots & \frac{\partial a_{n}}{\partial x_{1}} \\ \frac{\partial a_{1}}{\partial x_{2}} & \frac{\partial a_{2}}{\partial x_{2}} & \dots & \frac{\partial a_{n}}{\partial x_{2}} \\ \vdots & \vdots & \ddots & \vdots \\ \frac{\partial a_{1}}{\partial x_{n}} & \frac{\partial a_{2}}{\partial x_{n}} & \dots & \frac{\partial a_{n}}{\partial x_{n}} \\ \end{bmatrix} \end{aligned} $$

Above the diagonal of J are the only terms that can be nonzero:

$$ \begin{aligned} J = \begin{bmatrix} \frac{\partial a_{1}}{\partial x_{1}} & 0 & \dots & 0 \\ 0 & \frac{\partial a_{2}}{\partial x_{2}} & \dots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \dots & \frac{\partial a_{n}}{\partial x_{n}} \\ \end{bmatrix} \end{aligned} $$

$$ \text{ As } (\frac{\partial a}{\partial x})_{ij} = \frac{\partial a_i}{\partial x_j} = \frac { \partial f(x_i)}{ \partial x_j} = \begin{cases} f'(x_i) & \text{if $i=j$} \\ 0 & \text{otherwise} \end{cases} $$ And the authors go on to explain that $\frac{\partial a}{\partial x}$ can be written as $\text{diag}(f'(x))$ and the hardmarad or elementwise multiplication ($\odot$ or $\circ$) can be applied instead of matrix multiplication to this Jacobian matrix like $\odot f'(x)$ when applying the Chain Rule and converting from index notation to matrix notation.

Sorry for the rather long explanation. This was mostly to make clear the context. On to the real question.

While implementing the neural network practically the input is not a Vector but an $M*N$ dimensional Matrix ; $M, N > 1$.

Taking a simple $2*2$ input matrix on which the sigmoid activation function is done; the Jacobian of the same is a $8*2$ matrix and no longer a square matrix.

Does it make sense to say the derivative of Matrix $a_{i,j}$ - where an element-wise function is applied; over the input matrix $x_{i,j}$ as a Jacobian.

$$ \frac{\partial a_{i,j}}{\partial x_{i,j}} = J_{k,l} $$

Even if so, there is no certainty that this will be a square matrix and we can generalize to the diagonal ? Am I correct in these above statements?

However, all articles treat this matrix case as a generalization of the Vector case and write $\frac{\partial a}{\partial x}$ as the $\text{diag}(f'(x))$, and then use the element-wise/Hadamard product for the Chain Rule. This way also in implementation. But there is no meaning of diagonal in a non-square matrix; What am I missing ?

Also asked here https://math.stackexchange.com/questions/4397390/jacobian-matrix-of-an-element-wise-operation-on-a-matrix.

  • $\begingroup$ A matrix simply is a vector, written in a rectangular format rather than a linear format. Your problem setting has nothing to do with any actual properties of the matrix per se: "elementwise" refers only to its vector nature. $\endgroup$
    – whuber
    Mar 7, 2022 at 3:39
  • $\begingroup$ I had thought the same, however when implementing taking say a $2*2$ input matrix and writing out the Jacobian matrix in long-form, the dimension of the Jacobian matrix is $8*2$; so the diagonal does not make sense here $\endgroup$ Mar 7, 2022 at 4:14
  • 1
    $\begingroup$ That makes no sense, because all Jacobians are (by construction) square matrices. It sounds like you might have reformatted a $4\times 4$ matrix into an $8\times 2$ rectangle. $\endgroup$
    – whuber
    Mar 7, 2022 at 13:24
  • 1
    $\begingroup$ I think that was the problem $\endgroup$ Mar 8, 2022 at 4:31
  • $\begingroup$ At stats.stackexchange.com/a/257616/919 I outline a rigorous approach to resolving such issues. It helps avoid issues of notation and how vectors might be written, to focus on what the derivative is really doing. It can be a useful way to investigate questions like this one. $\endgroup$
    – whuber
    Mar 8, 2022 at 19:02

1 Answer 1


Adding the answer as per the explanation by @whuber

$$ \begin{aligned} \text{Let's take a 2x2 matrix , X } = \begin{bmatrix} x_{{1}{1}} & x_{{1}{2}} \\ x_{{2}{1}} & x_{{2}{2}} \\ \end{bmatrix} \end{aligned} $$ On which an element wise operation is done $a_{{i}{j}} = \sigma ({x_{{i}{j}}})$

Writing that out as matrix $A$ $$ \begin{aligned} A = \begin{bmatrix} a_{{1}{1}} & a_{{1}{2}} \\ a_{{2}{1}} & a_{{2}{2}} \\ \end{bmatrix} \end{aligned} $$

The partial derivative of the elements of A with its inputs is $\frac {\partial A }{\partial x_{{i}{j}}}$

$$ \begin{aligned} \frac {\partial \vec A }{\partial X} = \begin{bmatrix} a_{{1}{1}} & a_{{1}{2}} & a_{{2}{1}} & a_{{2}{2}} \\ \end{bmatrix} \end{aligned} $$ We vectorized the matrix; Now we need to take the partial derivative of the vector with each element of the matrix $X$

$$ \begin{aligned} \frac {\partial \vec A }{\partial X} = \begin{bmatrix} \frac{\partial a_{{1}{1}} }{\partial x_{{1}{1}}} & \frac{\partial a_{{1}{2}} }{\partial x_{{1}{1}}} & \frac{\partial a_{{2}{1}} }{\partial x_{{1}{1}}} & \frac{\partial a_{{2}{2}} }{\partial x_{{1}{1}}} \\ \\ \frac{\partial a_{{1}{1}} }{\partial x_{{1}{2}}} & \frac{\partial a_{{1}{2}} }{\partial x_{{1}{2}}} & \frac{\partial a_{{2}{1}} }{\partial x_{{1}{2}}} & \frac{\partial a_{{2}{2}} }{\partial x_{{1}{2}}} \\ \\ \frac{\partial a_{{1}{1}} }{\partial x_{{2}{1}}} & \frac{\partial a_{{1}{2}} }{\partial x_{{2}{1}}} & \frac{\partial a_{{2}{1}} }{\partial x_{{2}{1}}} & \frac{\partial a_{{2}{2}} }{\partial x_{{2}{1}}} \\ \\ \frac{\partial a_{{1}{1}} }{\partial x_{{2}{2}}} & \frac{\partial a_{{1}{2}} }{\partial x_{{2}{2}}} & \frac{\partial a_{{2}{1}} }{\partial x_{{2}{2}}} & \frac{\partial a_{{2}{2}} }{\partial x_{{2}{2}}} \\ \\ \end{bmatrix} \end{aligned} $$ The non diagonal terms are of the form $\frac{\partial a_{{i}{j}} }{\partial x_{{k}{k}}}$ and reduce to 0 and we get the resultant Jacobian Matrix as

$$ \begin{aligned} \frac {\partial \vec A }{\partial X} = \begin{bmatrix} \frac{\partial a_{{i}{j}} }{\partial x_{{i}{i}}} & \cdot \cdot \cdot & 0 \\ 0 & \frac{\partial a_{{i}{j}} }{\partial x_{{i}{i}}} & \cdot \cdot \cdot \\ \cdot \cdot \cdot \\ \cdot \cdot \cdot & \cdot \cdot \cdot & \frac{\partial a_{{N}{N}} }{\partial x_{{N}{N}}} \end{bmatrix} \end{aligned} $$

Hence $\frac{\partial a_{{i}{j}}}{\partial X}$ can be written as $\text{ diag}(f'(X))$ ; $(A =f(X))$


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