1
$\begingroup$

I'm new to Matrix Calculus. Recently I've been working on that and have a question. Please see the following:

$J=J(\mathbf{z})$

$\mathbf{z}=\mathbf{W}\mathbf{a}$

Where $J: R^m \rightarrow R$, $\mathbf{z}$ is $m\times1$ vector, $\mathbf{a}$ is $n\times1$ vector and $\mathbf{W}$ is $m\times n$ matrix.

I want to calculate $\frac{\partial{J}}{\partial{\mathbf{W}}}$. A reference paper tells I need to turn $\mathbf{W}$ to vector by stacking the column:

$\frac{\partial{J}}{\partial{vec(\mathbf{W})}} = \frac{\partial J}{\partial \mathbf{z}}\cdot\frac{\partial \mathbf{z}}{\partial vec(\mathbf{W})}$

let $\delta^T = \frac{\partial J}{\partial \mathbf{z}}$ and it is a $1\times m$ vector (numerator layout).

$\frac{\partial \mathbf{z}}{\partial vec(\mathbf{W})} = \frac{\partial \mathbf{W}\mathbf{a}}{\partial vec(\mathbf{W})}=\frac{\partial vec(\mathbf{W}\mathbf{a})}{\partial vec(\mathbf{W})} = \frac{\partial (\mathbf{a}^T \otimes I_{mm})vec(\mathbf{W})}{\partial vec(\mathbf{W})}=\mathbf{a}^T \otimes I_{mm}$, $\otimes$ is Kronecker product.

$\mathbf{a}^T \otimes I_{mm}$ is $m\times mn$ matrix.

$\delta^T\cdot(\mathbf{a}^T \otimes I_{mm}) = [\delta_1\cdot \mathbf{a}^T, \delta_2\cdot \mathbf{a}^T, ..., \delta_m\cdot \mathbf{a}^T]$. If I recover this to matrix by invert stacking column, the result is strage: $$ \begin{matrix} \delta_1a_1 & \delta_{1}a_{m+1} & \dots \\ \delta_1a_2 & \delta_{1}a_{m+2} & \dots \\ \vdots & \vdots & \dots \\ \delta_1a_m & \delta_{2}a_{2m-n} & \dots \end{matrix} $$

This is obviously wrong. It looks strange. And I find another reference, the result should be $\mathbf{\delta}\cdot \mathbf{a}^T$. So I think the $invert\ vec(\cdot)$ should be row stacking, is it right?

$\endgroup$
0
$\begingroup$

emm.... I know where I am wrong. Actually, I took a shower before went to bed yesterday, and during the shower, I knew the mistake.

This result, $\delta^T\cdot(\mathbf{a}^T \otimes I_{mm})$ is right, while this equation $\delta^T\cdot(\mathbf{a}^T \otimes I_{mm}) = [\delta_1\cdot \mathbf{a}^T, \delta_2\cdot \mathbf{a}^T, ..., \delta_m\cdot \mathbf{a}^T]$ is wrong.

It should be:

$\delta^T\cdot(\mathbf{a}^T \otimes I_{mm}) = [a_1\cdot \mathbf{\delta}^T, a_2\cdot \mathbf{\delta}^T, ..., a_m\cdot \mathbf{\delta}^T]$

After inverting this row vector (numerator layout) back to matrix, the final result is: $\delta\cdot\mathbf{a}^T$.

I misunderstood the definition of Kronecker product.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.