Is $\frac{\partial X}{\partial I_{rs}}=x_{rs}$ right?Proof of $\frac{\partial y_{mn}}{\partial X}=\mathbf A^T \mathbf I_{mn} \mathbf B^T$

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The book said that $\mathbf X$ is a real matrix,and $\frac{\partial\mathbf X}{\partial x_{rs}}=\mathbf I_{rs}$.Can i think as because $\frac{\partial\mathbf X}{\partial \mathbf I_{rs}}=x_{rs}$,so $\frac{\partial\mathbf X}{\partial x_{rs}}=\mathbf I_{rs}$ ?

I mean ,is $\frac{\partial\mathbf X}{\partial \mathbf I_{rs}}=x_{rs}$ right?Because i think this formula$,\frac{\partial\mathbf X}{\partial \mathbf I_{rs}}=x_{rs}$, is easier to explain.

Explaination:If you want to get an element,which is at the $r$ row and $s$ clomn , in the $\mathbf X$ matrix ,then derivative the $\mathbf I_{rs}$ so that you can extract the $x_{rs}$ from $\mathbf X$ matrix

The another question is according to $\frac{\partial\mathbf X}{\partial x_{rs}}=\mathbf I_{rs}$,and $\mathbf Y =\mathbf A \mathbf X \mathbf B$,then $\frac{\partial\mathbf Y}{\partial x_{rs}}=\mathbf A \mathbf I_{rs} \mathbf B$ if only if $\frac{\partial y_{mn}}{\partial X}=\mathbf A^T \mathbf I_{mn} \mathbf B^T$.

I can understand why is $\frac{\partial\mathbf Y}{\partial x_{rs}}=\mathbf A \mathbf I_{rs} \mathbf B$,but i can't understans why is $\frac{\partial y_{mn}}{\partial \mathbf X}=\mathbf A^T \mathbf I_{mn} \mathbf B^T$,can anyone explain this to me?