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I am very confused reading all those symbols I do not understand, I am just trying to find partial derivative when two matrices are multiplied.

For an example, suppose that $A$ is a $3\times 4$ matrix and that $B$ is a $4\times 2$.

After I do multiplication $f(A,B) = AB$, I get a matrix $3\times 2$ matrix $C$.

Now my question is, if I try to find the partial derivative of $f(A,B)$ with respect to $B$, do I use the same rule as with scalars? I guess not, correct, so would anyone be so nice and tell to me a simple person how I could solve this. Please do not write in equations, because I can not read them! I understand mathematics only from intuition.

diligar
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filtertips
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  • I'm sorry, I don't understand what you are asking. What is $f$? What is $F$? How are you defining the derivative of a function with respect to a matrix? Where are you seeing this notation? – Xander Henderson Mar 22 '18 at 22:32
  • F is f = function. f(A,B) = A dot B Hopefully you understand me know:) – filtertips Mar 22 '18 at 22:35
  • So $f$ is just multiplication of the two functions? And $F$ is the same thing? Why have you used different symbols for them, then? Also, what about the rest of my questions? – Xander Henderson Mar 22 '18 at 22:38
  • like I said... I can not read symbols. I am sorry. I don't understand what you mean, but I guess, if we have partial derivatives with ordinary numbers, we must have it with matrixes as well, right? – filtertips Mar 22 '18 at 22:42
  • @XanderHenderson And when we try to find partial derivatives with scalars... like x,y,z, we have rules.. is there something with matrixes... thanks – filtertips Mar 22 '18 at 22:44
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    Symbols are the language of mathematics. If you can't understand the basic notation, then I highly doubt that you are going to make any progress on the question that you are asking. As was said in response to another question you asked on the same topic, this is "Utterly incomprehensible". – Xander Henderson Mar 22 '18 at 22:46
  • ahh... I understand math in full depth, I just can't read the symbols...ah never mind. Thank you for your time. – filtertips Mar 22 '18 at 22:51
  • Short solution based on the most elementary definitions of the subject (but nothing "full depth", sorry about that... or not): let $g(B)=f(A,B)$, then $g(B+tH)=A(B+tH)=AB+tAH=g(B)+tAH$ hence $\langle\nabla g(B),H\rangle=AH$, which some would shorten into $\frac{\partial g(B)}{\partial B}=A$, I guess. – Did Mar 23 '18 at 06:12
  • @Did Thanks my friend :) – filtertips Mar 23 '18 at 09:23

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Each entry of the $3 \times 2$ matrix $C$ is a function of the entries of $B$, so we can take the partial derivative with respect to each entry of $B$.

Since $c_{ij} = a_{i1} b_{1j} + a_{i2} b_{2j} + a_{i3} b_{3j} + a_{i4} b_{4j}$, one could write $\frac{\partial c_{ij}}{\partial b_{kj}} = a_{ik}$ and $\frac{\partial c_{ij}}{\partial b_{k\ell}} = 0$ for $\ell \ne j$. Then you can somehow arrange this into a $3 \times 2 \times 4 \times 2$ array of partial derivatives and call this the "partial derivative of $f$ with respect to $B$."

angryavian
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