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 Oct5 awarded Commentator Oct5 comment Dimensions of matrix derivative and chain rule I'm a bit confused by notation. What is $D_A Inv \cdot A$? And what is matrix $H$? Oct4 asked Dimensions of matrix derivative and chain rule Jun18 comment Solve matrix equation $XB + CX^{-1} = aI$ Oh, I just meant that my initial problem can be reduced to another equation, of course X, B and C would be different. Regarding 3) is it somehow connected to the my last question? I cannot get that equality without commutativity of matrices B and X and matrices B^(1/2) and X. But it seems that if X is supposed to be > 0 then B and X must commute (without requiring B to be > 0), but why B^(1/2) must commute? Jun18 comment Solve matrix equation $XB + CX^{-1} = aI$ how did you use the fact that $B$ is strictly positive definite? Isn't positive semi-definiteness sufficient here? BTW how to show that $(B^{1/2}XB^{1/2})^2 = BX^2B$? Jun18 comment Solve matrix equation $XB + CX^{-1} = aI$ I've just checked: 1) I can formulate my equation as $BX + CX^{-1} = I$ as well. 2) Matrix B is difference of two positive semidefinite matrices and matrix C is positive semidefinite too. Unfortunately I have no guarantee that either B or C is positive definite Jun17 comment Solve matrix equation $XB + CX^{-1} = aI$ Wow, very neat solution. Could you please explain how to get the expression for $Y$ for the last equation? Jun17 awarded Supporter Jun17 comment Solve matrix equation $XB + CX^{-1} = aI$ Thank you for your answer, Robert. In my case B and C are not even diagonal matrices, probably I should mention this in my question Jun17 comment Solve matrix equation $XB + CX^{-1} = aI$ @SebastienB, unfortunately no Jun17 revised Solve matrix equation $XB + CX^{-1} = aI$ deleted 95 characters in body Jun17 asked Solve matrix equation $XB + CX^{-1} = aI$ Jun16 revised Derivative of column-row multiplication added 201 characters in body Jun16 awarded Student Jun16 awarded Editor Jun16 revised Derivative of column-row multiplication edited title Jun16 comment Derivative of column-row multiplication Thank you for your answer, Matt. Could you please explain the first step? How can you rewrite $(x - Ab)(x - Ab)^T$ as $(x - Ab)^T(x - Ab)$ and what is the reason for doing this? It seems that I miss something Jun16 comment Derivative of column-row multiplication @par, $x x^T$ is not scalar, it's a matrix, so I'm not sure that this would help me Jun16 comment Derivative of column-row multiplication @TZakrevskiy, not yet, but it seems that I have to as I couldn't find any standard related formula Jun16 asked Derivative of column-row multiplication