Timeline for Solve for $A$ and $B$ in $AXB=Y$
Current License: CC BY-SA 3.0
10 events
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| May 15, 2013 at 8:51 | comment | added | Tom | @ Peter. I was never claiming that I could solve the actual question. In fact, I stated that I can't. I said in the second line that I can give a linear polynomial (in an answer as I haven't got enough rep for commenting) and was then asked to elaborate which I did. | |
| May 15, 2013 at 5:29 | comment | added | Peter Michor | @Tom: Your last example is a sum of matrix products, not one product. Note that the rank of $AXB$ is $\le$ the minimum of the ranks, since it is the dimension of the image. | |
| May 14, 2013 at 20:07 | comment | added | Tom | I guess we are not on the same page. But, if I take $X$ to be a non-zero number -denoted by $x$- and $Y$ to be $x \cdot I_n$, then there is a solution, although $x$ has rank 1 and $Y$ rank n. Indeed, denoting by $e_i$ the i-th basis vector we have $$Y=\sum_i e_i \cdot x \cdot e_i^t=\sum_i x \cdot e_i \cdot e_i^t.$$ I hope haven't made new mistakes now. | |
| May 14, 2013 at 13:05 | comment | added | Tom | I have to admit that I am puzzled now. Peter, could you be so kind to give a short example ? Furthermore, wouldn't that make your comment the desired answer ? btw I just clarified the notation above. | |
| May 14, 2013 at 12:59 | history | edited | Tom | CC BY-SA 3.0 |
clarified what x_{ij} means
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| May 14, 2013 at 12:13 | comment | added | Peter Michor | If (after evaluating $x_1,\dots,x_r$ at some points in $\mathbb Z$ or $\mathbb R$) the rank of $X$ is smaller than the rank of $Y$ there can be no solution. So there is a gap in your proof. | |
| May 13, 2013 at 18:50 | history | edited | Emil Jeřábek | CC BY-SA 3.0 |
fix markup
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| May 13, 2013 at 17:52 | history | edited | Tom | CC BY-SA 3.0 |
I elaborated as asked for
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| May 11, 2013 at 5:47 | comment | added | Turbo | Could you elaborate? | |
| May 9, 2013 at 14:16 | history | answered | Tom | CC BY-SA 3.0 |