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Qustion:

if Matrix equation $AY = C$ and $ZB = C$ has solution,

show that:

the equation of $AXB= C$ has solution

This problem is from this PDF(page 3) problem 2 :http://wenku.baidu.com/view/d6625d1cff00bed5b9f31dba.html

My idea:we have $$\left(A\bigotimes I\right)\overline{Y}=\overline{C}$$ $$\left(I\bigotimes B^T\right)\overline{Z}=\overline{C}$$ we only prove $$\left(A\bigotimes B^T\right)\overline{X}=\overline{C}$$ has solution.

since $$\left(A\bigotimes I\right)\left(I\bigotimes B^T\right)=A\bigotimes B^T$$ then ony prove follow $$I_{m}\left(A\bigotimes I\right)\bigcap I_{m}\left(I\bigotimes B^T\right)\subset I_{m}\left(A\bigotimes B^T\right)$$ My idea is wrong? If is true,and follow How works? Thank you

and maybe have other methods.Thank you very much!

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  • $\begingroup$ For matrices $A$, $B$, it is not always true that $Im(A)\cap Im(B)\subset Im(AB)$. So, I wouldn't try this approach. $\endgroup$ – Sungjin Kim Nov 29 '13 at 9:18
  • $\begingroup$ Why $Im(A)\bigcap Im(B)\subset Im(AB)$ this is not always true? $\endgroup$ – user94270 Nov 29 '13 at 9:24
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    $\begingroup$ Consider nilpotent matrix $A$. In particular, if $A$ is a nonzero matrix with $A^2=0$, then $Im(A)=Im(B)\neq (0)$, but $Im(A^2)=Im(0)=(0)$. $\endgroup$ – Sungjin Kim Nov 29 '13 at 9:28
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    $\begingroup$ This exercise shows that if $im(C)\subset im(A)$ and $\ker(B)\subset\ker(C)$, then the equation $AXB=C$ has at least one solution. Moreover, the converse is obviously true. $\endgroup$ – loup blanc Dec 6 '13 at 19:27
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I will assume that your matrices are over a regular ring (maybe even over a field). Then each of these is regular. By this I mean for $A$ there is matrix $A^-$ such that $AA^-A=A$.

Then the first two equations you have imply that

$AA^-C=C$ and

$CB^-B=C$.

Now you may like to guess what a solution of $AXB=C$ may be. If not see bellow.

Answer $X=A^- CB^-$.

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  • $\begingroup$ $CB^{-}B=C$ a typo. $\endgroup$ – Sungjin Kim Nov 29 '13 at 9:16

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