Suppose $ABC = ABD$ and $\text{rank}(AB) = \text{rank}(B)$.

It is said that $\text{rank}(AB) =\text{rank}(B)$ would imply $B = EAB$.

However, I fail to understand this implication as I understand that the rank gives us information about the dimension of column/rank space but not the values in the matrix.

  • $\begingroup$ Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. $\endgroup$ – José Carlos Santos Oct 3 '18 at 12:47
  • $\begingroup$ @user593721: You should provide more information. In particular, what is $E$? Also, when you say "It is said . . .", you should give a reference. Is the problem from a book? If so, which book, which problem? The additional information should be included as part of your post. $\endgroup$ – quasi Oct 3 '18 at 12:48
  • $\begingroup$ E is just another matrix. It is not a problem of a book but part of the answer that was given which I do not understand $\endgroup$ – user593721 Oct 3 '18 at 12:57

Let $U$ be the image of $B$ and $W$ the image of $AB$. We are given that $\dim U=\dim W$ and conclude that $A|_U\colon U\to W$ is an isomorphism. Let $E$ be the inverse of $A$ on $W$ and arbitrary on a complement of $W$. Then $EAB=B$.


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