Given 2 matrixes A and B: Assuming that A and B are both nxn and that ≠ 0 and ≠ 0. Does rank()=rank()? I’ve tried like 8 different examples and they all turned out to be equal but I’m not sure whether it actually is or not, can someone help me prove/disprove this?
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2You can see counterexamples here: https://math.stackexchange.com/questions/77650/is-the-rank-of-ab-always-equal-to-the-rank-of-ba – Snowball May 06 '21 at 06:36
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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. – José Carlos Santos May 06 '21 at 06:38
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1Just an explanation about why all your examples turn out to work : if you take "random" matrices, they have a high probability to be invertible. And of course, if $A$ and $B$ are invertible, then $AB$ and $BA$ do have the same rank. So if you want a counterexample, you must use singular matrices. – TheSilverDoe May 06 '21 at 07:16