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I know that if A and B are square nxn matrices, then AB and BA have the same characteristic polynomial and thus the same eigenvalues (and same algebraïc multiplicity).

I'm wondering though if this can be generalized: if A is a nxm matrix and B a mxn matrix, then AB is a nxn matrix and BA a mxm matrix. So my question is: will the eigenvalues of AB and BA, that differ from zero, have the same algebraïc multiplicity?

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How on earth can you compare two polynomials while "ignoring the exponent of x"? –  Chris Eagle Mar 17 '13 at 10:56
    
@ChrisEagle He clarified at the end what he actually meant. –  Git Gud Mar 17 '13 at 10:57
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You should use a title that reflects your actual question. The answer to your title question is obviously no, and you seem to realise this. –  Marc van Leeuwen Mar 17 '13 at 10:59
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I guess this theorem will be helpful. –  sos440 Mar 17 '13 at 11:04
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Yes, I messed a little up in formulating the question. I edited my post now ;). –  yarnamc Mar 17 '13 at 11:04

1 Answer 1

Suppose $A$ has $d$ more rows than columns, and therefore that $B$ has $d$ more columns than rows. Add $d$ zero columns to $A$, and $d$ zero rows to $B$, to get square matrices $A',B'$. The product $A'B'$ is identical to $AB$, while $B'A'$ is obtained from $BA$ by adding $d$ zero rows and $d$ zero columns. Since $B'A'$ is block diagonal (actually block-triangular would have sufficed), the characterisitic polynomial of $B'A'$, which is equal to that of $A'B'$ by the result for square matrices, is $X^d$ times the characteristic polynomial of $BA$. Therefore what you guessed is indeed true: $$\chi_{BA}=X^d\chi_{AB}.$$

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Thank you very much, this was exactly what i was looking for :D! –  yarnamc Mar 17 '13 at 11:31

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