Spectrum of matrix product

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Let $A,BinmathbbR^mtimes n$ be real-valued $mtimes n$-matrices. What can be said about the relation of the spectra of $A^TB$ and $AB^T$? Is it true that these spectra coincide apart from zero?










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    Let $A,BinmathbbR^mtimes n$ be real-valued $mtimes n$-matrices. What can be said about the relation of the spectra of $A^TB$ and $AB^T$? Is it true that these spectra coincide apart from zero?










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      favorite









      up vote
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      favorite











      Let $A,BinmathbbR^mtimes n$ be real-valued $mtimes n$-matrices. What can be said about the relation of the spectra of $A^TB$ and $AB^T$? Is it true that these spectra coincide apart from zero?










      share|cite|improve this question













      Let $A,BinmathbbR^mtimes n$ be real-valued $mtimes n$-matrices. What can be said about the relation of the spectra of $A^TB$ and $AB^T$? Is it true that these spectra coincide apart from zero?







      linear-algebra matrices






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      asked Sep 6 at 9:34









      julian

      314110




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          Using singular value decomposition or by comparing characteristic polynomials you can show that the eigenvalues of $M$ and $M^T$ agree for a square matrix $M$.



          Hence you can go from $A^T B$ to $B^T A$ without changing the eigenvalues and your question comes down to comparing the eigenvalues of a product $AB$ with those of $BA$. Their eigenvalues also agree, see for example this answer.






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            1 Answer
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            active

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            1 Answer
            1






            active

            oldest

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            active

            oldest

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            active

            oldest

            votes








            up vote
            1
            down vote



            accepted










            Using singular value decomposition or by comparing characteristic polynomials you can show that the eigenvalues of $M$ and $M^T$ agree for a square matrix $M$.



            Hence you can go from $A^T B$ to $B^T A$ without changing the eigenvalues and your question comes down to comparing the eigenvalues of a product $AB$ with those of $BA$. Their eigenvalues also agree, see for example this answer.






            share|cite|improve this answer


























              up vote
              1
              down vote



              accepted










              Using singular value decomposition or by comparing characteristic polynomials you can show that the eigenvalues of $M$ and $M^T$ agree for a square matrix $M$.



              Hence you can go from $A^T B$ to $B^T A$ without changing the eigenvalues and your question comes down to comparing the eigenvalues of a product $AB$ with those of $BA$. Their eigenvalues also agree, see for example this answer.






              share|cite|improve this answer
























                up vote
                1
                down vote



                accepted







                up vote
                1
                down vote



                accepted






                Using singular value decomposition or by comparing characteristic polynomials you can show that the eigenvalues of $M$ and $M^T$ agree for a square matrix $M$.



                Hence you can go from $A^T B$ to $B^T A$ without changing the eigenvalues and your question comes down to comparing the eigenvalues of a product $AB$ with those of $BA$. Their eigenvalues also agree, see for example this answer.






                share|cite|improve this answer














                Using singular value decomposition or by comparing characteristic polynomials you can show that the eigenvalues of $M$ and $M^T$ agree for a square matrix $M$.



                Hence you can go from $A^T B$ to $B^T A$ without changing the eigenvalues and your question comes down to comparing the eigenvalues of a product $AB$ with those of $BA$. Their eigenvalues also agree, see for example this answer.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Sep 6 at 10:37

























                answered Sep 6 at 10:30









                Christoph

                11k1240




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