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?
linear-algebra matrices
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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?
linear-algebra matrices
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up vote
0
down vote
favorite
up vote
0
down vote
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?
linear-algebra matrices
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
linear-algebra matrices
asked Sep 6 at 9:34
julian
314110
314110
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1 Answer
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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
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
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.
add a comment |Â
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.
add a comment |Â
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.
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.
edited Sep 6 at 10:37
answered Sep 6 at 10:30
Christoph
11k1240
11k1240
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