Eigenvalues of a matrix product
$begingroup$
it is a well know result that for two $ntimes n$ matrices $A$ and $B$ the eigenvalues of the product $AB$ are equal to those of the product $BA$.
Are there similar results for a product of $m>2$ matrices? E.g. does the product of four $ntimes n$ matrices $ABCD$ have the same eigenvalues as $ACBD$?
Thanks for your help!
linear-algebra matrices eigenvalues-eigenvectors
asked Nov 14 '18 at 9:54
maxmilgrammaxmilgram
6577
$endgroup$
add a comment |
$begingroup$
it is a well know result that for two $ntimes n$ matrices $A$ and $B$ the eigenvalues of the product $AB$ are equal to those of the product $BA$.
Are there similar results for a product of $m>2$ matrices? E.g. does the product of four $ntimes n$ matrices $ABCD$ have the same eigenvalues as $ACBD$?
Thanks for your help!
linear-algebra matrices eigenvalues-eigenvectors
asked Nov 14 '18 at 9:54
maxmilgrammaxmilgram
6577
$endgroup$
2
$begingroup$
The only thing that can be said is what can be deduced from the first result, which is that eigenvalues are invariant under cyclic permutations, so $ABCD$ has the same eigenvalues as $DABC, CDAB$, and $BCDA$, and that's it.
$endgroup$
– Qiaochu Yuan
Nov 14 '18 at 10:16
add a comment |
$begingroup$
it is a well know result that for two $ntimes n$ matrices $A$ and $B$ the eigenvalues of the product $AB$ are equal to those of the product $BA$.
Are there similar results for a product of $m>2$ matrices? E.g. does the product of four $ntimes n$ matrices $ABCD$ have the same eigenvalues as $ACBD$?
Thanks for your help!
linear-algebra matrices eigenvalues-eigenvectors
asked Nov 14 '18 at 9:54
maxmilgrammaxmilgram
6577
$endgroup$
it is a well know result that for two $ntimes n$ matrices $A$ and $B$ the eigenvalues of the product $AB$ are equal to those of the product $BA$.
Are there similar results for a product of $m>2$ matrices? E.g. does the product of four $ntimes n$ matrices $ABCD$ have the same eigenvalues as $ACBD$?
Thanks for your help!
linear-algebra matrices eigenvalues-eigenvectors
linear-algebra matrices eigenvalues-eigenvectors
asked Nov 14 '18 at 9:54
maxmilgrammaxmilgram
6577
asked Nov 14 '18 at 9:54
maxmilgrammaxmilgram
6577
asked Nov 14 '18 at 9:54
maxmilgrammaxmilgram
6577
asked Nov 14 '18 at 9:54
maxmilgrammaxmilgram
6577
asked Nov 14 '18 at 9:54
maxmilgrammaxmilgram
6577
6577
2
$begingroup$
The only thing that can be said is what can be deduced from the first result, which is that eigenvalues are invariant under cyclic permutations, so $ABCD$ has the same eigenvalues as $DABC, CDAB$, and $BCDA$, and that's it.
$endgroup$
– Qiaochu Yuan
Nov 14 '18 at 10:16
add a comment |
2
$begingroup$
The only thing that can be said is what can be deduced from the first result, which is that eigenvalues are invariant under cyclic permutations, so $ABCD$ has the same eigenvalues as $DABC, CDAB$, and $BCDA$, and that's it.
$endgroup$
– Qiaochu Yuan
Nov 14 '18 at 10:16
2
2
$begingroup$
The only thing that can be said is what can be deduced from the first result, which is that eigenvalues are invariant under cyclic permutations, so $ABCD$ has the same eigenvalues as $DABC, CDAB$, and $BCDA$, and that's it.
$endgroup$
– Qiaochu Yuan
Nov 14 '18 at 10:16
$begingroup$
The only thing that can be said is what can be deduced from the first result, which is that eigenvalues are invariant under cyclic permutations, so $ABCD$ has the same eigenvalues as $DABC, CDAB$, and $BCDA$, and that's it.
$endgroup$
– Qiaochu Yuan
Nov 14 '18 at 10:16
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let check this on orthogonal matrices (e.g. rotations dim. $3 times 3$).
We have for example $R_1R_1^TR_2R_2^T=I$ with $1$ as eigenvalues. If we change the order of rotations for example $R_1R_2(R_2R_1)^T=R_1R_2R_1^TR_2^T$ certainly if $R_1$ and $R_2$ are not commutative we would not obtain identity matrix and consequently the eigenvalues are different.
answered Nov 14 '18 at 10:24
WidawensenWidawensen
4,50321446
$endgroup$
add a comment |
$begingroup$
In fact, $AB$ and $BA$ have the same characteristic polynomial.
The problem with expecting switches in between various other matrices to preserve eigenvalues, is that the characteristic polynomial does not behave well with multiplication of matrices from left/right.
In other words, if $A$ and $B$ have the same characteristic polynomial, then this may not be true of $CA$ and $CB$, where $C$ is some other matrix. This is because the trace(sum of eigenvalues) is not multiplicative in general, and we can use this to produce counterexamples.
To give an example, take $A = beginbmatrix1 0 \ 0 2 endbmatrix$ and $B = beginbmatrix 1 1 \ 0 2endbmatrix$. These matrices have the same characteristic polynomial. For yourself, find a $C$ such that $CA$ and $CB$ have different trace, and therefore different eigenvalues. Find a $D$ such that $AD$ and $BD$ have different trace. It follows that $CA,CB$ and $AD,BD$ don't have the same characteristic polynomials.
The stumbling block in trying to prove, say , that $CBA$ and $CAB$ have the same eigenvalues, is the inability to assure that multiplying by $C$, after knowing that the eigenvalues of $AB$ and $BA$ are the same, retains the equality of eigenvalues. This is why you can use the $C,A,B,D$ above to find counterexamples of the assertions you were making.
However, cyclic permutations retain eigenvalues, because we are just applying the "two-matrix" case to deduce this. For example, $ABCD = (AB)(CD)$ has the same eigenvalues as $(CD)(AB) = CDAB$ and so on. So while this is a generalization, it may not feel satisfactory.
answered Nov 14 '18 at 10:33
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.6k33376
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add a comment |
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2 Answers
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2 Answers
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active
oldest
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votes
$begingroup$
Let check this on orthogonal matrices (e.g. rotations dim. $3 times 3$).
We have for example $R_1R_1^TR_2R_2^T=I$ with $1$ as eigenvalues. If we change the order of rotations for example $R_1R_2(R_2R_1)^T=R_1R_2R_1^TR_2^T$ certainly if $R_1$ and $R_2$ are not commutative we would not obtain identity matrix and consequently the eigenvalues are different.
answered Nov 14 '18 at 10:24
WidawensenWidawensen
4,50321446
$endgroup$
add a comment |
$begingroup$
Let check this on orthogonal matrices (e.g. rotations dim. $3 times 3$).
We have for example $R_1R_1^TR_2R_2^T=I$ with $1$ as eigenvalues. If we change the order of rotations for example $R_1R_2(R_2R_1)^T=R_1R_2R_1^TR_2^T$ certainly if $R_1$ and $R_2$ are not commutative we would not obtain identity matrix and consequently the eigenvalues are different.
answered Nov 14 '18 at 10:24
WidawensenWidawensen
4,50321446
$endgroup$
add a comment |
$begingroup$
Let check this on orthogonal matrices (e.g. rotations dim. $3 times 3$).
We have for example $R_1R_1^TR_2R_2^T=I$ with $1$ as eigenvalues. If we change the order of rotations for example $R_1R_2(R_2R_1)^T=R_1R_2R_1^TR_2^T$ certainly if $R_1$ and $R_2$ are not commutative we would not obtain identity matrix and consequently the eigenvalues are different.
answered Nov 14 '18 at 10:24
WidawensenWidawensen
4,50321446
$endgroup$
Let check this on orthogonal matrices (e.g. rotations dim. $3 times 3$).
We have for example $R_1R_1^TR_2R_2^T=I$ with $1$ as eigenvalues. If we change the order of rotations for example $R_1R_2(R_2R_1)^T=R_1R_2R_1^TR_2^T$ certainly if $R_1$ and $R_2$ are not commutative we would not obtain identity matrix and consequently the eigenvalues are different.
answered Nov 14 '18 at 10:24
WidawensenWidawensen
4,50321446
answered Nov 14 '18 at 10:24
WidawensenWidawensen
4,50321446
answered Nov 14 '18 at 10:24
WidawensenWidawensen
4,50321446
answered Nov 14 '18 at 10:24
WidawensenWidawensen
4,50321446
4,50321446
add a comment |
add a comment |
$begingroup$
In fact, $AB$ and $BA$ have the same characteristic polynomial.
The problem with expecting switches in between various other matrices to preserve eigenvalues, is that the characteristic polynomial does not behave well with multiplication of matrices from left/right.
In other words, if $A$ and $B$ have the same characteristic polynomial, then this may not be true of $CA$ and $CB$, where $C$ is some other matrix. This is because the trace(sum of eigenvalues) is not multiplicative in general, and we can use this to produce counterexamples.
To give an example, take $A = beginbmatrix1 0 \ 0 2 endbmatrix$ and $B = beginbmatrix 1 1 \ 0 2endbmatrix$. These matrices have the same characteristic polynomial. For yourself, find a $C$ such that $CA$ and $CB$ have different trace, and therefore different eigenvalues. Find a $D$ such that $AD$ and $BD$ have different trace. It follows that $CA,CB$ and $AD,BD$ don't have the same characteristic polynomials.
The stumbling block in trying to prove, say , that $CBA$ and $CAB$ have the same eigenvalues, is the inability to assure that multiplying by $C$, after knowing that the eigenvalues of $AB$ and $BA$ are the same, retains the equality of eigenvalues. This is why you can use the $C,A,B,D$ above to find counterexamples of the assertions you were making.
However, cyclic permutations retain eigenvalues, because we are just applying the "two-matrix" case to deduce this. For example, $ABCD = (AB)(CD)$ has the same eigenvalues as $(CD)(AB) = CDAB$ and so on. So while this is a generalization, it may not feel satisfactory.
answered Nov 14 '18 at 10:33
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.6k33376
$endgroup$
add a comment |
$begingroup$
In fact, $AB$ and $BA$ have the same characteristic polynomial.
The problem with expecting switches in between various other matrices to preserve eigenvalues, is that the characteristic polynomial does not behave well with multiplication of matrices from left/right.
In other words, if $A$ and $B$ have the same characteristic polynomial, then this may not be true of $CA$ and $CB$, where $C$ is some other matrix. This is because the trace(sum of eigenvalues) is not multiplicative in general, and we can use this to produce counterexamples.
To give an example, take $A = beginbmatrix1 0 \ 0 2 endbmatrix$ and $B = beginbmatrix 1 1 \ 0 2endbmatrix$. These matrices have the same characteristic polynomial. For yourself, find a $C$ such that $CA$ and $CB$ have different trace, and therefore different eigenvalues. Find a $D$ such that $AD$ and $BD$ have different trace. It follows that $CA,CB$ and $AD,BD$ don't have the same characteristic polynomials.
The stumbling block in trying to prove, say , that $CBA$ and $CAB$ have the same eigenvalues, is the inability to assure that multiplying by $C$, after knowing that the eigenvalues of $AB$ and $BA$ are the same, retains the equality of eigenvalues. This is why you can use the $C,A,B,D$ above to find counterexamples of the assertions you were making.
However, cyclic permutations retain eigenvalues, because we are just applying the "two-matrix" case to deduce this. For example, $ABCD = (AB)(CD)$ has the same eigenvalues as $(CD)(AB) = CDAB$ and so on. So while this is a generalization, it may not feel satisfactory.
answered Nov 14 '18 at 10:33
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.6k33376
$endgroup$
add a comment |
$begingroup$
In fact, $AB$ and $BA$ have the same characteristic polynomial.
The problem with expecting switches in between various other matrices to preserve eigenvalues, is that the characteristic polynomial does not behave well with multiplication of matrices from left/right.
In other words, if $A$ and $B$ have the same characteristic polynomial, then this may not be true of $CA$ and $CB$, where $C$ is some other matrix. This is because the trace(sum of eigenvalues) is not multiplicative in general, and we can use this to produce counterexamples.
To give an example, take $A = beginbmatrix1 0 \ 0 2 endbmatrix$ and $B = beginbmatrix 1 1 \ 0 2endbmatrix$. These matrices have the same characteristic polynomial. For yourself, find a $C$ such that $CA$ and $CB$ have different trace, and therefore different eigenvalues. Find a $D$ such that $AD$ and $BD$ have different trace. It follows that $CA,CB$ and $AD,BD$ don't have the same characteristic polynomials.
The stumbling block in trying to prove, say , that $CBA$ and $CAB$ have the same eigenvalues, is the inability to assure that multiplying by $C$, after knowing that the eigenvalues of $AB$ and $BA$ are the same, retains the equality of eigenvalues. This is why you can use the $C,A,B,D$ above to find counterexamples of the assertions you were making.
However, cyclic permutations retain eigenvalues, because we are just applying the "two-matrix" case to deduce this. For example, $ABCD = (AB)(CD)$ has the same eigenvalues as $(CD)(AB) = CDAB$ and so on. So while this is a generalization, it may not feel satisfactory.
answered Nov 14 '18 at 10:33
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.6k33376
$endgroup$
In fact, $AB$ and $BA$ have the same characteristic polynomial.
The problem with expecting switches in between various other matrices to preserve eigenvalues, is that the characteristic polynomial does not behave well with multiplication of matrices from left/right.
In other words, if $A$ and $B$ have the same characteristic polynomial, then this may not be true of $CA$ and $CB$, where $C$ is some other matrix. This is because the trace(sum of eigenvalues) is not multiplicative in general, and we can use this to produce counterexamples.
To give an example, take $A = beginbmatrix1 0 \ 0 2 endbmatrix$ and $B = beginbmatrix 1 1 \ 0 2endbmatrix$. These matrices have the same characteristic polynomial. For yourself, find a $C$ such that $CA$ and $CB$ have different trace, and therefore different eigenvalues. Find a $D$ such that $AD$ and $BD$ have different trace. It follows that $CA,CB$ and $AD,BD$ don't have the same characteristic polynomials.
The stumbling block in trying to prove, say , that $CBA$ and $CAB$ have the same eigenvalues, is the inability to assure that multiplying by $C$, after knowing that the eigenvalues of $AB$ and $BA$ are the same, retains the equality of eigenvalues. This is why you can use the $C,A,B,D$ above to find counterexamples of the assertions you were making.
However, cyclic permutations retain eigenvalues, because we are just applying the "two-matrix" case to deduce this. For example, $ABCD = (AB)(CD)$ has the same eigenvalues as $(CD)(AB) = CDAB$ and so on. So while this is a generalization, it may not feel satisfactory.
answered Nov 14 '18 at 10:33
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.6k33376
answered Nov 14 '18 at 10:33
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.6k33376
answered Nov 14 '18 at 10:33
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.6k33376
answered Nov 14 '18 at 10:33
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.6k33376
38.6k33376
add a comment |
add a comment |
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$begingroup$
The only thing that can be said is what can be deduced from the first result, which is that eigenvalues are invariant under cyclic permutations, so $ABCD$ has the same eigenvalues as $DABC, CDAB$, and $BCDA$, and that's it.
$endgroup$
– Qiaochu Yuan
Nov 14 '18 at 10:16