Singular Value Decomposition of Commuting Matrices
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If two square matrices $M_1$ and $M_2$ commute, does it mean that the $U$ and $V^dagger$ appearing in their singular value decompositions are the same?
Specifically, does it imply that
$$M_1 = U Sigma_1 V^dagger$$
$$M_2 = U Sigma_2 V^dagger$$
for two possibly different $Sigma_{1,2}$?
As $M_1$ and $M_2$ commute, I know that they can be simultaneously diagonalized, which means there's an orthogonal transformation $mathcal{O}$ which can diagonalize both matrices. But does this have any bearing on $U$ and $V$ of the singular value decomposition?
linear-algebra matrices eigenvalues-eigenvectors matrix-decomposition
asked Jul 26 '16 at 13:33
leastaction
278111
add a comment |
up vote
1
down vote
favorite
If two square matrices $M_1$ and $M_2$ commute, does it mean that the $U$ and $V^dagger$ appearing in their singular value decompositions are the same?
Specifically, does it imply that
$$M_1 = U Sigma_1 V^dagger$$
$$M_2 = U Sigma_2 V^dagger$$
for two possibly different $Sigma_{1,2}$?
As $M_1$ and $M_2$ commute, I know that they can be simultaneously diagonalized, which means there's an orthogonal transformation $mathcal{O}$ which can diagonalize both matrices. But does this have any bearing on $U$ and $V$ of the singular value decomposition?
linear-algebra matrices eigenvalues-eigenvectors matrix-decomposition
asked Jul 26 '16 at 13:33
leastaction
278111
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
If two square matrices $M_1$ and $M_2$ commute, does it mean that the $U$ and $V^dagger$ appearing in their singular value decompositions are the same?
Specifically, does it imply that
$$M_1 = U Sigma_1 V^dagger$$
$$M_2 = U Sigma_2 V^dagger$$
for two possibly different $Sigma_{1,2}$?
As $M_1$ and $M_2$ commute, I know that they can be simultaneously diagonalized, which means there's an orthogonal transformation $mathcal{O}$ which can diagonalize both matrices. But does this have any bearing on $U$ and $V$ of the singular value decomposition?
linear-algebra matrices eigenvalues-eigenvectors matrix-decomposition
asked Jul 26 '16 at 13:33
leastaction
278111
If two square matrices $M_1$ and $M_2$ commute, does it mean that the $U$ and $V^dagger$ appearing in their singular value decompositions are the same?
Specifically, does it imply that
$$M_1 = U Sigma_1 V^dagger$$
$$M_2 = U Sigma_2 V^dagger$$
for two possibly different $Sigma_{1,2}$?
As $M_1$ and $M_2$ commute, I know that they can be simultaneously diagonalized, which means there's an orthogonal transformation $mathcal{O}$ which can diagonalize both matrices. But does this have any bearing on $U$ and $V$ of the singular value decomposition?
linear-algebra matrices eigenvalues-eigenvectors matrix-decomposition
linear-algebra matrices eigenvalues-eigenvectors matrix-decomposition
asked Jul 26 '16 at 13:33
leastaction
278111
asked Jul 26 '16 at 13:33
leastaction
278111
asked Jul 26 '16 at 13:33
leastaction
278111
asked Jul 26 '16 at 13:33
leastaction
278111
asked Jul 26 '16 at 13:33
leastaction
278111
278111
add a comment |
add a comment |
1 Answer
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Counter-example: take $M_1 = I$, the identity matrix, and $M_2$ as any matrix of the same dimension (not a scalar multiple of the identity). You can see the above statement fails to hold.
answered Jul 26 '16 at 18:50
harvey
1455
Thanks. Is there any related statement for commuting matrices though?
– leastaction
Jul 26 '16 at 18:51
add a comment |
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1 Answer
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1 Answer
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active
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up vote
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Counter-example: take $M_1 = I$, the identity matrix, and $M_2$ as any matrix of the same dimension (not a scalar multiple of the identity). You can see the above statement fails to hold.
answered Jul 26 '16 at 18:50
harvey
1455
Thanks. Is there any related statement for commuting matrices though?
– leastaction
Jul 26 '16 at 18:51
add a comment |
up vote
2
down vote
Counter-example: take $M_1 = I$, the identity matrix, and $M_2$ as any matrix of the same dimension (not a scalar multiple of the identity). You can see the above statement fails to hold.
answered Jul 26 '16 at 18:50
harvey
1455
Thanks. Is there any related statement for commuting matrices though?
– leastaction
Jul 26 '16 at 18:51
add a comment |
up vote
2
down vote
up vote
2
down vote
Counter-example: take $M_1 = I$, the identity matrix, and $M_2$ as any matrix of the same dimension (not a scalar multiple of the identity). You can see the above statement fails to hold.
answered Jul 26 '16 at 18:50
harvey
1455
Counter-example: take $M_1 = I$, the identity matrix, and $M_2$ as any matrix of the same dimension (not a scalar multiple of the identity). You can see the above statement fails to hold.
answered Jul 26 '16 at 18:50
harvey
1455
answered Jul 26 '16 at 18:50
harvey
1455
answered Jul 26 '16 at 18:50
harvey
1455
answered Jul 26 '16 at 18:50
harvey
1455
1455
Thanks. Is there any related statement for commuting matrices though?
– leastaction
Jul 26 '16 at 18:51
add a comment |
Thanks. Is there any related statement for commuting matrices though?
– leastaction
Jul 26 '16 at 18:51
Thanks. Is there any related statement for commuting matrices though?
– leastaction
Jul 26 '16 at 18:51
Thanks. Is there any related statement for commuting matrices though?
– leastaction
Jul 26 '16 at 18:51
add a comment |
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