Is the product of two Smith Normal Forms a the Smith normal form of the product?
0
$begingroup$
Suppose A and B are square matrices of the same size over a PID R. Does the Smith Normal form of AB equal the product of the Smith normal form of A and B?
I think this should be false. However, I can't quite find an example.
modules principal-ideal-domains smith-normal-form
asked Dec 2 '18 at 18:19
justanothermathstudentjustanothermathstudent
1018
$endgroup$
add a comment |
0
$begingroup$
Suppose A and B are square matrices of the same size over a PID R. Does the Smith Normal form of AB equal the product of the Smith normal form of A and B?
I think this should be false. However, I can't quite find an example.
modules principal-ideal-domains smith-normal-form
asked Dec 2 '18 at 18:19
justanothermathstudentjustanothermathstudent
1018
$endgroup$
1
$begingroup$
It is false. Try $A = begin{pmatrix} 1 & 0 \ 0 & 2 end{pmatrix}$ and $B = begin{pmatrix} 2 & 0 \ 0 & 1 end{pmatrix}$. Their Smith normal forms are both $A$, but their product's Smith normal form is not $A^2$.
$endgroup$
– darij grinberg
Dec 2 '18 at 18:40
$begingroup$
Wow! Thanks a lot.
$endgroup$
– justanothermathstudent
Dec 2 '18 at 19:28
add a comment |
0
0
0
$begingroup$
Suppose A and B are square matrices of the same size over a PID R. Does the Smith Normal form of AB equal the product of the Smith normal form of A and B?
I think this should be false. However, I can't quite find an example.
modules principal-ideal-domains smith-normal-form
asked Dec 2 '18 at 18:19
justanothermathstudentjustanothermathstudent
1018
$endgroup$
Suppose A and B are square matrices of the same size over a PID R. Does the Smith Normal form of AB equal the product of the Smith normal form of A and B?
I think this should be false. However, I can't quite find an example.
modules principal-ideal-domains smith-normal-form
modules principal-ideal-domains smith-normal-form
asked Dec 2 '18 at 18:19
justanothermathstudentjustanothermathstudent
1018
asked Dec 2 '18 at 18:19
justanothermathstudentjustanothermathstudent
1018
asked Dec 2 '18 at 18:19
justanothermathstudentjustanothermathstudent
1018
asked Dec 2 '18 at 18:19
justanothermathstudentjustanothermathstudent
1018
asked Dec 2 '18 at 18:19
justanothermathstudentjustanothermathstudent
1018
1018
1
$begingroup$
It is false. Try $A = begin{pmatrix} 1 & 0 \ 0 & 2 end{pmatrix}$ and $B = begin{pmatrix} 2 & 0 \ 0 & 1 end{pmatrix}$. Their Smith normal forms are both $A$, but their product's Smith normal form is not $A^2$.
$endgroup$
– darij grinberg
Dec 2 '18 at 18:40
$begingroup$
Wow! Thanks a lot.
$endgroup$
– justanothermathstudent
Dec 2 '18 at 19:28
add a comment |
1
$begingroup$
It is false. Try $A = begin{pmatrix} 1 & 0 \ 0 & 2 end{pmatrix}$ and $B = begin{pmatrix} 2 & 0 \ 0 & 1 end{pmatrix}$. Their Smith normal forms are both $A$, but their product's Smith normal form is not $A^2$.
$endgroup$
– darij grinberg
Dec 2 '18 at 18:40
$begingroup$
Wow! Thanks a lot.
$endgroup$
– justanothermathstudent
Dec 2 '18 at 19:28
1
1
$begingroup$
It is false. Try $A = begin{pmatrix} 1 & 0 \ 0 & 2 end{pmatrix}$ and $B = begin{pmatrix} 2 & 0 \ 0 & 1 end{pmatrix}$. Their Smith normal forms are both $A$, but their product's Smith normal form is not $A^2$.
$endgroup$
– darij grinberg
Dec 2 '18 at 18:40
$begingroup$
It is false. Try $A = begin{pmatrix} 1 & 0 \ 0 & 2 end{pmatrix}$ and $B = begin{pmatrix} 2 & 0 \ 0 & 1 end{pmatrix}$. Their Smith normal forms are both $A$, but their product's Smith normal form is not $A^2$.
$endgroup$
– darij grinberg
Dec 2 '18 at 18:40
$begingroup$
Wow! Thanks a lot.
$endgroup$
– justanothermathstudent
Dec 2 '18 at 19:28
$begingroup$
Wow! Thanks a lot.
$endgroup$
– justanothermathstudent
Dec 2 '18 at 19:28
add a comment |
0
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1
$begingroup$
It is false. Try $A = begin{pmatrix} 1 & 0 \ 0 & 2 end{pmatrix}$ and $B = begin{pmatrix} 2 & 0 \ 0 & 1 end{pmatrix}$. Their Smith normal forms are both $A$, but their product's Smith normal form is not $A^2$.
$endgroup$
– darij grinberg
Dec 2 '18 at 18:40
$begingroup$
Wow! Thanks a lot.
$endgroup$
– justanothermathstudent
Dec 2 '18 at 19:28