Smith Form of a matrix which is already diagonal
0
$begingroup$
Find the Smith form of
$$
pmatrix{12 & 0 & 0 \0 & 18 & 0 }$$
I don't find a simple explanation of this algorithm.
Could you please help?
Thank you.
linear-algebra abstract-algebra
asked Dec 18 '18 at 15:17
PerelManPerelMan
709313
$endgroup$
add a comment |
0
$begingroup$
Find the Smith form of
$$
pmatrix{12 & 0 & 0 \0 & 18 & 0 }$$
I don't find a simple explanation of this algorithm.
Could you please help?
Thank you.
linear-algebra abstract-algebra
asked Dec 18 '18 at 15:17
PerelManPerelMan
709313
$endgroup$
1
$begingroup$
I think you're going to have to be specific about what you don't understand. You're only likely to get an explanation no clearer than this which is already very clear. People have already tried to be very clear when writing up such things. So it doesn't make sense to ask in the way you have asked. If you specify, however, you'll have better luck.
$endgroup$
– rschwieb
Dec 18 '18 at 15:24
$begingroup$
Actually I've already checked the wikipedia page and I think step II is too heavy. It is suitable for a computer program, but to do the euclid algorithm by hand for each pivot to find bezout coefficients is a tedious task and I think there is a simpler way to do it bu hand?
$endgroup$
– PerelMan
Dec 18 '18 at 15:40
$begingroup$
OK, well that is definitely more specific. You find the Euclidean algorithm for two small numbers tedious? Seems pretty easy, especially in this case.
$endgroup$
– rschwieb
Dec 18 '18 at 16:03
$begingroup$
Here's a start: $pmatrix{12 & 0 & 0 \0 & 18 & 0 } leadsto pmatrix{12 & 18 & 0 \0 & 18 & 0 } leadsto pmatrix{-6 & 18 & 0 \-18 & 18 & 0 }$
$endgroup$
– André 3000
Dec 18 '18 at 16:47
add a comment |
0
0
0
1
$begingroup$
Find the Smith form of
$$
pmatrix{12 & 0 & 0 \0 & 18 & 0 }$$
I don't find a simple explanation of this algorithm.
Could you please help?
Thank you.
linear-algebra abstract-algebra
asked Dec 18 '18 at 15:17
PerelManPerelMan
709313
$endgroup$
Find the Smith form of
$$
pmatrix{12 & 0 & 0 \0 & 18 & 0 }$$
I don't find a simple explanation of this algorithm.
Could you please help?
Thank you.
linear-algebra abstract-algebra
linear-algebra abstract-algebra
asked Dec 18 '18 at 15:17
PerelManPerelMan
709313
asked Dec 18 '18 at 15:17
PerelManPerelMan
709313
asked Dec 18 '18 at 15:17
PerelManPerelMan
709313
asked Dec 18 '18 at 15:17
PerelManPerelMan
709313
asked Dec 18 '18 at 15:17
PerelManPerelMan
709313
709313
1
$begingroup$
I think you're going to have to be specific about what you don't understand. You're only likely to get an explanation no clearer than this which is already very clear. People have already tried to be very clear when writing up such things. So it doesn't make sense to ask in the way you have asked. If you specify, however, you'll have better luck.
$endgroup$
– rschwieb
Dec 18 '18 at 15:24
$begingroup$
Actually I've already checked the wikipedia page and I think step II is too heavy. It is suitable for a computer program, but to do the euclid algorithm by hand for each pivot to find bezout coefficients is a tedious task and I think there is a simpler way to do it bu hand?
$endgroup$
– PerelMan
Dec 18 '18 at 15:40
$begingroup$
OK, well that is definitely more specific. You find the Euclidean algorithm for two small numbers tedious? Seems pretty easy, especially in this case.
$endgroup$
– rschwieb
Dec 18 '18 at 16:03
$begingroup$
Here's a start: $pmatrix{12 & 0 & 0 \0 & 18 & 0 } leadsto pmatrix{12 & 18 & 0 \0 & 18 & 0 } leadsto pmatrix{-6 & 18 & 0 \-18 & 18 & 0 }$
$endgroup$
– André 3000
Dec 18 '18 at 16:47
add a comment |
1
$begingroup$
I think you're going to have to be specific about what you don't understand. You're only likely to get an explanation no clearer than this which is already very clear. People have already tried to be very clear when writing up such things. So it doesn't make sense to ask in the way you have asked. If you specify, however, you'll have better luck.
$endgroup$
– rschwieb
Dec 18 '18 at 15:24
$begingroup$
Actually I've already checked the wikipedia page and I think step II is too heavy. It is suitable for a computer program, but to do the euclid algorithm by hand for each pivot to find bezout coefficients is a tedious task and I think there is a simpler way to do it bu hand?
$endgroup$
– PerelMan
Dec 18 '18 at 15:40
$begingroup$
OK, well that is definitely more specific. You find the Euclidean algorithm for two small numbers tedious? Seems pretty easy, especially in this case.
$endgroup$
– rschwieb
Dec 18 '18 at 16:03
$begingroup$
Here's a start: $pmatrix{12 & 0 & 0 \0 & 18 & 0 } leadsto pmatrix{12 & 18 & 0 \0 & 18 & 0 } leadsto pmatrix{-6 & 18 & 0 \-18 & 18 & 0 }$
$endgroup$
– André 3000
Dec 18 '18 at 16:47
1
1
$begingroup$
I think you're going to have to be specific about what you don't understand. You're only likely to get an explanation no clearer than this which is already very clear. People have already tried to be very clear when writing up such things. So it doesn't make sense to ask in the way you have asked. If you specify, however, you'll have better luck.
$endgroup$
– rschwieb
Dec 18 '18 at 15:24
$begingroup$
I think you're going to have to be specific about what you don't understand. You're only likely to get an explanation no clearer than this which is already very clear. People have already tried to be very clear when writing up such things. So it doesn't make sense to ask in the way you have asked. If you specify, however, you'll have better luck.
$endgroup$
– rschwieb
Dec 18 '18 at 15:24
$begingroup$
Actually I've already checked the wikipedia page and I think step II is too heavy. It is suitable for a computer program, but to do the euclid algorithm by hand for each pivot to find bezout coefficients is a tedious task and I think there is a simpler way to do it bu hand?
$endgroup$
– PerelMan
Dec 18 '18 at 15:40
$begingroup$
Actually I've already checked the wikipedia page and I think step II is too heavy. It is suitable for a computer program, but to do the euclid algorithm by hand for each pivot to find bezout coefficients is a tedious task and I think there is a simpler way to do it bu hand?
$endgroup$
– PerelMan
Dec 18 '18 at 15:40
$begingroup$
OK, well that is definitely more specific. You find the Euclidean algorithm for two small numbers tedious? Seems pretty easy, especially in this case.
$endgroup$
– rschwieb
Dec 18 '18 at 16:03
$begingroup$
OK, well that is definitely more specific. You find the Euclidean algorithm for two small numbers tedious? Seems pretty easy, especially in this case.
$endgroup$
– rschwieb
Dec 18 '18 at 16:03
$begingroup$
Here's a start: $pmatrix{12 & 0 & 0 \0 & 18 & 0 } leadsto pmatrix{12 & 18 & 0 \0 & 18 & 0 } leadsto pmatrix{-6 & 18 & 0 \-18 & 18 & 0 }$
$endgroup$
– André 3000
Dec 18 '18 at 16:47
$begingroup$
Here's a start: $pmatrix{12 & 0 & 0 \0 & 18 & 0 } leadsto pmatrix{12 & 18 & 0 \0 & 18 & 0 } leadsto pmatrix{-6 & 18 & 0 \-18 & 18 & 0 }$
$endgroup$
– André 3000
Dec 18 '18 at 16:47
add a comment |
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1
$begingroup$
I think you're going to have to be specific about what you don't understand. You're only likely to get an explanation no clearer than this which is already very clear. People have already tried to be very clear when writing up such things. So it doesn't make sense to ask in the way you have asked. If you specify, however, you'll have better luck.
$endgroup$
– rschwieb
Dec 18 '18 at 15:24
$begingroup$
Actually I've already checked the wikipedia page and I think step II is too heavy. It is suitable for a computer program, but to do the euclid algorithm by hand for each pivot to find bezout coefficients is a tedious task and I think there is a simpler way to do it bu hand?
$endgroup$
– PerelMan
Dec 18 '18 at 15:40
$begingroup$
OK, well that is definitely more specific. You find the Euclidean algorithm for two small numbers tedious? Seems pretty easy, especially in this case.
$endgroup$
– rschwieb
Dec 18 '18 at 16:03
$begingroup$
Here's a start: $pmatrix{12 & 0 & 0 \0 & 18 & 0 } leadsto pmatrix{12 & 18 & 0 \0 & 18 & 0 } leadsto pmatrix{-6 & 18 & 0 \-18 & 18 & 0 }$
$endgroup$
– André 3000
Dec 18 '18 at 16:47