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Is Frobenius norm constant?
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I came across an article that makes me doubt what I have learnt so far.
For example, let us decompose matrix A into L and U (LU factorization)
$A=left(
begin{array}{cccc}
5 & 4 & 1 & 1 \
4 & 5 & 1 & 1 \
1&1&4&2 \
1&1&2&4
end{array}right)$
The Frobenius norm should be $||A-LU||$. Now, the authors gave different answers when they implement this on MATLAB 2010 and 2012 and also different answers on AMD and Intel processor. Why should we have different answers for a consistent norm?
matrices norm matrix-decomposition
asked 2 days ago
dupsy
475
add a comment |
up vote
0
down vote
favorite
I came across an article that makes me doubt what I have learnt so far.
For example, let us decompose matrix A into L and U (LU factorization)
$A=left(
begin{array}{cccc}
5 & 4 & 1 & 1 \
4 & 5 & 1 & 1 \
1&1&4&2 \
1&1&2&4
end{array}right)$
The Frobenius norm should be $||A-LU||$. Now, the authors gave different answers when they implement this on MATLAB 2010 and 2012 and also different answers on AMD and Intel processor. Why should we have different answers for a consistent norm?
matrices norm matrix-decomposition
asked 2 days ago
dupsy
475
I think something is not quite right with your question. The Frobenius norm of A is the sum of squared entries of $A$. Is $| cdot |$ denoting the Frobenius norm in your question? We certainly wouldn't have $|A|=|A-LU|$. As for why the norm may be different on different machines or versions of matlab, it may have to do with the precision the computation was run at. If the difference is very small this is likely the case. Otherwise something is wrong.
– Eric
2 days ago
I do not still get it. When L and U are multiplied and then subtracted from A, the result is A-LU. When we take the Frobenius norm of this result, should we have different answers on different machine or versions of MATLAb?
– dupsy
yesterday
Your question is somewhat implicitly assuming the existence of an LU decomposition. That means that theoretically A=LU, so and we have $|A-LU|_F= |0|_F=0$. However, you might not get exactly zero in a computer. This is because the entries of $L$ and $U$ may be irrational numbers. A computer cannot store an irrational number perfectly, meaning the $L$ and $U$ stored in your machine will be very slightly different than the correct ones. This leads to small amounts of round off error when computing $A-LU$. So if all answers are very close, the error is probably just round off error.
– Eric
yesterday
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I came across an article that makes me doubt what I have learnt so far.
For example, let us decompose matrix A into L and U (LU factorization)
$A=left(
begin{array}{cccc}
5 & 4 & 1 & 1 \
4 & 5 & 1 & 1 \
1&1&4&2 \
1&1&2&4
end{array}right)$
The Frobenius norm should be $||A-LU||$. Now, the authors gave different answers when they implement this on MATLAB 2010 and 2012 and also different answers on AMD and Intel processor. Why should we have different answers for a consistent norm?
matrices norm matrix-decomposition
asked 2 days ago
dupsy
475
I came across an article that makes me doubt what I have learnt so far.
For example, let us decompose matrix A into L and U (LU factorization)
$A=left(
begin{array}{cccc}
5 & 4 & 1 & 1 \
4 & 5 & 1 & 1 \
1&1&4&2 \
1&1&2&4
end{array}right)$
The Frobenius norm should be $||A-LU||$. Now, the authors gave different answers when they implement this on MATLAB 2010 and 2012 and also different answers on AMD and Intel processor. Why should we have different answers for a consistent norm?
matrices norm matrix-decomposition
matrices norm matrix-decomposition
asked 2 days ago
dupsy
475
asked 2 days ago
dupsy
475
asked 2 days ago
dupsy
475
asked 2 days ago
dupsy
475
asked 2 days ago
dupsy
475
475
I think something is not quite right with your question. The Frobenius norm of A is the sum of squared entries of $A$. Is $| cdot |$ denoting the Frobenius norm in your question? We certainly wouldn't have $|A|=|A-LU|$. As for why the norm may be different on different machines or versions of matlab, it may have to do with the precision the computation was run at. If the difference is very small this is likely the case. Otherwise something is wrong.
– Eric
2 days ago
I do not still get it. When L and U are multiplied and then subtracted from A, the result is A-LU. When we take the Frobenius norm of this result, should we have different answers on different machine or versions of MATLAb?
– dupsy
yesterday
Your question is somewhat implicitly assuming the existence of an LU decomposition. That means that theoretically A=LU, so and we have $|A-LU|_F= |0|_F=0$. However, you might not get exactly zero in a computer. This is because the entries of $L$ and $U$ may be irrational numbers. A computer cannot store an irrational number perfectly, meaning the $L$ and $U$ stored in your machine will be very slightly different than the correct ones. This leads to small amounts of round off error when computing $A-LU$. So if all answers are very close, the error is probably just round off error.
– Eric
yesterday
add a comment |
I think something is not quite right with your question. The Frobenius norm of A is the sum of squared entries of $A$. Is $| cdot |$ denoting the Frobenius norm in your question? We certainly wouldn't have $|A|=|A-LU|$. As for why the norm may be different on different machines or versions of matlab, it may have to do with the precision the computation was run at. If the difference is very small this is likely the case. Otherwise something is wrong.
– Eric
2 days ago
I do not still get it. When L and U are multiplied and then subtracted from A, the result is A-LU. When we take the Frobenius norm of this result, should we have different answers on different machine or versions of MATLAb?
– dupsy
yesterday
Your question is somewhat implicitly assuming the existence of an LU decomposition. That means that theoretically A=LU, so and we have $|A-LU|_F= |0|_F=0$. However, you might not get exactly zero in a computer. This is because the entries of $L$ and $U$ may be irrational numbers. A computer cannot store an irrational number perfectly, meaning the $L$ and $U$ stored in your machine will be very slightly different than the correct ones. This leads to small amounts of round off error when computing $A-LU$. So if all answers are very close, the error is probably just round off error.
– Eric
yesterday
I think something is not quite right with your question. The Frobenius norm of A is the sum of squared entries of $A$. Is $| cdot |$ denoting the Frobenius norm in your question? We certainly wouldn't have $|A|=|A-LU|$. As for why the norm may be different on different machines or versions of matlab, it may have to do with the precision the computation was run at. If the difference is very small this is likely the case. Otherwise something is wrong.
– Eric
2 days ago
I think something is not quite right with your question. The Frobenius norm of A is the sum of squared entries of $A$. Is $| cdot |$ denoting the Frobenius norm in your question? We certainly wouldn't have $|A|=|A-LU|$. As for why the norm may be different on different machines or versions of matlab, it may have to do with the precision the computation was run at. If the difference is very small this is likely the case. Otherwise something is wrong.
– Eric
2 days ago
I do not still get it. When L and U are multiplied and then subtracted from A, the result is A-LU. When we take the Frobenius norm of this result, should we have different answers on different machine or versions of MATLAb?
– dupsy
yesterday
I do not still get it. When L and U are multiplied and then subtracted from A, the result is A-LU. When we take the Frobenius norm of this result, should we have different answers on different machine or versions of MATLAb?
– dupsy
yesterday
Your question is somewhat implicitly assuming the existence of an LU decomposition. That means that theoretically A=LU, so and we have $|A-LU|_F= |0|_F=0$. However, you might not get exactly zero in a computer. This is because the entries of $L$ and $U$ may be irrational numbers. A computer cannot store an irrational number perfectly, meaning the $L$ and $U$ stored in your machine will be very slightly different than the correct ones. This leads to small amounts of round off error when computing $A-LU$. So if all answers are very close, the error is probably just round off error.
– Eric
yesterday
Your question is somewhat implicitly assuming the existence of an LU decomposition. That means that theoretically A=LU, so and we have $|A-LU|_F= |0|_F=0$. However, you might not get exactly zero in a computer. This is because the entries of $L$ and $U$ may be irrational numbers. A computer cannot store an irrational number perfectly, meaning the $L$ and $U$ stored in your machine will be very slightly different than the correct ones. This leads to small amounts of round off error when computing $A-LU$. So if all answers are very close, the error is probably just round off error.
– Eric
yesterday
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I think something is not quite right with your question. The Frobenius norm of A is the sum of squared entries of $A$. Is $| cdot |$ denoting the Frobenius norm in your question? We certainly wouldn't have $|A|=|A-LU|$. As for why the norm may be different on different machines or versions of matlab, it may have to do with the precision the computation was run at. If the difference is very small this is likely the case. Otherwise something is wrong.
– Eric
2 days ago
I do not still get it. When L and U are multiplied and then subtracted from A, the result is A-LU. When we take the Frobenius norm of this result, should we have different answers on different machine or versions of MATLAb?
– dupsy
yesterday
Your question is somewhat implicitly assuming the existence of an LU decomposition. That means that theoretically A=LU, so and we have $|A-LU|_F= |0|_F=0$. However, you might not get exactly zero in a computer. This is because the entries of $L$ and $U$ may be irrational numbers. A computer cannot store an irrational number perfectly, meaning the $L$ and $U$ stored in your machine will be very slightly different than the correct ones. This leads to small amounts of round off error when computing $A-LU$. So if all answers are very close, the error is probably just round off error.
– Eric
yesterday