Mixing
Efficiency in matrix multiplication with diagonalizable matrix. [closed]
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Are there efficient algorithms to compute Ax in the case of A being diagonalizable instead of not being?
numerical-linear-algebra
asked Jan 17 at 6:56
Santiago AranguriSantiago Aranguri
12
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closed as off-topic by José Carlos Santos, Shailesh, Lord_Farin, Henrik, Cesareo Jan 17 at 9:49
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If this question can be reworded to fit the rules in the help center, please edit the question.
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$begingroup$
Are there efficient algorithms to compute Ax in the case of A being diagonalizable instead of not being?
numerical-linear-algebra
asked Jan 17 at 6:56
Santiago AranguriSantiago Aranguri
12
$endgroup$
closed as off-topic by José Carlos Santos, Shailesh, Lord_Farin, Henrik, Cesareo Jan 17 at 9:49
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Shailesh, Lord_Farin, Henrik, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Are there efficient algorithms to compute Ax in the case of A being diagonalizable instead of not being?
numerical-linear-algebra
asked Jan 17 at 6:56
Santiago AranguriSantiago Aranguri
12
$endgroup$
Are there efficient algorithms to compute Ax in the case of A being diagonalizable instead of not being?
numerical-linear-algebra
numerical-linear-algebra
asked Jan 17 at 6:56
Santiago AranguriSantiago Aranguri
12
asked Jan 17 at 6:56
Santiago AranguriSantiago Aranguri
12
asked Jan 17 at 6:56
Santiago AranguriSantiago Aranguri
12
asked Jan 17 at 6:56
Santiago AranguriSantiago Aranguri
12
asked Jan 17 at 6:56
Santiago AranguriSantiago Aranguri
12
12
closed as off-topic by José Carlos Santos, Shailesh, Lord_Farin, Henrik, Cesareo Jan 17 at 9:49
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Shailesh, Lord_Farin, Henrik, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by José Carlos Santos, Shailesh, Lord_Farin, Henrik, Cesareo Jan 17 at 9:49
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Shailesh, Lord_Farin, Henrik, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
1 Answer
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$begingroup$
It's just not going to matter. A randomly chosen complex matrix is diagonalizable with probability $1$ - and finding that diagonalization is considerably more involved than just doing the multiplication. Sure, multiplying by a diagonal matrix is nice and simple, but if we didn't start there, converting to (diagonal * something) form requires two matrix multiplications - one by the similarity matrix, one by its inverse. Two matrix multiplications in order to compute one - not worth it.
answered Jan 17 at 7:31
jmerryjmerry
10.1k1225
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$begingroup$
I think you are confusing two things here. He did not ask for the efficiency of diagonalizing a matrix in order to perform a mutliplication, but if the fact that a matrix is diagonalizable can lead to an algorithm simplifying the multiplication.
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– nicomezi
Jan 17 at 10:15
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It's just not going to matter. A randomly chosen complex matrix is diagonalizable with probability $1$ - and finding that diagonalization is considerably more involved than just doing the multiplication. Sure, multiplying by a diagonal matrix is nice and simple, but if we didn't start there, converting to (diagonal * something) form requires two matrix multiplications - one by the similarity matrix, one by its inverse. Two matrix multiplications in order to compute one - not worth it.
answered Jan 17 at 7:31
jmerryjmerry
10.1k1225
$endgroup$
$begingroup$
I think you are confusing two things here. He did not ask for the efficiency of diagonalizing a matrix in order to perform a mutliplication, but if the fact that a matrix is diagonalizable can lead to an algorithm simplifying the multiplication.
$endgroup$
– nicomezi
Jan 17 at 10:15
add a comment |
$begingroup$
It's just not going to matter. A randomly chosen complex matrix is diagonalizable with probability $1$ - and finding that diagonalization is considerably more involved than just doing the multiplication. Sure, multiplying by a diagonal matrix is nice and simple, but if we didn't start there, converting to (diagonal * something) form requires two matrix multiplications - one by the similarity matrix, one by its inverse. Two matrix multiplications in order to compute one - not worth it.
answered Jan 17 at 7:31
jmerryjmerry
10.1k1225
$endgroup$
$begingroup$
I think you are confusing two things here. He did not ask for the efficiency of diagonalizing a matrix in order to perform a mutliplication, but if the fact that a matrix is diagonalizable can lead to an algorithm simplifying the multiplication.
$endgroup$
– nicomezi
Jan 17 at 10:15
add a comment |
$begingroup$
It's just not going to matter. A randomly chosen complex matrix is diagonalizable with probability $1$ - and finding that diagonalization is considerably more involved than just doing the multiplication. Sure, multiplying by a diagonal matrix is nice and simple, but if we didn't start there, converting to (diagonal * something) form requires two matrix multiplications - one by the similarity matrix, one by its inverse. Two matrix multiplications in order to compute one - not worth it.
answered Jan 17 at 7:31
jmerryjmerry
10.1k1225
$endgroup$
It's just not going to matter. A randomly chosen complex matrix is diagonalizable with probability $1$ - and finding that diagonalization is considerably more involved than just doing the multiplication. Sure, multiplying by a diagonal matrix is nice and simple, but if we didn't start there, converting to (diagonal * something) form requires two matrix multiplications - one by the similarity matrix, one by its inverse. Two matrix multiplications in order to compute one - not worth it.
answered Jan 17 at 7:31
jmerryjmerry
10.1k1225
answered Jan 17 at 7:31
jmerryjmerry
10.1k1225
answered Jan 17 at 7:31
jmerryjmerry
10.1k1225
answered Jan 17 at 7:31
jmerryjmerry
10.1k1225
10.1k1225
$begingroup$
I think you are confusing two things here. He did not ask for the efficiency of diagonalizing a matrix in order to perform a mutliplication, but if the fact that a matrix is diagonalizable can lead to an algorithm simplifying the multiplication.
$endgroup$
– nicomezi
Jan 17 at 10:15
add a comment |
$begingroup$
I think you are confusing two things here. He did not ask for the efficiency of diagonalizing a matrix in order to perform a mutliplication, but if the fact that a matrix is diagonalizable can lead to an algorithm simplifying the multiplication.
$endgroup$
– nicomezi
Jan 17 at 10:15
$begingroup$
I think you are confusing two things here. He did not ask for the efficiency of diagonalizing a matrix in order to perform a mutliplication, but if the fact that a matrix is diagonalizable can lead to an algorithm simplifying the multiplication.
$endgroup$
– nicomezi
Jan 17 at 10:15
$begingroup$
I think you are confusing two things here. He did not ask for the efficiency of diagonalizing a matrix in order to perform a mutliplication, but if the fact that a matrix is diagonalizable can lead to an algorithm simplifying the multiplication.
$endgroup$
– nicomezi
Jan 17 at 10:15
add a comment |
