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

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asked Jan 17 at 6:56

Santiago AranguriSantiago Aranguri

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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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-1

$begingroup$

Are there efficient algorithms to compute Ax in the case of A being diagonalizable instead of not being?

numerical-linear-algebra

share|cite|improve this question

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 | 

-1

-1

-1

$begingroup$

Are there efficient algorithms to compute Ax in the case of A being diagonalizable instead of not being?

numerical-linear-algebra

share|cite|improve this question

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

share|cite|improve this question

asked Jan 17 at 6:56

Santiago AranguriSantiago Aranguri

12

share|cite|improve this question

asked Jan 17 at 6:56

Santiago AranguriSantiago Aranguri

12

share|cite|improve this question

share|cite|improve this question

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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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.

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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.
  $endgroup$
  – 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

1

$begingroup$

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.

share|cite|improve this answer

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 | 

1

$begingroup$

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.

share|cite|improve this answer

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 | 

1

1

1

$begingroup$

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.

share|cite|improve this answer

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.

share|cite|improve this answer

answered Jan 17 at 7:31

jmerryjmerry

10.1k1225

share|cite|improve this answer

share|cite|improve this answer

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 | 

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