What is the best algorithm to find the inverse of matrix $A$












-1












$begingroup$


I'm going to build a C-library so I can do linear algebra at embedded systems. It's most for machine learning. https://github.com/DanielMartensson/EmbeddedAlgebra/



Anyway, I need to compute inverse of matrix $A$. What is the absolute best algorithm to use for that? Remember it must be a numerical algorithm.



Which one should I use?



Analytic solution



$mathbf{A}^{-1}={1 over begin{vmatrix}mathbf{A}end{vmatrix}}mathbf{C}^{mathrm{T}}={1 over begin{vmatrix}mathbf{A}end{vmatrix}}
begin{pmatrix}
mathbf{C}_{11} & mathbf{C}_{21} & cdots & mathbf{C}_{n1} \
mathbf{C}_{12} & mathbf{C}_{22} & cdots & mathbf{C}_{n2} \
vdots & vdots & ddots & vdots \
mathbf{C}_{1n} & mathbf{C}_{2n} & cdots & mathbf{C}_{nn} \
end{pmatrix}$



Cholesky decomposition



$mathbf{A}^{-1} = (mathbf{L}^{*})^{-1} mathbf{L}^{-1}$



Eigendecomposition



$mathbf{A}^{-1}=mathbf{Q}mathbf{Lambda}^{-1}mathbf{Q}^{-1}$



Cayley–Hamilton method



$mathbf{A}^{-1} = frac{1}{det (mathbf{A})}sum_{s=0}^{n-1}A^s$



Newton's method



$X_{k+1} = 2X_k - X_k A X_k$



Neuman series
$lim_{n to infty} (mathbf I - mathbf A)^n = 0$



Gaussian elimination



$A =
begin{bmatrix}
2 & -1 & 0 \
-1 & 2 & -1 \
0 & -1 & 2
end{bmatrix}$



To find the inverse of this matrix, one takes the following matrix augmented by the identity, and row reduces it as a 3 × 6 matrix:
$
[ A | I ] =
left[ begin{array}{rrr|rrr}
2 & -1 & 0 & 1 & 0 & 0\
-1 & 2 & -1 & 0 & 1 & 0\
0 & -1 & 2 & 0 & 0 & 1
end{array} right]$



By performing row operations, one can check that the reduced row echelon form of this augmented matrix is:



$[ I | B ] =
left[ begin{array}{rrr|rrr}
1 & 0 & 0 & frac34 & frac12 & frac14\[3pt]
0 & 1 & 0 & frac12 & 1 & frac12\[3pt]
0 & 0 & 1 & frac14 & frac12 & frac34
end{array} right]$






matrices inverse numerical-linear-algebra







share|cite|improve this question









$endgroup$








  • 5




    $begingroup$
    As a non-mathematical comment: for the love of god, don't. Firstly, you probably don't want to invert a matrix at all (see, for example, here). Secondly, use one of the many impementations that already exist.
    $endgroup$
    – user3482749
    Jan 11 at 20:16










  • $begingroup$
    So what should I do?
    $endgroup$
    – Daniel Mårtensson
    Jan 11 at 21:24










  • $begingroup$
    In almost all cases, you should just solve $Ax = b$, since that's probably what you want to do anyway, and it's vastly quicker than computing $A^{-1}$ then calculating $A^{-1}b$ (indeed, it's quicker than either of those steps alone). If you really must invert $A$, just use one of the implementations that already exist in other people's algorithms.
    $endgroup$
    – user3482749
    Jan 11 at 21:33






  • 1




    $begingroup$
    See here. The correct, answer, however, is almost certainly "just use somebody else's solution".
    $endgroup$
    – user3482749
    Jan 11 at 22:35








  • 1




    $begingroup$
    All six of them are in C.
    $endgroup$
    – user3482749
    Jan 12 at 11:27
















-1












$begingroup$


I'm going to build a C-library so I can do linear algebra at embedded systems. It's most for machine learning. https://github.com/DanielMartensson/EmbeddedAlgebra/



Anyway, I need to compute inverse of matrix $A$. What is the absolute best algorithm to use for that? Remember it must be a numerical algorithm.



Which one should I use?



Analytic solution



$mathbf{A}^{-1}={1 over begin{vmatrix}mathbf{A}end{vmatrix}}mathbf{C}^{mathrm{T}}={1 over begin{vmatrix}mathbf{A}end{vmatrix}}
begin{pmatrix}
mathbf{C}_{11} & mathbf{C}_{21} & cdots & mathbf{C}_{n1} \
mathbf{C}_{12} & mathbf{C}_{22} & cdots & mathbf{C}_{n2} \
vdots & vdots & ddots & vdots \
mathbf{C}_{1n} & mathbf{C}_{2n} & cdots & mathbf{C}_{nn} \
end{pmatrix}$



Cholesky decomposition



$mathbf{A}^{-1} = (mathbf{L}^{*})^{-1} mathbf{L}^{-1}$



Eigendecomposition



$mathbf{A}^{-1}=mathbf{Q}mathbf{Lambda}^{-1}mathbf{Q}^{-1}$



Cayley–Hamilton method



$mathbf{A}^{-1} = frac{1}{det (mathbf{A})}sum_{s=0}^{n-1}A^s$



Newton's method



$X_{k+1} = 2X_k - X_k A X_k$



Neuman series
$lim_{n to infty} (mathbf I - mathbf A)^n = 0$



Gaussian elimination



$A =
begin{bmatrix}
2 & -1 & 0 \
-1 & 2 & -1 \
0 & -1 & 2
end{bmatrix}$



To find the inverse of this matrix, one takes the following matrix augmented by the identity, and row reduces it as a 3 × 6 matrix:
$
[ A | I ] =
left[ begin{array}{rrr|rrr}
2 & -1 & 0 & 1 & 0 & 0\
-1 & 2 & -1 & 0 & 1 & 0\
0 & -1 & 2 & 0 & 0 & 1
end{array} right]$



By performing row operations, one can check that the reduced row echelon form of this augmented matrix is:



$[ I | B ] =
left[ begin{array}{rrr|rrr}
1 & 0 & 0 & frac34 & frac12 & frac14\[3pt]
0 & 1 & 0 & frac12 & 1 & frac12\[3pt]
0 & 0 & 1 & frac14 & frac12 & frac34
end{array} right]$






matrices inverse numerical-linear-algebra







share|cite|improve this question









$endgroup$








  • 5




    $begingroup$
    As a non-mathematical comment: for the love of god, don't. Firstly, you probably don't want to invert a matrix at all (see, for example, here). Secondly, use one of the many impementations that already exist.
    $endgroup$
    – user3482749
    Jan 11 at 20:16










  • $begingroup$
    So what should I do?
    $endgroup$
    – Daniel Mårtensson
    Jan 11 at 21:24










  • $begingroup$
    In almost all cases, you should just solve $Ax = b$, since that's probably what you want to do anyway, and it's vastly quicker than computing $A^{-1}$ then calculating $A^{-1}b$ (indeed, it's quicker than either of those steps alone). If you really must invert $A$, just use one of the implementations that already exist in other people's algorithms.
    $endgroup$
    – user3482749
    Jan 11 at 21:33






  • 1




    $begingroup$
    See here. The correct, answer, however, is almost certainly "just use somebody else's solution".
    $endgroup$
    – user3482749
    Jan 11 at 22:35








  • 1




    $begingroup$
    All six of them are in C.
    $endgroup$
    – user3482749
    Jan 12 at 11:27














-1












-1








-1





$begingroup$


I'm going to build a C-library so I can do linear algebra at embedded systems. It's most for machine learning. https://github.com/DanielMartensson/EmbeddedAlgebra/



Anyway, I need to compute inverse of matrix $A$. What is the absolute best algorithm to use for that? Remember it must be a numerical algorithm.



Which one should I use?



Analytic solution



$mathbf{A}^{-1}={1 over begin{vmatrix}mathbf{A}end{vmatrix}}mathbf{C}^{mathrm{T}}={1 over begin{vmatrix}mathbf{A}end{vmatrix}}
begin{pmatrix}
mathbf{C}_{11} & mathbf{C}_{21} & cdots & mathbf{C}_{n1} \
mathbf{C}_{12} & mathbf{C}_{22} & cdots & mathbf{C}_{n2} \
vdots & vdots & ddots & vdots \
mathbf{C}_{1n} & mathbf{C}_{2n} & cdots & mathbf{C}_{nn} \
end{pmatrix}$



Cholesky decomposition



$mathbf{A}^{-1} = (mathbf{L}^{*})^{-1} mathbf{L}^{-1}$



Eigendecomposition



$mathbf{A}^{-1}=mathbf{Q}mathbf{Lambda}^{-1}mathbf{Q}^{-1}$



Cayley–Hamilton method



$mathbf{A}^{-1} = frac{1}{det (mathbf{A})}sum_{s=0}^{n-1}A^s$



Newton's method



$X_{k+1} = 2X_k - X_k A X_k$



Neuman series
$lim_{n to infty} (mathbf I - mathbf A)^n = 0$



Gaussian elimination



$A =
begin{bmatrix}
2 & -1 & 0 \
-1 & 2 & -1 \
0 & -1 & 2
end{bmatrix}$



To find the inverse of this matrix, one takes the following matrix augmented by the identity, and row reduces it as a 3 × 6 matrix:
$
[ A | I ] =
left[ begin{array}{rrr|rrr}
2 & -1 & 0 & 1 & 0 & 0\
-1 & 2 & -1 & 0 & 1 & 0\
0 & -1 & 2 & 0 & 0 & 1
end{array} right]$



By performing row operations, one can check that the reduced row echelon form of this augmented matrix is:



$[ I | B ] =
left[ begin{array}{rrr|rrr}
1 & 0 & 0 & frac34 & frac12 & frac14\[3pt]
0 & 1 & 0 & frac12 & 1 & frac12\[3pt]
0 & 0 & 1 & frac14 & frac12 & frac34
end{array} right]$






matrices inverse numerical-linear-algebra







share|cite|improve this question









$endgroup$




I'm going to build a C-library so I can do linear algebra at embedded systems. It's most for machine learning. https://github.com/DanielMartensson/EmbeddedAlgebra/



Anyway, I need to compute inverse of matrix $A$. What is the absolute best algorithm to use for that? Remember it must be a numerical algorithm.



Which one should I use?



Analytic solution



$mathbf{A}^{-1}={1 over begin{vmatrix}mathbf{A}end{vmatrix}}mathbf{C}^{mathrm{T}}={1 over begin{vmatrix}mathbf{A}end{vmatrix}}
begin{pmatrix}
mathbf{C}_{11} & mathbf{C}_{21} & cdots & mathbf{C}_{n1} \
mathbf{C}_{12} & mathbf{C}_{22} & cdots & mathbf{C}_{n2} \
vdots & vdots & ddots & vdots \
mathbf{C}_{1n} & mathbf{C}_{2n} & cdots & mathbf{C}_{nn} \
end{pmatrix}$



Cholesky decomposition



$mathbf{A}^{-1} = (mathbf{L}^{*})^{-1} mathbf{L}^{-1}$



Eigendecomposition



$mathbf{A}^{-1}=mathbf{Q}mathbf{Lambda}^{-1}mathbf{Q}^{-1}$



Cayley–Hamilton method



$mathbf{A}^{-1} = frac{1}{det (mathbf{A})}sum_{s=0}^{n-1}A^s$



Newton's method



$X_{k+1} = 2X_k - X_k A X_k$



Neuman series
$lim_{n to infty} (mathbf I - mathbf A)^n = 0$



Gaussian elimination



$A =
begin{bmatrix}
2 & -1 & 0 \
-1 & 2 & -1 \
0 & -1 & 2
end{bmatrix}$



To find the inverse of this matrix, one takes the following matrix augmented by the identity, and row reduces it as a 3 × 6 matrix:
$
[ A | I ] =
left[ begin{array}{rrr|rrr}
2 & -1 & 0 & 1 & 0 & 0\
-1 & 2 & -1 & 0 & 1 & 0\
0 & -1 & 2 & 0 & 0 & 1
end{array} right]$



By performing row operations, one can check that the reduced row echelon form of this augmented matrix is:



$[ I | B ] =
left[ begin{array}{rrr|rrr}
1 & 0 & 0 & frac34 & frac12 & frac14\[3pt]
0 & 1 & 0 & frac12 & 1 & frac12\[3pt]
0 & 0 & 1 & frac14 & frac12 & frac34
end{array} right]$






matrices inverse numerical-linear-algebra




matrices inverse numerical-linear-algebra






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 11 at 20:03









Daniel MårtenssonDaniel Mårtensson

993419




993419








  • 5




    $begingroup$
    As a non-mathematical comment: for the love of god, don't. Firstly, you probably don't want to invert a matrix at all (see, for example, here). Secondly, use one of the many impementations that already exist.
    $endgroup$
    – user3482749
    Jan 11 at 20:16










  • $begingroup$
    So what should I do?
    $endgroup$
    – Daniel Mårtensson
    Jan 11 at 21:24










  • $begingroup$
    In almost all cases, you should just solve $Ax = b$, since that's probably what you want to do anyway, and it's vastly quicker than computing $A^{-1}$ then calculating $A^{-1}b$ (indeed, it's quicker than either of those steps alone). If you really must invert $A$, just use one of the implementations that already exist in other people's algorithms.
    $endgroup$
    – user3482749
    Jan 11 at 21:33






  • 1




    $begingroup$
    See here. The correct, answer, however, is almost certainly "just use somebody else's solution".
    $endgroup$
    – user3482749
    Jan 11 at 22:35








  • 1




    $begingroup$
    All six of them are in C.
    $endgroup$
    – user3482749
    Jan 12 at 11:27














  • 5




    $begingroup$
    As a non-mathematical comment: for the love of god, don't. Firstly, you probably don't want to invert a matrix at all (see, for example, here). Secondly, use one of the many impementations that already exist.
    $endgroup$
    – user3482749
    Jan 11 at 20:16










  • $begingroup$
    So what should I do?
    $endgroup$
    – Daniel Mårtensson
    Jan 11 at 21:24










  • $begingroup$
    In almost all cases, you should just solve $Ax = b$, since that's probably what you want to do anyway, and it's vastly quicker than computing $A^{-1}$ then calculating $A^{-1}b$ (indeed, it's quicker than either of those steps alone). If you really must invert $A$, just use one of the implementations that already exist in other people's algorithms.
    $endgroup$
    – user3482749
    Jan 11 at 21:33






  • 1




    $begingroup$
    See here. The correct, answer, however, is almost certainly "just use somebody else's solution".
    $endgroup$
    – user3482749
    Jan 11 at 22:35








  • 1




    $begingroup$
    All six of them are in C.
    $endgroup$
    – user3482749
    Jan 12 at 11:27








5




5




$begingroup$
As a non-mathematical comment: for the love of god, don't. Firstly, you probably don't want to invert a matrix at all (see, for example, here). Secondly, use one of the many impementations that already exist.
$endgroup$
– user3482749
Jan 11 at 20:16




$begingroup$
As a non-mathematical comment: for the love of god, don't. Firstly, you probably don't want to invert a matrix at all (see, for example, here). Secondly, use one of the many impementations that already exist.
$endgroup$
– user3482749
Jan 11 at 20:16












$begingroup$
So what should I do?
$endgroup$
– Daniel Mårtensson
Jan 11 at 21:24




$begingroup$
So what should I do?
$endgroup$
– Daniel Mårtensson
Jan 11 at 21:24












$begingroup$
In almost all cases, you should just solve $Ax = b$, since that's probably what you want to do anyway, and it's vastly quicker than computing $A^{-1}$ then calculating $A^{-1}b$ (indeed, it's quicker than either of those steps alone). If you really must invert $A$, just use one of the implementations that already exist in other people's algorithms.
$endgroup$
– user3482749
Jan 11 at 21:33




$begingroup$
In almost all cases, you should just solve $Ax = b$, since that's probably what you want to do anyway, and it's vastly quicker than computing $A^{-1}$ then calculating $A^{-1}b$ (indeed, it's quicker than either of those steps alone). If you really must invert $A$, just use one of the implementations that already exist in other people's algorithms.
$endgroup$
– user3482749
Jan 11 at 21:33




1




1




$begingroup$
See here. The correct, answer, however, is almost certainly "just use somebody else's solution".
$endgroup$
– user3482749
Jan 11 at 22:35






$begingroup$
See here. The correct, answer, however, is almost certainly "just use somebody else's solution".
$endgroup$
– user3482749
Jan 11 at 22:35






1




1




$begingroup$
All six of them are in C.
$endgroup$
– user3482749
Jan 12 at 11:27




$begingroup$
All six of them are in C.
$endgroup$
– user3482749
Jan 12 at 11:27










1 Answer
1






active

oldest

votes


















3












$begingroup$

It truly depends on the type of matrix you're going to compute the inverse from. Some methods are better for some classes of matrices than other.



But more importantly, why do you want to invert matrices? In many problems, you don't need to invert matrices, but only need to apply the inverse to some vectors. The latter problem is much easier to tackle, especially from a computational complexity standpoint (e.g. if your matrix is very large) and stability point of view (a matrix could have bad conditioning, be numerically close to non-invertible etc...). Does your application require that compute the matrix inverse?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    No. But I need to find the inverse. Compute it or not.
    $endgroup$
    – Daniel Mårtensson
    Jan 11 at 21:26












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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

It truly depends on the type of matrix you're going to compute the inverse from. Some methods are better for some classes of matrices than other.



But more importantly, why do you want to invert matrices? In many problems, you don't need to invert matrices, but only need to apply the inverse to some vectors. The latter problem is much easier to tackle, especially from a computational complexity standpoint (e.g. if your matrix is very large) and stability point of view (a matrix could have bad conditioning, be numerically close to non-invertible etc...). Does your application require that compute the matrix inverse?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    No. But I need to find the inverse. Compute it or not.
    $endgroup$
    – Daniel Mårtensson
    Jan 11 at 21:26
















3












$begingroup$

It truly depends on the type of matrix you're going to compute the inverse from. Some methods are better for some classes of matrices than other.



But more importantly, why do you want to invert matrices? In many problems, you don't need to invert matrices, but only need to apply the inverse to some vectors. The latter problem is much easier to tackle, especially from a computational complexity standpoint (e.g. if your matrix is very large) and stability point of view (a matrix could have bad conditioning, be numerically close to non-invertible etc...). Does your application require that compute the matrix inverse?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    No. But I need to find the inverse. Compute it or not.
    $endgroup$
    – Daniel Mårtensson
    Jan 11 at 21:26














3












3








3





$begingroup$

It truly depends on the type of matrix you're going to compute the inverse from. Some methods are better for some classes of matrices than other.



But more importantly, why do you want to invert matrices? In many problems, you don't need to invert matrices, but only need to apply the inverse to some vectors. The latter problem is much easier to tackle, especially from a computational complexity standpoint (e.g. if your matrix is very large) and stability point of view (a matrix could have bad conditioning, be numerically close to non-invertible etc...). Does your application require that compute the matrix inverse?






share|cite|improve this answer









$endgroup$



It truly depends on the type of matrix you're going to compute the inverse from. Some methods are better for some classes of matrices than other.



But more importantly, why do you want to invert matrices? In many problems, you don't need to invert matrices, but only need to apply the inverse to some vectors. The latter problem is much easier to tackle, especially from a computational complexity standpoint (e.g. if your matrix is very large) and stability point of view (a matrix could have bad conditioning, be numerically close to non-invertible etc...). Does your application require that compute the matrix inverse?







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 11 at 20:14









Stefan LafonStefan Lafon

3,005212




3,005212












  • $begingroup$
    No. But I need to find the inverse. Compute it or not.
    $endgroup$
    – Daniel Mårtensson
    Jan 11 at 21:26


















  • $begingroup$
    No. But I need to find the inverse. Compute it or not.
    $endgroup$
    – Daniel Mårtensson
    Jan 11 at 21:26
















$begingroup$
No. But I need to find the inverse. Compute it or not.
$endgroup$
– Daniel Mårtensson
Jan 11 at 21:26




$begingroup$
No. But I need to find the inverse. Compute it or not.
$endgroup$
– Daniel Mårtensson
Jan 11 at 21:26


















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.mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
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