Associative property in matrix multiplication

Clash Royale CLAN TAG#URR8PPP
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Does the associative property of matrix multiplication hold when one of the matrices is a permutation matrix?
For example consider the following:
X = [ 1 2 3 ] (a 1*3 matrix)
P = [ 0 1 0;
0 0 1;
1 0 0 ] (a 3 * 3 permutation matrix)
Y = transpose([1 4 2] ) (a 3 * 1 matrix)
Here (XP)Y = 9 and X(PY) = 11...
Can I conclude that the associative property of matrix multiplication holds only when all matrices have been reduced their "static" form after row and column exchanges - which is essentially what happens when you multiply by a permutation matrix?
linear-algebra
add a comment |Â
up vote
1
down vote
favorite
Does the associative property of matrix multiplication hold when one of the matrices is a permutation matrix?
For example consider the following:
X = [ 1 2 3 ] (a 1*3 matrix)
P = [ 0 1 0;
0 0 1;
1 0 0 ] (a 3 * 3 permutation matrix)
Y = transpose([1 4 2] ) (a 3 * 1 matrix)
Here (XP)Y = 9 and X(PY) = 11...
Can I conclude that the associative property of matrix multiplication holds only when all matrices have been reduced their "static" form after row and column exchanges - which is essentially what happens when you multiply by a permutation matrix?
linear-algebra
3
Matrix multiplication is emphalways associative.
â Igor Rivin
Jan 4 '14 at 3:05
Igor - in the example above, (XP)Y = 9 and X(PY) = 11.
â Student_forever
Jan 4 '14 at 3:20
2
Um, you made a miscalculation....
â Gina
Jan 4 '14 at 3:25
@IgorRivin What about ABC, where A:row matrix, B:column matrix and C:row matrix?
â jinawee
Mar 13 '14 at 17:40
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Does the associative property of matrix multiplication hold when one of the matrices is a permutation matrix?
For example consider the following:
X = [ 1 2 3 ] (a 1*3 matrix)
P = [ 0 1 0;
0 0 1;
1 0 0 ] (a 3 * 3 permutation matrix)
Y = transpose([1 4 2] ) (a 3 * 1 matrix)
Here (XP)Y = 9 and X(PY) = 11...
Can I conclude that the associative property of matrix multiplication holds only when all matrices have been reduced their "static" form after row and column exchanges - which is essentially what happens when you multiply by a permutation matrix?
linear-algebra
Does the associative property of matrix multiplication hold when one of the matrices is a permutation matrix?
For example consider the following:
X = [ 1 2 3 ] (a 1*3 matrix)
P = [ 0 1 0;
0 0 1;
1 0 0 ] (a 3 * 3 permutation matrix)
Y = transpose([1 4 2] ) (a 3 * 1 matrix)
Here (XP)Y = 9 and X(PY) = 11...
Can I conclude that the associative property of matrix multiplication holds only when all matrices have been reduced their "static" form after row and column exchanges - which is essentially what happens when you multiply by a permutation matrix?
linear-algebra
edited Jan 4 '14 at 3:33
user940
asked Jan 4 '14 at 3:00
Student_forever
61
61
3
Matrix multiplication is emphalways associative.
â Igor Rivin
Jan 4 '14 at 3:05
Igor - in the example above, (XP)Y = 9 and X(PY) = 11.
â Student_forever
Jan 4 '14 at 3:20
2
Um, you made a miscalculation....
â Gina
Jan 4 '14 at 3:25
@IgorRivin What about ABC, where A:row matrix, B:column matrix and C:row matrix?
â jinawee
Mar 13 '14 at 17:40
add a comment |Â
3
Matrix multiplication is emphalways associative.
â Igor Rivin
Jan 4 '14 at 3:05
Igor - in the example above, (XP)Y = 9 and X(PY) = 11.
â Student_forever
Jan 4 '14 at 3:20
2
Um, you made a miscalculation....
â Gina
Jan 4 '14 at 3:25
@IgorRivin What about ABC, where A:row matrix, B:column matrix and C:row matrix?
â jinawee
Mar 13 '14 at 17:40
3
3
Matrix multiplication is emphalways associative.
â Igor Rivin
Jan 4 '14 at 3:05
Matrix multiplication is emphalways associative.
â Igor Rivin
Jan 4 '14 at 3:05
Igor - in the example above, (XP)Y = 9 and X(PY) = 11.
â Student_forever
Jan 4 '14 at 3:20
Igor - in the example above, (XP)Y = 9 and X(PY) = 11.
â Student_forever
Jan 4 '14 at 3:20
2
2
Um, you made a miscalculation....
â Gina
Jan 4 '14 at 3:25
Um, you made a miscalculation....
â Gina
Jan 4 '14 at 3:25
@IgorRivin What about ABC, where A:row matrix, B:column matrix and C:row matrix?
â jinawee
Mar 13 '14 at 17:40
@IgorRivin What about ABC, where A:row matrix, B:column matrix and C:row matrix?
â jinawee
Mar 13 '14 at 17:40
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
1
down vote
$XP=[3 1 2]$ and $(XP)Y=3cdot1+1cdot4+2cdot2=11$.
Like Igor said, matrix multiplication is always associative.
add a comment |Â
up vote
1
down vote
Matrix multiplication is in fact just a composition of linear transformation. Just like function in general, composition of function is associative (intuitively, in does not even make sense to even talk about bracketing a function composition). So yes, multiplication is probably a somewhat misleading term.
add a comment |Â
up vote
0
down vote
I was not satisfied with any explanation that I've read in Internet about associativity of the matrices product.
Algebric proof based on matrices product calculation is cumbersome, but it works. I like to consider matrices as a linear transformation operator.
Let's do it.
For me it's very intuitive that the functions are associative.
$ colorblue f(g(x)) = (fcirc g)x $
For instance
$ colorbluef(x) = 2x+1 quad and quad g(x) = x^2$
$ colorbluef(2) = 5 qquad rightarrow qquad g(5) = 25$
also
$colorblueg(f(x)) = 4x^2 + 4x + 1 quad rightarrow quad f(g(2)) = 25$
Notice that we make the composition, using the operands like $colorbluex$ as a placeholder in place of the actual operand (in this case, $colorblue2$)
Imagine $colorblueF$, $colorblueX$, $colorblueM$, $colorblueT$, $colorblueC$ and $colorblueU$ as matrices.
Consider $colorblueR(M)$ as the number of $colorblueM$ rows and $colorblueC(M)$ the number of $colorblueM$ columns.
We prove intuitively that $colorblueT(UV) = (TU)V$
We know that, by the algebric definition or linear transformation paradigm $colorblueC(U) = R (V)$ and $colorblueC(V) = R(U)$
Because in a multiplication of matrices, the left matrix $colorblueF$ represents in practice $colorblueR (F)$ functions composed of $colorblueC(F)$ coefficients, which will be the parameters of a scalar product, which will result in a single number (for each function execution)
The right-hand matrix $colorblueX$, represents a collection of $colorblueC(X)$ operands, where each has $colorblueR(X)$ values âÂÂâÂÂfor each of the $colorblueR(F)$ functions to act. So in the end we will have $colorblueR(X) * C(X)$ values
Thus in $colorblueUV$ we produce a matrix $colorblueX$, with $colorblueR(U) * C(V)$ values, which will be the target of the functions represented by the left matrix $colorblueT$, generating final values, one for each function execution. In short, $colorblueT$ acts on the effect it has on $colorblueUV$,
When we first make $colorblueTU$ we produce $colorblueL (T) * R (U)$ values, but, in fact, values are more than numbers, they are a functional composition that will act together on the values of $colorblueV$, making the same effect of the function composition.
It is similar than function composition, the function composition can use placeholders in place of the actual operands in $colorblueV$. In the end, one can get the final result, replacing the placeholders for the real $colorblueV$ values.
The only difference, as described, is that an array involves several function executions and each function has several parameters.
Apart from this multiplicity, the associativity it is the same property that holds for functions.
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
$XP=[3 1 2]$ and $(XP)Y=3cdot1+1cdot4+2cdot2=11$.
Like Igor said, matrix multiplication is always associative.
add a comment |Â
up vote
1
down vote
$XP=[3 1 2]$ and $(XP)Y=3cdot1+1cdot4+2cdot2=11$.
Like Igor said, matrix multiplication is always associative.
add a comment |Â
up vote
1
down vote
up vote
1
down vote
$XP=[3 1 2]$ and $(XP)Y=3cdot1+1cdot4+2cdot2=11$.
Like Igor said, matrix multiplication is always associative.
$XP=[3 1 2]$ and $(XP)Y=3cdot1+1cdot4+2cdot2=11$.
Like Igor said, matrix multiplication is always associative.
answered Jan 4 '14 at 3:21
bof
46.1k348110
46.1k348110
add a comment |Â
add a comment |Â
up vote
1
down vote
Matrix multiplication is in fact just a composition of linear transformation. Just like function in general, composition of function is associative (intuitively, in does not even make sense to even talk about bracketing a function composition). So yes, multiplication is probably a somewhat misleading term.
add a comment |Â
up vote
1
down vote
Matrix multiplication is in fact just a composition of linear transformation. Just like function in general, composition of function is associative (intuitively, in does not even make sense to even talk about bracketing a function composition). So yes, multiplication is probably a somewhat misleading term.
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Matrix multiplication is in fact just a composition of linear transformation. Just like function in general, composition of function is associative (intuitively, in does not even make sense to even talk about bracketing a function composition). So yes, multiplication is probably a somewhat misleading term.
Matrix multiplication is in fact just a composition of linear transformation. Just like function in general, composition of function is associative (intuitively, in does not even make sense to even talk about bracketing a function composition). So yes, multiplication is probably a somewhat misleading term.
answered Jan 4 '14 at 3:28
Gina
4,5871033
4,5871033
add a comment |Â
add a comment |Â
up vote
0
down vote
I was not satisfied with any explanation that I've read in Internet about associativity of the matrices product.
Algebric proof based on matrices product calculation is cumbersome, but it works. I like to consider matrices as a linear transformation operator.
Let's do it.
For me it's very intuitive that the functions are associative.
$ colorblue f(g(x)) = (fcirc g)x $
For instance
$ colorbluef(x) = 2x+1 quad and quad g(x) = x^2$
$ colorbluef(2) = 5 qquad rightarrow qquad g(5) = 25$
also
$colorblueg(f(x)) = 4x^2 + 4x + 1 quad rightarrow quad f(g(2)) = 25$
Notice that we make the composition, using the operands like $colorbluex$ as a placeholder in place of the actual operand (in this case, $colorblue2$)
Imagine $colorblueF$, $colorblueX$, $colorblueM$, $colorblueT$, $colorblueC$ and $colorblueU$ as matrices.
Consider $colorblueR(M)$ as the number of $colorblueM$ rows and $colorblueC(M)$ the number of $colorblueM$ columns.
We prove intuitively that $colorblueT(UV) = (TU)V$
We know that, by the algebric definition or linear transformation paradigm $colorblueC(U) = R (V)$ and $colorblueC(V) = R(U)$
Because in a multiplication of matrices, the left matrix $colorblueF$ represents in practice $colorblueR (F)$ functions composed of $colorblueC(F)$ coefficients, which will be the parameters of a scalar product, which will result in a single number (for each function execution)
The right-hand matrix $colorblueX$, represents a collection of $colorblueC(X)$ operands, where each has $colorblueR(X)$ values âÂÂâÂÂfor each of the $colorblueR(F)$ functions to act. So in the end we will have $colorblueR(X) * C(X)$ values
Thus in $colorblueUV$ we produce a matrix $colorblueX$, with $colorblueR(U) * C(V)$ values, which will be the target of the functions represented by the left matrix $colorblueT$, generating final values, one for each function execution. In short, $colorblueT$ acts on the effect it has on $colorblueUV$,
When we first make $colorblueTU$ we produce $colorblueL (T) * R (U)$ values, but, in fact, values are more than numbers, they are a functional composition that will act together on the values of $colorblueV$, making the same effect of the function composition.
It is similar than function composition, the function composition can use placeholders in place of the actual operands in $colorblueV$. In the end, one can get the final result, replacing the placeholders for the real $colorblueV$ values.
The only difference, as described, is that an array involves several function executions and each function has several parameters.
Apart from this multiplicity, the associativity it is the same property that holds for functions.
add a comment |Â
up vote
0
down vote
I was not satisfied with any explanation that I've read in Internet about associativity of the matrices product.
Algebric proof based on matrices product calculation is cumbersome, but it works. I like to consider matrices as a linear transformation operator.
Let's do it.
For me it's very intuitive that the functions are associative.
$ colorblue f(g(x)) = (fcirc g)x $
For instance
$ colorbluef(x) = 2x+1 quad and quad g(x) = x^2$
$ colorbluef(2) = 5 qquad rightarrow qquad g(5) = 25$
also
$colorblueg(f(x)) = 4x^2 + 4x + 1 quad rightarrow quad f(g(2)) = 25$
Notice that we make the composition, using the operands like $colorbluex$ as a placeholder in place of the actual operand (in this case, $colorblue2$)
Imagine $colorblueF$, $colorblueX$, $colorblueM$, $colorblueT$, $colorblueC$ and $colorblueU$ as matrices.
Consider $colorblueR(M)$ as the number of $colorblueM$ rows and $colorblueC(M)$ the number of $colorblueM$ columns.
We prove intuitively that $colorblueT(UV) = (TU)V$
We know that, by the algebric definition or linear transformation paradigm $colorblueC(U) = R (V)$ and $colorblueC(V) = R(U)$
Because in a multiplication of matrices, the left matrix $colorblueF$ represents in practice $colorblueR (F)$ functions composed of $colorblueC(F)$ coefficients, which will be the parameters of a scalar product, which will result in a single number (for each function execution)
The right-hand matrix $colorblueX$, represents a collection of $colorblueC(X)$ operands, where each has $colorblueR(X)$ values âÂÂâÂÂfor each of the $colorblueR(F)$ functions to act. So in the end we will have $colorblueR(X) * C(X)$ values
Thus in $colorblueUV$ we produce a matrix $colorblueX$, with $colorblueR(U) * C(V)$ values, which will be the target of the functions represented by the left matrix $colorblueT$, generating final values, one for each function execution. In short, $colorblueT$ acts on the effect it has on $colorblueUV$,
When we first make $colorblueTU$ we produce $colorblueL (T) * R (U)$ values, but, in fact, values are more than numbers, they are a functional composition that will act together on the values of $colorblueV$, making the same effect of the function composition.
It is similar than function composition, the function composition can use placeholders in place of the actual operands in $colorblueV$. In the end, one can get the final result, replacing the placeholders for the real $colorblueV$ values.
The only difference, as described, is that an array involves several function executions and each function has several parameters.
Apart from this multiplicity, the associativity it is the same property that holds for functions.
add a comment |Â
up vote
0
down vote
up vote
0
down vote
I was not satisfied with any explanation that I've read in Internet about associativity of the matrices product.
Algebric proof based on matrices product calculation is cumbersome, but it works. I like to consider matrices as a linear transformation operator.
Let's do it.
For me it's very intuitive that the functions are associative.
$ colorblue f(g(x)) = (fcirc g)x $
For instance
$ colorbluef(x) = 2x+1 quad and quad g(x) = x^2$
$ colorbluef(2) = 5 qquad rightarrow qquad g(5) = 25$
also
$colorblueg(f(x)) = 4x^2 + 4x + 1 quad rightarrow quad f(g(2)) = 25$
Notice that we make the composition, using the operands like $colorbluex$ as a placeholder in place of the actual operand (in this case, $colorblue2$)
Imagine $colorblueF$, $colorblueX$, $colorblueM$, $colorblueT$, $colorblueC$ and $colorblueU$ as matrices.
Consider $colorblueR(M)$ as the number of $colorblueM$ rows and $colorblueC(M)$ the number of $colorblueM$ columns.
We prove intuitively that $colorblueT(UV) = (TU)V$
We know that, by the algebric definition or linear transformation paradigm $colorblueC(U) = R (V)$ and $colorblueC(V) = R(U)$
Because in a multiplication of matrices, the left matrix $colorblueF$ represents in practice $colorblueR (F)$ functions composed of $colorblueC(F)$ coefficients, which will be the parameters of a scalar product, which will result in a single number (for each function execution)
The right-hand matrix $colorblueX$, represents a collection of $colorblueC(X)$ operands, where each has $colorblueR(X)$ values âÂÂâÂÂfor each of the $colorblueR(F)$ functions to act. So in the end we will have $colorblueR(X) * C(X)$ values
Thus in $colorblueUV$ we produce a matrix $colorblueX$, with $colorblueR(U) * C(V)$ values, which will be the target of the functions represented by the left matrix $colorblueT$, generating final values, one for each function execution. In short, $colorblueT$ acts on the effect it has on $colorblueUV$,
When we first make $colorblueTU$ we produce $colorblueL (T) * R (U)$ values, but, in fact, values are more than numbers, they are a functional composition that will act together on the values of $colorblueV$, making the same effect of the function composition.
It is similar than function composition, the function composition can use placeholders in place of the actual operands in $colorblueV$. In the end, one can get the final result, replacing the placeholders for the real $colorblueV$ values.
The only difference, as described, is that an array involves several function executions and each function has several parameters.
Apart from this multiplicity, the associativity it is the same property that holds for functions.
I was not satisfied with any explanation that I've read in Internet about associativity of the matrices product.
Algebric proof based on matrices product calculation is cumbersome, but it works. I like to consider matrices as a linear transformation operator.
Let's do it.
For me it's very intuitive that the functions are associative.
$ colorblue f(g(x)) = (fcirc g)x $
For instance
$ colorbluef(x) = 2x+1 quad and quad g(x) = x^2$
$ colorbluef(2) = 5 qquad rightarrow qquad g(5) = 25$
also
$colorblueg(f(x)) = 4x^2 + 4x + 1 quad rightarrow quad f(g(2)) = 25$
Notice that we make the composition, using the operands like $colorbluex$ as a placeholder in place of the actual operand (in this case, $colorblue2$)
Imagine $colorblueF$, $colorblueX$, $colorblueM$, $colorblueT$, $colorblueC$ and $colorblueU$ as matrices.
Consider $colorblueR(M)$ as the number of $colorblueM$ rows and $colorblueC(M)$ the number of $colorblueM$ columns.
We prove intuitively that $colorblueT(UV) = (TU)V$
We know that, by the algebric definition or linear transformation paradigm $colorblueC(U) = R (V)$ and $colorblueC(V) = R(U)$
Because in a multiplication of matrices, the left matrix $colorblueF$ represents in practice $colorblueR (F)$ functions composed of $colorblueC(F)$ coefficients, which will be the parameters of a scalar product, which will result in a single number (for each function execution)
The right-hand matrix $colorblueX$, represents a collection of $colorblueC(X)$ operands, where each has $colorblueR(X)$ values âÂÂâÂÂfor each of the $colorblueR(F)$ functions to act. So in the end we will have $colorblueR(X) * C(X)$ values
Thus in $colorblueUV$ we produce a matrix $colorblueX$, with $colorblueR(U) * C(V)$ values, which will be the target of the functions represented by the left matrix $colorblueT$, generating final values, one for each function execution. In short, $colorblueT$ acts on the effect it has on $colorblueUV$,
When we first make $colorblueTU$ we produce $colorblueL (T) * R (U)$ values, but, in fact, values are more than numbers, they are a functional composition that will act together on the values of $colorblueV$, making the same effect of the function composition.
It is similar than function composition, the function composition can use placeholders in place of the actual operands in $colorblueV$. In the end, one can get the final result, replacing the placeholders for the real $colorblueV$ values.
The only difference, as described, is that an array involves several function executions and each function has several parameters.
Apart from this multiplicity, the associativity it is the same property that holds for functions.
edited Aug 14 at 2:45
answered Aug 14 at 2:29
Paulo Buchsbaum
1013
1013
add a comment |Â
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3
Matrix multiplication is emphalways associative.
â Igor Rivin
Jan 4 '14 at 3:05
Igor - in the example above, (XP)Y = 9 and X(PY) = 11.
â Student_forever
Jan 4 '14 at 3:20
2
Um, you made a miscalculation....
â Gina
Jan 4 '14 at 3:25
@IgorRivin What about ABC, where A:row matrix, B:column matrix and C:row matrix?
â jinawee
Mar 13 '14 at 17:40