Distributing a matrix The 2019 Stack Overflow Developer Survey Results Are InOn multiplying...
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Distributing a matrix
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Distributing a matrix
The 2019 Stack Overflow Developer Survey Results Are InOn multiplying quaternion matricesWhen is matrix multiplication commutative?Matrix multiplicationWhy aren't all matrices diagonalisable?Linear Transformation vs Matrixhow many ways is there to factor matrix?Can an arbitrary matrix represent any linear map just by changing the basis?Inverse matrix confusionA question matrix multiplication commutative?Joint Matrices Factorization
$begingroup$
Since matrix mutiplication is not commutative, the two ways in which you can factorize matrices makes a difference in which side the factor goes on.
In particular, if I want to distribute
$$((I - A) + A)(I - A)^{-1},$$
would it become
$$(I - A)(I - A)^{-1} + A(I - A)^{-1} $$
OR would it be
$$(I - A)^{-1}(I - A) + (I - A)^{-1}A?$$
How do I know which side it goes on? I think the first one is correct.
linear-algebra
asked 4 hours ago
redblacktreesredblacktrees
424
$endgroup$
add a comment |
$begingroup$
Since matrix mutiplication is not commutative, the two ways in which you can factorize matrices makes a difference in which side the factor goes on.
In particular, if I want to distribute
$$((I - A) + A)(I - A)^{-1},$$
would it become
$$(I - A)(I - A)^{-1} + A(I - A)^{-1} $$
OR would it be
$$(I - A)^{-1}(I - A) + (I - A)^{-1}A?$$
How do I know which side it goes on? I think the first one is correct.
linear-algebra
asked 4 hours ago
redblacktreesredblacktrees
424
$endgroup$
add a comment |
$begingroup$
Since matrix mutiplication is not commutative, the two ways in which you can factorize matrices makes a difference in which side the factor goes on.
In particular, if I want to distribute
$$((I - A) + A)(I - A)^{-1},$$
would it become
$$(I - A)(I - A)^{-1} + A(I - A)^{-1} $$
OR would it be
$$(I - A)^{-1}(I - A) + (I - A)^{-1}A?$$
How do I know which side it goes on? I think the first one is correct.
linear-algebra
asked 4 hours ago
redblacktreesredblacktrees
424
$endgroup$
Since matrix mutiplication is not commutative, the two ways in which you can factorize matrices makes a difference in which side the factor goes on.
In particular, if I want to distribute
$$((I - A) + A)(I - A)^{-1},$$
would it become
$$(I - A)(I - A)^{-1} + A(I - A)^{-1} $$
OR would it be
$$(I - A)^{-1}(I - A) + (I - A)^{-1}A?$$
How do I know which side it goes on? I think the first one is correct.
linear-algebra
linear-algebra
asked 4 hours ago
redblacktreesredblacktrees
424
asked 4 hours ago
redblacktreesredblacktrees
424
asked 4 hours ago
redblacktreesredblacktrees
424
asked 4 hours ago
redblacktreesredblacktrees
424
asked 4 hours ago
redblacktreesredblacktrees
424
424
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Your first answer is correct. There are two distributive laws for matrices,
$$A(B+C)=AB+ACquadhbox{and}quad (A+B)C=AC+BC ,$$
but not $A(B+C)=BA+CA$ or $(A+B)C=AC+CB$ or.....
answered 3 hours ago
DavidDavid
69.8k668131
$endgroup$
add a comment |
$begingroup$
In general, this is what we call "right distributivity" - I usually hear the context for this in the sense of ring axioms. Let's sojourn into this a bit - though if you're not familiar with abstract algebra, this won't be particularly enlightening, and you might be better off skipping to the very end.
Let $(R,+,cdot,0,1)$ be a ring; then we call left-distributivity and define it by
$$a cdot (b+c) = acdot b + a cdot c$$
Similarly, right-distributivity is given by
$$(b+c)cdot a = bcdot a + ccdot a$$
Note: we are not guaranteed that $acdot b = bcdot a$ unless $R$ is a commutative ring.
In the context of matrices over rings, for which I reference Wikipedia, you can define $M_n(R)$ as the $ntimes n$ matrices over a ring $R$ (i.e. its elements come from the ring, and the addition and multiplication of elements are shared). Notably, we have that $M_n(R)$ is a commutative ring if and only if $R$ is a commutative ring and $n=1$ (so basically effectively no different from working in the ring in question).
So what does this mean? This means, in your case, you probably do not have $AB=BA$ (of course, I imagine you know this). And thus in the context of the distributivity thigns above, you would have
$$(B+C)A = BA + CA$$
Your example has $B = I-A$ and $C=A$. And thus, your first example is correct: if you are distributing something on the right side, and cannot ensure commutativity, you should multiply that element by everything in the brackets on the right side.
answered 3 hours ago
Eevee TrainerEevee Trainer
10.4k31742
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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active
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votes
$begingroup$
Your first answer is correct. There are two distributive laws for matrices,
$$A(B+C)=AB+ACquadhbox{and}quad (A+B)C=AC+BC ,$$
but not $A(B+C)=BA+CA$ or $(A+B)C=AC+CB$ or.....
answered 3 hours ago
DavidDavid
69.8k668131
$endgroup$
add a comment |
$begingroup$
Your first answer is correct. There are two distributive laws for matrices,
$$A(B+C)=AB+ACquadhbox{and}quad (A+B)C=AC+BC ,$$
but not $A(B+C)=BA+CA$ or $(A+B)C=AC+CB$ or.....
answered 3 hours ago
DavidDavid
69.8k668131
$endgroup$
add a comment |
$begingroup$
Your first answer is correct. There are two distributive laws for matrices,
$$A(B+C)=AB+ACquadhbox{and}quad (A+B)C=AC+BC ,$$
but not $A(B+C)=BA+CA$ or $(A+B)C=AC+CB$ or.....
answered 3 hours ago
DavidDavid
69.8k668131
$endgroup$
Your first answer is correct. There are two distributive laws for matrices,
$$A(B+C)=AB+ACquadhbox{and}quad (A+B)C=AC+BC ,$$
but not $A(B+C)=BA+CA$ or $(A+B)C=AC+CB$ or.....
answered 3 hours ago
DavidDavid
69.8k668131
answered 3 hours ago
DavidDavid
69.8k668131
answered 3 hours ago
DavidDavid
69.8k668131
answered 3 hours ago
DavidDavid
69.8k668131
69.8k668131
add a comment |
add a comment |
$begingroup$
In general, this is what we call "right distributivity" - I usually hear the context for this in the sense of ring axioms. Let's sojourn into this a bit - though if you're not familiar with abstract algebra, this won't be particularly enlightening, and you might be better off skipping to the very end.
Let $(R,+,cdot,0,1)$ be a ring; then we call left-distributivity and define it by
$$a cdot (b+c) = acdot b + a cdot c$$
Similarly, right-distributivity is given by
$$(b+c)cdot a = bcdot a + ccdot a$$
Note: we are not guaranteed that $acdot b = bcdot a$ unless $R$ is a commutative ring.
In the context of matrices over rings, for which I reference Wikipedia, you can define $M_n(R)$ as the $ntimes n$ matrices over a ring $R$ (i.e. its elements come from the ring, and the addition and multiplication of elements are shared). Notably, we have that $M_n(R)$ is a commutative ring if and only if $R$ is a commutative ring and $n=1$ (so basically effectively no different from working in the ring in question).
So what does this mean? This means, in your case, you probably do not have $AB=BA$ (of course, I imagine you know this). And thus in the context of the distributivity thigns above, you would have
$$(B+C)A = BA + CA$$
Your example has $B = I-A$ and $C=A$. And thus, your first example is correct: if you are distributing something on the right side, and cannot ensure commutativity, you should multiply that element by everything in the brackets on the right side.
answered 3 hours ago
Eevee TrainerEevee Trainer
10.4k31742
$endgroup$
add a comment |
$begingroup$
In general, this is what we call "right distributivity" - I usually hear the context for this in the sense of ring axioms. Let's sojourn into this a bit - though if you're not familiar with abstract algebra, this won't be particularly enlightening, and you might be better off skipping to the very end.
Let $(R,+,cdot,0,1)$ be a ring; then we call left-distributivity and define it by
$$a cdot (b+c) = acdot b + a cdot c$$
Similarly, right-distributivity is given by
$$(b+c)cdot a = bcdot a + ccdot a$$
Note: we are not guaranteed that $acdot b = bcdot a$ unless $R$ is a commutative ring.
In the context of matrices over rings, for which I reference Wikipedia, you can define $M_n(R)$ as the $ntimes n$ matrices over a ring $R$ (i.e. its elements come from the ring, and the addition and multiplication of elements are shared). Notably, we have that $M_n(R)$ is a commutative ring if and only if $R$ is a commutative ring and $n=1$ (so basically effectively no different from working in the ring in question).
So what does this mean? This means, in your case, you probably do not have $AB=BA$ (of course, I imagine you know this). And thus in the context of the distributivity thigns above, you would have
$$(B+C)A = BA + CA$$
Your example has $B = I-A$ and $C=A$. And thus, your first example is correct: if you are distributing something on the right side, and cannot ensure commutativity, you should multiply that element by everything in the brackets on the right side.
answered 3 hours ago
Eevee TrainerEevee Trainer
10.4k31742
$endgroup$
add a comment |
$begingroup$
In general, this is what we call "right distributivity" - I usually hear the context for this in the sense of ring axioms. Let's sojourn into this a bit - though if you're not familiar with abstract algebra, this won't be particularly enlightening, and you might be better off skipping to the very end.
Let $(R,+,cdot,0,1)$ be a ring; then we call left-distributivity and define it by
$$a cdot (b+c) = acdot b + a cdot c$$
Similarly, right-distributivity is given by
$$(b+c)cdot a = bcdot a + ccdot a$$
Note: we are not guaranteed that $acdot b = bcdot a$ unless $R$ is a commutative ring.
In the context of matrices over rings, for which I reference Wikipedia, you can define $M_n(R)$ as the $ntimes n$ matrices over a ring $R$ (i.e. its elements come from the ring, and the addition and multiplication of elements are shared). Notably, we have that $M_n(R)$ is a commutative ring if and only if $R$ is a commutative ring and $n=1$ (so basically effectively no different from working in the ring in question).
So what does this mean? This means, in your case, you probably do not have $AB=BA$ (of course, I imagine you know this). And thus in the context of the distributivity thigns above, you would have
$$(B+C)A = BA + CA$$
Your example has $B = I-A$ and $C=A$. And thus, your first example is correct: if you are distributing something on the right side, and cannot ensure commutativity, you should multiply that element by everything in the brackets on the right side.
answered 3 hours ago
Eevee TrainerEevee Trainer
10.4k31742
$endgroup$
In general, this is what we call "right distributivity" - I usually hear the context for this in the sense of ring axioms. Let's sojourn into this a bit - though if you're not familiar with abstract algebra, this won't be particularly enlightening, and you might be better off skipping to the very end.
Let $(R,+,cdot,0,1)$ be a ring; then we call left-distributivity and define it by
$$a cdot (b+c) = acdot b + a cdot c$$
Similarly, right-distributivity is given by
$$(b+c)cdot a = bcdot a + ccdot a$$
Note: we are not guaranteed that $acdot b = bcdot a$ unless $R$ is a commutative ring.
In the context of matrices over rings, for which I reference Wikipedia, you can define $M_n(R)$ as the $ntimes n$ matrices over a ring $R$ (i.e. its elements come from the ring, and the addition and multiplication of elements are shared). Notably, we have that $M_n(R)$ is a commutative ring if and only if $R$ is a commutative ring and $n=1$ (so basically effectively no different from working in the ring in question).
So what does this mean? This means, in your case, you probably do not have $AB=BA$ (of course, I imagine you know this). And thus in the context of the distributivity thigns above, you would have
$$(B+C)A = BA + CA$$
Your example has $B = I-A$ and $C=A$. And thus, your first example is correct: if you are distributing something on the right side, and cannot ensure commutativity, you should multiply that element by everything in the brackets on the right side.
answered 3 hours ago
Eevee TrainerEevee Trainer
10.4k31742
answered 3 hours ago
Eevee TrainerEevee Trainer
10.4k31742
answered 3 hours ago
Eevee TrainerEevee Trainer
10.4k31742
answered 3 hours ago
Eevee TrainerEevee Trainer
10.4k31742
10.4k31742
add a comment |
add a comment |
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