Distributing a matrix
$begingroup$
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
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$begingroup$
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
$endgroup$
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
$endgroup$
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 Apr 11 at 2:15
user646175
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2 Answers
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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 Apr 11 at 2:26
Eevee TrainerEevee Trainer
10.5k31842
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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 Apr 11 at 2:27
DavidDavid
69.9k668131
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2 Answers
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2 Answers
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$begingroup$
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 Apr 11 at 2:26
Eevee TrainerEevee Trainer
10.5k31842
$endgroup$
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 Apr 11 at 2:26
Eevee TrainerEevee Trainer
10.5k31842
$endgroup$
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 Apr 11 at 2:26
Eevee TrainerEevee Trainer
10.5k31842
$endgroup$
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 Apr 11 at 2:26
Eevee TrainerEevee Trainer
10.5k31842
answered Apr 11 at 2:26
Eevee TrainerEevee Trainer
10.5k31842
answered Apr 11 at 2:26
Eevee TrainerEevee Trainer
10.5k31842
answered Apr 11 at 2:26
Eevee TrainerEevee Trainer
10.5k31842
10.5k31842
add a comment |
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$begingroup$
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 Apr 11 at 2:27
DavidDavid
69.9k668131
$endgroup$
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 Apr 11 at 2:27
DavidDavid
69.9k668131
$endgroup$
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 Apr 11 at 2:27
DavidDavid
69.9k668131
$endgroup$
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 Apr 11 at 2:27
DavidDavid
69.9k668131
answered Apr 11 at 2:27
DavidDavid
69.9k668131
answered Apr 11 at 2:27
DavidDavid
69.9k668131
answered Apr 11 at 2:27
DavidDavid
69.9k668131
69.9k668131
add a comment |
add a comment |
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