Distributing a matrix The 2019 Stack Overflow Developer Survey Results Are InOn multiplying...
Am I thawing this London Broil safely?
Why did Acorn's A3000 have red function keys?
Distributing a matrix
How to save as into a customized destination on macOS?
"as much details as you can remember"
Resizing object distorts it (Illustrator CC 2018)
One word riddle: Vowel in the middle
Is flight data recorder erased after every flight?
Why can Shazam fly?
Is this app Icon Browser Safe/Legit?
Why is the maximum length of OpenWrt’s root password 8 characters?
Output the Arecibo Message
Origin of "cooter" meaning "vagina"
Can a flute soloist sit?
Apparent duplicates between Haynes service instructions and MOT
Is a "Democratic" Oligarchy-Style System Possible?
Is there a symbol for a right arrow with a square in the middle?
How to type this arrow in math mode?
Feature engineering suggestion required
Pokemon Turn Based battle (Python)
Why was M87 targetted for the Event Horizon Telescope instead of Sagittarius A*?
Why didn't the Event Horizon Telescope team mention Sagittarius A*?
If I score a critical hit on an 18 or higher, what are my chances of getting a critical hit if I roll 3d20?
Right tool to dig six foot holes?
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 |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3183231%2fdistributing-a-matrix%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
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 |
$begingroup$
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 |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3183231%2fdistributing-a-matrix%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
