Help understanding proof for: B is row equivalent to A iff B=PA where P is product of elementary matrices.
$begingroup$
In Hoffman and Kunze's Linear Algebra, following proof is given for this corollary:
Let A and B be m x n matrices over the field F. Then B is row-equivalent to A if and only if B=PA, where P is a product of m x m elementary matrices.
Proof. (1)Suppose B=PA where P= Es · · · E2E1 and the Ei are m x m elementary matrices. Then E1A is row-equivalent to A, and E2(E1A ) is row-equivalent to E1A . So E2E1A is row-equivalent to A ; and continuing in this way we see that (Es . . . E1)A is row-equivalent to A.
(2)Now suppose that B is row-equivalent to A. Let E1, E2,...,Es
the elementary matrices corresponding to some sequence of elementary row operations which carries A into B. Then B = (Es . . . E1)A
Although I get the corollary intuitively, as each elementary matrix is equivalent to elementary row operation, and thus multiplication of matrix with A is equivalent to series of applications of such elementary row operations. But I don't get how it's being proved here.
- Particularly, how (1) and (2) are connected? Moreover, in (2) we are supposing what we had to actually prove, so how does it help? Also, it sees quite like re-statement of corollary itself.
- In general, what constitutes a mathematical proof? In other words, how do we judge whether it's sufficient and correct? I did go through few questions here, like this one, but how to validate this proof specifically (especially differentiating it with intuition or re-statement)
linear-algebra proof-explanation
asked Dec 29 '18 at 19:42
dheeraj suthardheeraj suthar
72
$endgroup$
add a comment |
$begingroup$
In Hoffman and Kunze's Linear Algebra, following proof is given for this corollary:
Let A and B be m x n matrices over the field F. Then B is row-equivalent to A if and only if B=PA, where P is a product of m x m elementary matrices.
Proof. (1)Suppose B=PA where P= Es · · · E2E1 and the Ei are m x m elementary matrices. Then E1A is row-equivalent to A, and E2(E1A ) is row-equivalent to E1A . So E2E1A is row-equivalent to A ; and continuing in this way we see that (Es . . . E1)A is row-equivalent to A.
(2)Now suppose that B is row-equivalent to A. Let E1, E2,...,Es
the elementary matrices corresponding to some sequence of elementary row operations which carries A into B. Then B = (Es . . . E1)A
Although I get the corollary intuitively, as each elementary matrix is equivalent to elementary row operation, and thus multiplication of matrix with A is equivalent to series of applications of such elementary row operations. But I don't get how it's being proved here.
- Particularly, how (1) and (2) are connected? Moreover, in (2) we are supposing what we had to actually prove, so how does it help? Also, it sees quite like re-statement of corollary itself.
- In general, what constitutes a mathematical proof? In other words, how do we judge whether it's sufficient and correct? I did go through few questions here, like this one, but how to validate this proof specifically (especially differentiating it with intuition or re-statement)
linear-algebra proof-explanation
asked Dec 29 '18 at 19:42
dheeraj suthardheeraj suthar
72
$endgroup$
1
$begingroup$
In an "$X$ if and only if $Y$" proof, you have to prove two separate things: that $X$ implies $Y$, and that $Y$ implies $X$. This is what (1) and (2) are doing, and indeed they are separate proofs. In (1) you assume $X$ and prove $Y$, while in (2) you assume $Y$ and prove $X$.
$endgroup$
– angryavian
Dec 29 '18 at 20:02
$begingroup$
@angryavian Ahh.. Thanks. Got it. I was confused with the abrupt conclusion in (2). I now got it (it comes from the theorem which this corollary belongs to. e(A) = EA). Will be more patient in further reading. Should I delete this question, or any way I can accept your comment as answer?
$endgroup$
– dheeraj suthar
Dec 29 '18 at 20:24
$begingroup$
I copied my comment as an answer.
$endgroup$
– angryavian
Dec 29 '18 at 20:40
add a comment |
$begingroup$
In Hoffman and Kunze's Linear Algebra, following proof is given for this corollary:
Let A and B be m x n matrices over the field F. Then B is row-equivalent to A if and only if B=PA, where P is a product of m x m elementary matrices.
Proof. (1)Suppose B=PA where P= Es · · · E2E1 and the Ei are m x m elementary matrices. Then E1A is row-equivalent to A, and E2(E1A ) is row-equivalent to E1A . So E2E1A is row-equivalent to A ; and continuing in this way we see that (Es . . . E1)A is row-equivalent to A.
(2)Now suppose that B is row-equivalent to A. Let E1, E2,...,Es
the elementary matrices corresponding to some sequence of elementary row operations which carries A into B. Then B = (Es . . . E1)A
Although I get the corollary intuitively, as each elementary matrix is equivalent to elementary row operation, and thus multiplication of matrix with A is equivalent to series of applications of such elementary row operations. But I don't get how it's being proved here.
- Particularly, how (1) and (2) are connected? Moreover, in (2) we are supposing what we had to actually prove, so how does it help? Also, it sees quite like re-statement of corollary itself.
- In general, what constitutes a mathematical proof? In other words, how do we judge whether it's sufficient and correct? I did go through few questions here, like this one, but how to validate this proof specifically (especially differentiating it with intuition or re-statement)
linear-algebra proof-explanation
asked Dec 29 '18 at 19:42
dheeraj suthardheeraj suthar
72
$endgroup$
In Hoffman and Kunze's Linear Algebra, following proof is given for this corollary:
Let A and B be m x n matrices over the field F. Then B is row-equivalent to A if and only if B=PA, where P is a product of m x m elementary matrices.
Proof. (1)Suppose B=PA where P= Es · · · E2E1 and the Ei are m x m elementary matrices. Then E1A is row-equivalent to A, and E2(E1A ) is row-equivalent to E1A . So E2E1A is row-equivalent to A ; and continuing in this way we see that (Es . . . E1)A is row-equivalent to A.
(2)Now suppose that B is row-equivalent to A. Let E1, E2,...,Es
the elementary matrices corresponding to some sequence of elementary row operations which carries A into B. Then B = (Es . . . E1)A
Although I get the corollary intuitively, as each elementary matrix is equivalent to elementary row operation, and thus multiplication of matrix with A is equivalent to series of applications of such elementary row operations. But I don't get how it's being proved here.
- Particularly, how (1) and (2) are connected? Moreover, in (2) we are supposing what we had to actually prove, so how does it help? Also, it sees quite like re-statement of corollary itself.
- In general, what constitutes a mathematical proof? In other words, how do we judge whether it's sufficient and correct? I did go through few questions here, like this one, but how to validate this proof specifically (especially differentiating it with intuition or re-statement)
linear-algebra proof-explanation
linear-algebra proof-explanation
asked Dec 29 '18 at 19:42
dheeraj suthardheeraj suthar
72
asked Dec 29 '18 at 19:42
dheeraj suthardheeraj suthar
72
edited Dec 29 '18 at 20:19
dheeraj suthar
asked Dec 29 '18 at 19:42
dheeraj suthardheeraj suthar
72
asked Dec 29 '18 at 19:42
dheeraj suthardheeraj suthar
72
asked Dec 29 '18 at 19:42
dheeraj suthardheeraj suthar
72
72
1
$begingroup$
In an "$X$ if and only if $Y$" proof, you have to prove two separate things: that $X$ implies $Y$, and that $Y$ implies $X$. This is what (1) and (2) are doing, and indeed they are separate proofs. In (1) you assume $X$ and prove $Y$, while in (2) you assume $Y$ and prove $X$.
$endgroup$
– angryavian
Dec 29 '18 at 20:02
$begingroup$
@angryavian Ahh.. Thanks. Got it. I was confused with the abrupt conclusion in (2). I now got it (it comes from the theorem which this corollary belongs to. e(A) = EA). Will be more patient in further reading. Should I delete this question, or any way I can accept your comment as answer?
$endgroup$
– dheeraj suthar
Dec 29 '18 at 20:24
$begingroup$
I copied my comment as an answer.
$endgroup$
– angryavian
Dec 29 '18 at 20:40
add a comment |
1
$begingroup$
In an "$X$ if and only if $Y$" proof, you have to prove two separate things: that $X$ implies $Y$, and that $Y$ implies $X$. This is what (1) and (2) are doing, and indeed they are separate proofs. In (1) you assume $X$ and prove $Y$, while in (2) you assume $Y$ and prove $X$.
$endgroup$
– angryavian
Dec 29 '18 at 20:02
$begingroup$
@angryavian Ahh.. Thanks. Got it. I was confused with the abrupt conclusion in (2). I now got it (it comes from the theorem which this corollary belongs to. e(A) = EA). Will be more patient in further reading. Should I delete this question, or any way I can accept your comment as answer?
$endgroup$
– dheeraj suthar
Dec 29 '18 at 20:24
$begingroup$
I copied my comment as an answer.
$endgroup$
– angryavian
Dec 29 '18 at 20:40
1
1
$begingroup$
In an "$X$ if and only if $Y$" proof, you have to prove two separate things: that $X$ implies $Y$, and that $Y$ implies $X$. This is what (1) and (2) are doing, and indeed they are separate proofs. In (1) you assume $X$ and prove $Y$, while in (2) you assume $Y$ and prove $X$.
$endgroup$
– angryavian
Dec 29 '18 at 20:02
$begingroup$
In an "$X$ if and only if $Y$" proof, you have to prove two separate things: that $X$ implies $Y$, and that $Y$ implies $X$. This is what (1) and (2) are doing, and indeed they are separate proofs. In (1) you assume $X$ and prove $Y$, while in (2) you assume $Y$ and prove $X$.
$endgroup$
– angryavian
Dec 29 '18 at 20:02
$begingroup$
@angryavian Ahh.. Thanks. Got it. I was confused with the abrupt conclusion in (2). I now got it (it comes from the theorem which this corollary belongs to. e(A) = EA). Will be more patient in further reading. Should I delete this question, or any way I can accept your comment as answer?
$endgroup$
– dheeraj suthar
Dec 29 '18 at 20:24
$begingroup$
@angryavian Ahh.. Thanks. Got it. I was confused with the abrupt conclusion in (2). I now got it (it comes from the theorem which this corollary belongs to. e(A) = EA). Will be more patient in further reading. Should I delete this question, or any way I can accept your comment as answer?
$endgroup$
– dheeraj suthar
Dec 29 '18 at 20:24
$begingroup$
I copied my comment as an answer.
$endgroup$
– angryavian
Dec 29 '18 at 20:40
$begingroup$
I copied my comment as an answer.
$endgroup$
– angryavian
Dec 29 '18 at 20:40
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
In an "$X$ if and only if $Y$" proof, you have to prove two separate things: that $X$ implies $Y$, and that $Y$ implies $X$. This is what (1) and (2) are doing, and indeed they are separate proofs. In (1) you assume $X$ and prove $Y$, while in (2) you assume $Y$ and prove $X$.
answered Dec 29 '18 at 20:40
angryavianangryavian
42.2k23481
$endgroup$
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%2f3056193%2fhelp-understanding-proof-for-b-is-row-equivalent-to-a-iff-b-pa-where-p-is-produ%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
In an "$X$ if and only if $Y$" proof, you have to prove two separate things: that $X$ implies $Y$, and that $Y$ implies $X$. This is what (1) and (2) are doing, and indeed they are separate proofs. In (1) you assume $X$ and prove $Y$, while in (2) you assume $Y$ and prove $X$.
answered Dec 29 '18 at 20:40
angryavianangryavian
42.2k23481
$endgroup$
add a comment |
$begingroup$
In an "$X$ if and only if $Y$" proof, you have to prove two separate things: that $X$ implies $Y$, and that $Y$ implies $X$. This is what (1) and (2) are doing, and indeed they are separate proofs. In (1) you assume $X$ and prove $Y$, while in (2) you assume $Y$ and prove $X$.
answered Dec 29 '18 at 20:40
angryavianangryavian
42.2k23481
$endgroup$
add a comment |
$begingroup$
In an "$X$ if and only if $Y$" proof, you have to prove two separate things: that $X$ implies $Y$, and that $Y$ implies $X$. This is what (1) and (2) are doing, and indeed they are separate proofs. In (1) you assume $X$ and prove $Y$, while in (2) you assume $Y$ and prove $X$.
answered Dec 29 '18 at 20:40
angryavianangryavian
42.2k23481
$endgroup$
In an "$X$ if and only if $Y$" proof, you have to prove two separate things: that $X$ implies $Y$, and that $Y$ implies $X$. This is what (1) and (2) are doing, and indeed they are separate proofs. In (1) you assume $X$ and prove $Y$, while in (2) you assume $Y$ and prove $X$.
answered Dec 29 '18 at 20:40
angryavianangryavian
42.2k23481
answered Dec 29 '18 at 20:40
angryavianangryavian
42.2k23481
answered Dec 29 '18 at 20:40
angryavianangryavian
42.2k23481
answered Dec 29 '18 at 20:40
angryavianangryavian
42.2k23481
42.2k23481
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%2f3056193%2fhelp-understanding-proof-for-b-is-row-equivalent-to-a-iff-b-pa-where-p-is-produ%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
1
$begingroup$
In an "$X$ if and only if $Y$" proof, you have to prove two separate things: that $X$ implies $Y$, and that $Y$ implies $X$. This is what (1) and (2) are doing, and indeed they are separate proofs. In (1) you assume $X$ and prove $Y$, while in (2) you assume $Y$ and prove $X$.
$endgroup$
– angryavian
Dec 29 '18 at 20:02
$begingroup$
@angryavian Ahh.. Thanks. Got it. I was confused with the abrupt conclusion in (2). I now got it (it comes from the theorem which this corollary belongs to. e(A) = EA). Will be more patient in further reading. Should I delete this question, or any way I can accept your comment as answer?
$endgroup$
– dheeraj suthar
Dec 29 '18 at 20:24
$begingroup$
I copied my comment as an answer.
$endgroup$
– angryavian
Dec 29 '18 at 20:40