A, B and C are matrices. Prove that $A^{T}AB = A^{T}AC$ iff AB = AC












1












$begingroup$


Let $A in mathbb{R}^{m times n}$, $B in mathbb{R}^{n times p}$, and $C in mathbb{R}^{n times p}$. Show that $A^{T}AB = A^{T}AC$ iff $AB=AC$.



What I have done so far: The direction where I assume $AB=AC$ is trivial. So far, I want to claim that if $A^{T}AB = A^{T}AC$ holds for all $B$ and $C$, then A^{T}A must be a full rank matrix. The reason I suspect this is if $AB neq AC$ and $A^{T}AB = A^{T}AC$, I would have at least one column of $B$ and $C$, say $b_i$ and $c_i$, such that $Ab_i neq Ac_i$, but $A^{T}Ab_i = A^{T}Ac_i$. This isn't really too formal, but it gives me some intuition as to why I expect this result to be true.






linear-algebra matrices







share|cite|improve this question









$endgroup$












  • $begingroup$
    I'd try to show that $A^TA$ and $A$ have the same rank.
    $endgroup$
    – Lord Shark the Unknown
    Jan 2 at 15:59
















1












$begingroup$


Let $A in mathbb{R}^{m times n}$, $B in mathbb{R}^{n times p}$, and $C in mathbb{R}^{n times p}$. Show that $A^{T}AB = A^{T}AC$ iff $AB=AC$.



What I have done so far: The direction where I assume $AB=AC$ is trivial. So far, I want to claim that if $A^{T}AB = A^{T}AC$ holds for all $B$ and $C$, then A^{T}A must be a full rank matrix. The reason I suspect this is if $AB neq AC$ and $A^{T}AB = A^{T}AC$, I would have at least one column of $B$ and $C$, say $b_i$ and $c_i$, such that $Ab_i neq Ac_i$, but $A^{T}Ab_i = A^{T}Ac_i$. This isn't really too formal, but it gives me some intuition as to why I expect this result to be true.






linear-algebra matrices







share|cite|improve this question









$endgroup$












  • $begingroup$
    I'd try to show that $A^TA$ and $A$ have the same rank.
    $endgroup$
    – Lord Shark the Unknown
    Jan 2 at 15:59














1












1








1





$begingroup$


Let $A in mathbb{R}^{m times n}$, $B in mathbb{R}^{n times p}$, and $C in mathbb{R}^{n times p}$. Show that $A^{T}AB = A^{T}AC$ iff $AB=AC$.



What I have done so far: The direction where I assume $AB=AC$ is trivial. So far, I want to claim that if $A^{T}AB = A^{T}AC$ holds for all $B$ and $C$, then A^{T}A must be a full rank matrix. The reason I suspect this is if $AB neq AC$ and $A^{T}AB = A^{T}AC$, I would have at least one column of $B$ and $C$, say $b_i$ and $c_i$, such that $Ab_i neq Ac_i$, but $A^{T}Ab_i = A^{T}Ac_i$. This isn't really too formal, but it gives me some intuition as to why I expect this result to be true.






linear-algebra matrices







share|cite|improve this question









$endgroup$




Let $A in mathbb{R}^{m times n}$, $B in mathbb{R}^{n times p}$, and $C in mathbb{R}^{n times p}$. Show that $A^{T}AB = A^{T}AC$ iff $AB=AC$.



What I have done so far: The direction where I assume $AB=AC$ is trivial. So far, I want to claim that if $A^{T}AB = A^{T}AC$ holds for all $B$ and $C$, then A^{T}A must be a full rank matrix. The reason I suspect this is if $AB neq AC$ and $A^{T}AB = A^{T}AC$, I would have at least one column of $B$ and $C$, say $b_i$ and $c_i$, such that $Ab_i neq Ac_i$, but $A^{T}Ab_i = A^{T}Ac_i$. This isn't really too formal, but it gives me some intuition as to why I expect this result to be true.






linear-algebra matrices




linear-algebra matrices






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 2 at 15:55









BoboBobo

1226




1226












  • $begingroup$
    I'd try to show that $A^TA$ and $A$ have the same rank.
    $endgroup$
    – Lord Shark the Unknown
    Jan 2 at 15:59


















  • $begingroup$
    I'd try to show that $A^TA$ and $A$ have the same rank.
    $endgroup$
    – Lord Shark the Unknown
    Jan 2 at 15:59
















$begingroup$
I'd try to show that $A^TA$ and $A$ have the same rank.
$endgroup$
– Lord Shark the Unknown
Jan 2 at 15:59




$begingroup$
I'd try to show that $A^TA$ and $A$ have the same rank.
$endgroup$
– Lord Shark the Unknown
Jan 2 at 15:59










1 Answer
1






active

oldest

votes


















4












$begingroup$

Hint:
$$
A^T(AB-AC)=0qquadRightarrow qquad\
(B-C)^TA^T(AB-AC)=(AB-AC)^T(AB-AC)=0.
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    so are you saying that since $langle (AB-AC)x,(AB-AC)xrangle = 0$ for all $x$ that $AB=AC$?
    $endgroup$
    – Bobo
    Jan 2 at 16:06










  • $begingroup$
    @Bobo One can do it via that as $langle (AB-AC)x,(AB-AC)xrangle =|(AB-AC)x|^2$. If it is zero then $(AB-AC)x=0$ for all $x$.
    $endgroup$
    – A.Γ.
    Jan 2 at 16:08












  • $begingroup$
    <v,v> = 0 iff v =0. Thus every x is in the Null Space of AB-AC. Thus AB-AC = 0.
    $endgroup$
    – Joel Pereira
    Jan 2 at 16:08










  • $begingroup$
    @Bobo Another way is to take the trace $$operatorname{trace}((AB-AC)^T(AB-AC))=|AB-AC|_{text{frobenius}}^2=0.$$
    $endgroup$
    – A.Γ.
    Jan 2 at 16:09













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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3059640%2fa-b-and-c-are-matrices-prove-that-atab-atac-iff-ab-ac%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









4












$begingroup$

Hint:
$$
A^T(AB-AC)=0qquadRightarrow qquad\
(B-C)^TA^T(AB-AC)=(AB-AC)^T(AB-AC)=0.
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    so are you saying that since $langle (AB-AC)x,(AB-AC)xrangle = 0$ for all $x$ that $AB=AC$?
    $endgroup$
    – Bobo
    Jan 2 at 16:06










  • $begingroup$
    @Bobo One can do it via that as $langle (AB-AC)x,(AB-AC)xrangle =|(AB-AC)x|^2$. If it is zero then $(AB-AC)x=0$ for all $x$.
    $endgroup$
    – A.Γ.
    Jan 2 at 16:08












  • $begingroup$
    <v,v> = 0 iff v =0. Thus every x is in the Null Space of AB-AC. Thus AB-AC = 0.
    $endgroup$
    – Joel Pereira
    Jan 2 at 16:08










  • $begingroup$
    @Bobo Another way is to take the trace $$operatorname{trace}((AB-AC)^T(AB-AC))=|AB-AC|_{text{frobenius}}^2=0.$$
    $endgroup$
    – A.Γ.
    Jan 2 at 16:09


















4












$begingroup$

Hint:
$$
A^T(AB-AC)=0qquadRightarrow qquad\
(B-C)^TA^T(AB-AC)=(AB-AC)^T(AB-AC)=0.
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    so are you saying that since $langle (AB-AC)x,(AB-AC)xrangle = 0$ for all $x$ that $AB=AC$?
    $endgroup$
    – Bobo
    Jan 2 at 16:06










  • $begingroup$
    @Bobo One can do it via that as $langle (AB-AC)x,(AB-AC)xrangle =|(AB-AC)x|^2$. If it is zero then $(AB-AC)x=0$ for all $x$.
    $endgroup$
    – A.Γ.
    Jan 2 at 16:08












  • $begingroup$
    <v,v> = 0 iff v =0. Thus every x is in the Null Space of AB-AC. Thus AB-AC = 0.
    $endgroup$
    – Joel Pereira
    Jan 2 at 16:08










  • $begingroup$
    @Bobo Another way is to take the trace $$operatorname{trace}((AB-AC)^T(AB-AC))=|AB-AC|_{text{frobenius}}^2=0.$$
    $endgroup$
    – A.Γ.
    Jan 2 at 16:09
















4












4








4





$begingroup$

Hint:
$$
A^T(AB-AC)=0qquadRightarrow qquad\
(B-C)^TA^T(AB-AC)=(AB-AC)^T(AB-AC)=0.
$$






share|cite|improve this answer









$endgroup$



Hint:
$$
A^T(AB-AC)=0qquadRightarrow qquad\
(B-C)^TA^T(AB-AC)=(AB-AC)^T(AB-AC)=0.
$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 2 at 16:00









A.Γ.A.Γ.

22.8k32656




22.8k32656












  • $begingroup$
    so are you saying that since $langle (AB-AC)x,(AB-AC)xrangle = 0$ for all $x$ that $AB=AC$?
    $endgroup$
    – Bobo
    Jan 2 at 16:06










  • $begingroup$
    @Bobo One can do it via that as $langle (AB-AC)x,(AB-AC)xrangle =|(AB-AC)x|^2$. If it is zero then $(AB-AC)x=0$ for all $x$.
    $endgroup$
    – A.Γ.
    Jan 2 at 16:08












  • $begingroup$
    <v,v> = 0 iff v =0. Thus every x is in the Null Space of AB-AC. Thus AB-AC = 0.
    $endgroup$
    – Joel Pereira
    Jan 2 at 16:08










  • $begingroup$
    @Bobo Another way is to take the trace $$operatorname{trace}((AB-AC)^T(AB-AC))=|AB-AC|_{text{frobenius}}^2=0.$$
    $endgroup$
    – A.Γ.
    Jan 2 at 16:09




















  • $begingroup$
    so are you saying that since $langle (AB-AC)x,(AB-AC)xrangle = 0$ for all $x$ that $AB=AC$?
    $endgroup$
    – Bobo
    Jan 2 at 16:06










  • $begingroup$
    @Bobo One can do it via that as $langle (AB-AC)x,(AB-AC)xrangle =|(AB-AC)x|^2$. If it is zero then $(AB-AC)x=0$ for all $x$.
    $endgroup$
    – A.Γ.
    Jan 2 at 16:08












  • $begingroup$
    <v,v> = 0 iff v =0. Thus every x is in the Null Space of AB-AC. Thus AB-AC = 0.
    $endgroup$
    – Joel Pereira
    Jan 2 at 16:08










  • $begingroup$
    @Bobo Another way is to take the trace $$operatorname{trace}((AB-AC)^T(AB-AC))=|AB-AC|_{text{frobenius}}^2=0.$$
    $endgroup$
    – A.Γ.
    Jan 2 at 16:09


















$begingroup$
so are you saying that since $langle (AB-AC)x,(AB-AC)xrangle = 0$ for all $x$ that $AB=AC$?
$endgroup$
– Bobo
Jan 2 at 16:06




$begingroup$
so are you saying that since $langle (AB-AC)x,(AB-AC)xrangle = 0$ for all $x$ that $AB=AC$?
$endgroup$
– Bobo
Jan 2 at 16:06












$begingroup$
@Bobo One can do it via that as $langle (AB-AC)x,(AB-AC)xrangle =|(AB-AC)x|^2$. If it is zero then $(AB-AC)x=0$ for all $x$.
$endgroup$
– A.Γ.
Jan 2 at 16:08






$begingroup$
@Bobo One can do it via that as $langle (AB-AC)x,(AB-AC)xrangle =|(AB-AC)x|^2$. If it is zero then $(AB-AC)x=0$ for all $x$.
$endgroup$
– A.Γ.
Jan 2 at 16:08














$begingroup$
<v,v> = 0 iff v =0. Thus every x is in the Null Space of AB-AC. Thus AB-AC = 0.
$endgroup$
– Joel Pereira
Jan 2 at 16:08




$begingroup$
<v,v> = 0 iff v =0. Thus every x is in the Null Space of AB-AC. Thus AB-AC = 0.
$endgroup$
– Joel Pereira
Jan 2 at 16:08












$begingroup$
@Bobo Another way is to take the trace $$operatorname{trace}((AB-AC)^T(AB-AC))=|AB-AC|_{text{frobenius}}^2=0.$$
$endgroup$
– A.Γ.
Jan 2 at 16:09






$begingroup$
@Bobo Another way is to take the trace $$operatorname{trace}((AB-AC)^T(AB-AC))=|AB-AC|_{text{frobenius}}^2=0.$$
$endgroup$
– A.Γ.
Jan 2 at 16:09




















draft saved

draft discarded




















































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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3059640%2fa-b-and-c-are-matrices-prove-that-atab-atac-iff-ab-ac%23new-answer', 'question_page');
}
);

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







-

Popular posts from this blog

Bressuire

Image
.mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
Read more

Cabo Verde

Image
  Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
Read more

Gyllenstierna

Image
.mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
Read more