Prove that $p ◦ p = p$. (Representing a Linear Transformation as a Matrix)












0












$begingroup$


Let $B = (1, X, X^2)$ be an ordered basis for $mathbb{R}_2[X]$ and $p ∈ mathcal{L}(mathbb{R}_2[X])$ be the linear map defined by $p(1) = frac{1}{3}(2 − X − X^2)$, $p(X) = frac{1}{3}(−1 + 2X − X^2)$ and $p(X^2) = frac{1}{3} (−1 − X + 2X^2)$.



By definition of the task, I found that $A = M_B(p)$



Compute $A^2$. Prove that $p ◦ p = p$.



begin{bmatrix}
frac{2}{3} & -frac{1}{3} & -frac{1}{3} \
-frac{1}{3} & frac{2}{3} & -frac{1}{3} \
-frac{1}{3} & -frac{1}{3} &frac{2}{3} \
end{bmatrix}



I computed the multiplication and it appears that $A^2$ = $A$. How should I prove it?






linear-algebra matrices linear-transformations change-of-basis projection-matrices







share|cite|improve this question











$endgroup$












  • $begingroup$
    Does $M_B(p)$ mean the matrix of $p$ with respect to the basis $B$?
    $endgroup$
    – Shubham Johri
    Dec 13 '18 at 16:45










  • $begingroup$
    Sure, that's correct.
    $endgroup$
    – nasdaq
    Dec 13 '18 at 16:47
















0












$begingroup$


Let $B = (1, X, X^2)$ be an ordered basis for $mathbb{R}_2[X]$ and $p ∈ mathcal{L}(mathbb{R}_2[X])$ be the linear map defined by $p(1) = frac{1}{3}(2 − X − X^2)$, $p(X) = frac{1}{3}(−1 + 2X − X^2)$ and $p(X^2) = frac{1}{3} (−1 − X + 2X^2)$.



By definition of the task, I found that $A = M_B(p)$



Compute $A^2$. Prove that $p ◦ p = p$.



begin{bmatrix}
frac{2}{3} & -frac{1}{3} & -frac{1}{3} \
-frac{1}{3} & frac{2}{3} & -frac{1}{3} \
-frac{1}{3} & -frac{1}{3} &frac{2}{3} \
end{bmatrix}



I computed the multiplication and it appears that $A^2$ = $A$. How should I prove it?






linear-algebra matrices linear-transformations change-of-basis projection-matrices







share|cite|improve this question











$endgroup$












  • $begingroup$
    Does $M_B(p)$ mean the matrix of $p$ with respect to the basis $B$?
    $endgroup$
    – Shubham Johri
    Dec 13 '18 at 16:45










  • $begingroup$
    Sure, that's correct.
    $endgroup$
    – nasdaq
    Dec 13 '18 at 16:47














0












0








0





$begingroup$


Let $B = (1, X, X^2)$ be an ordered basis for $mathbb{R}_2[X]$ and $p ∈ mathcal{L}(mathbb{R}_2[X])$ be the linear map defined by $p(1) = frac{1}{3}(2 − X − X^2)$, $p(X) = frac{1}{3}(−1 + 2X − X^2)$ and $p(X^2) = frac{1}{3} (−1 − X + 2X^2)$.



By definition of the task, I found that $A = M_B(p)$



Compute $A^2$. Prove that $p ◦ p = p$.



begin{bmatrix}
frac{2}{3} & -frac{1}{3} & -frac{1}{3} \
-frac{1}{3} & frac{2}{3} & -frac{1}{3} \
-frac{1}{3} & -frac{1}{3} &frac{2}{3} \
end{bmatrix}



I computed the multiplication and it appears that $A^2$ = $A$. How should I prove it?






linear-algebra matrices linear-transformations change-of-basis projection-matrices







share|cite|improve this question











$endgroup$




Let $B = (1, X, X^2)$ be an ordered basis for $mathbb{R}_2[X]$ and $p ∈ mathcal{L}(mathbb{R}_2[X])$ be the linear map defined by $p(1) = frac{1}{3}(2 − X − X^2)$, $p(X) = frac{1}{3}(−1 + 2X − X^2)$ and $p(X^2) = frac{1}{3} (−1 − X + 2X^2)$.



By definition of the task, I found that $A = M_B(p)$



Compute $A^2$. Prove that $p ◦ p = p$.



begin{bmatrix}
frac{2}{3} & -frac{1}{3} & -frac{1}{3} \
-frac{1}{3} & frac{2}{3} & -frac{1}{3} \
-frac{1}{3} & -frac{1}{3} &frac{2}{3} \
end{bmatrix}



I computed the multiplication and it appears that $A^2$ = $A$. How should I prove it?






linear-algebra matrices linear-transformations change-of-basis projection-matrices




linear-algebra matrices linear-transformations change-of-basis projection-matrices






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 15 '18 at 12:18









Batominovski

1




1










asked Dec 13 '18 at 16:40









nasdaqnasdaq

62




62












  • $begingroup$
    Does $M_B(p)$ mean the matrix of $p$ with respect to the basis $B$?
    $endgroup$
    – Shubham Johri
    Dec 13 '18 at 16:45










  • $begingroup$
    Sure, that's correct.
    $endgroup$
    – nasdaq
    Dec 13 '18 at 16:47


















  • $begingroup$
    Does $M_B(p)$ mean the matrix of $p$ with respect to the basis $B$?
    $endgroup$
    – Shubham Johri
    Dec 13 '18 at 16:45










  • $begingroup$
    Sure, that's correct.
    $endgroup$
    – nasdaq
    Dec 13 '18 at 16:47
















$begingroup$
Does $M_B(p)$ mean the matrix of $p$ with respect to the basis $B$?
$endgroup$
– Shubham Johri
Dec 13 '18 at 16:45




$begingroup$
Does $M_B(p)$ mean the matrix of $p$ with respect to the basis $B$?
$endgroup$
– Shubham Johri
Dec 13 '18 at 16:45












$begingroup$
Sure, that's correct.
$endgroup$
– nasdaq
Dec 13 '18 at 16:47




$begingroup$
Sure, that's correct.
$endgroup$
– nasdaq
Dec 13 '18 at 16:47










1 Answer
1






active

oldest

votes


















2












$begingroup$

If $A$ is the matrix corresponding to the linear operator $p$, then $p(v)=Acdot v forall vinBbb R_2[X]$.



$implies(pcirc p)(v)=p(p(v))=Acdot(p(v))=Acdot(Av)=(Acdot A)v=A^2cdot v$



This means the matrix corresponding to the linear operator $pcirc p$ is $A^2$. Since $A^2=A$, the matrices corresponding to $p$ and $pcirc p$ are identical, which means $p=pcirc p$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks, that's very helpful!
    $endgroup$
    – nasdaq
    Dec 13 '18 at 16:52










  • $begingroup$
    You're welcome.
    $endgroup$
    – Shubham Johri
    Dec 13 '18 at 16:55











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%2f3038263%2fprove-that-p-p-p-representing-a-linear-transformation-as-a-matrix%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









2












$begingroup$

If $A$ is the matrix corresponding to the linear operator $p$, then $p(v)=Acdot v forall vinBbb R_2[X]$.



$implies(pcirc p)(v)=p(p(v))=Acdot(p(v))=Acdot(Av)=(Acdot A)v=A^2cdot v$



This means the matrix corresponding to the linear operator $pcirc p$ is $A^2$. Since $A^2=A$, the matrices corresponding to $p$ and $pcirc p$ are identical, which means $p=pcirc p$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks, that's very helpful!
    $endgroup$
    – nasdaq
    Dec 13 '18 at 16:52










  • $begingroup$
    You're welcome.
    $endgroup$
    – Shubham Johri
    Dec 13 '18 at 16:55
















2












$begingroup$

If $A$ is the matrix corresponding to the linear operator $p$, then $p(v)=Acdot v forall vinBbb R_2[X]$.



$implies(pcirc p)(v)=p(p(v))=Acdot(p(v))=Acdot(Av)=(Acdot A)v=A^2cdot v$



This means the matrix corresponding to the linear operator $pcirc p$ is $A^2$. Since $A^2=A$, the matrices corresponding to $p$ and $pcirc p$ are identical, which means $p=pcirc p$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks, that's very helpful!
    $endgroup$
    – nasdaq
    Dec 13 '18 at 16:52










  • $begingroup$
    You're welcome.
    $endgroup$
    – Shubham Johri
    Dec 13 '18 at 16:55














2












2








2





$begingroup$

If $A$ is the matrix corresponding to the linear operator $p$, then $p(v)=Acdot v forall vinBbb R_2[X]$.



$implies(pcirc p)(v)=p(p(v))=Acdot(p(v))=Acdot(Av)=(Acdot A)v=A^2cdot v$



This means the matrix corresponding to the linear operator $pcirc p$ is $A^2$. Since $A^2=A$, the matrices corresponding to $p$ and $pcirc p$ are identical, which means $p=pcirc p$.






share|cite|improve this answer











$endgroup$



If $A$ is the matrix corresponding to the linear operator $p$, then $p(v)=Acdot v forall vinBbb R_2[X]$.



$implies(pcirc p)(v)=p(p(v))=Acdot(p(v))=Acdot(Av)=(Acdot A)v=A^2cdot v$



This means the matrix corresponding to the linear operator $pcirc p$ is $A^2$. Since $A^2=A$, the matrices corresponding to $p$ and $pcirc p$ are identical, which means $p=pcirc p$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 13 '18 at 16:54

























answered Dec 13 '18 at 16:50









Shubham JohriShubham Johri

4,666717




4,666717












  • $begingroup$
    Thanks, that's very helpful!
    $endgroup$
    – nasdaq
    Dec 13 '18 at 16:52










  • $begingroup$
    You're welcome.
    $endgroup$
    – Shubham Johri
    Dec 13 '18 at 16:55


















  • $begingroup$
    Thanks, that's very helpful!
    $endgroup$
    – nasdaq
    Dec 13 '18 at 16:52










  • $begingroup$
    You're welcome.
    $endgroup$
    – Shubham Johri
    Dec 13 '18 at 16:55
















$begingroup$
Thanks, that's very helpful!
$endgroup$
– nasdaq
Dec 13 '18 at 16:52




$begingroup$
Thanks, that's very helpful!
$endgroup$
– nasdaq
Dec 13 '18 at 16:52












$begingroup$
You're welcome.
$endgroup$
– Shubham Johri
Dec 13 '18 at 16:55




$begingroup$
You're welcome.
$endgroup$
– Shubham Johri
Dec 13 '18 at 16:55


















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%2f3038263%2fprove-that-p-p-p-representing-a-linear-transformation-as-a-matrix%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