Artin's Algebra Exercise 1.1.16












2












$begingroup$


The question goes $A^k = 0$, for some $k>0$ ($A$ is square). Prove that $A+I$ is invertible.



I did $A^k = 0 implies A^k+I=I$. So, $(A+I)(A^{k-1} +.... + I) = I$. So it's invertible with the inverse given above.



I feel like this is wrong. I'm not very confident that I'm allowed to factorise it like that since $det{A^k}=(det{A})^k=0.$ And doing $A^{k-1}$ during factorisation doesn't feel right since you are multiplying with $A^{-1}$.



Can you please direct me to the correct direction and tell me why the thing I did is incorrect?



Thanks in advance.



Edit: I see that my factorisation is incorrect (plus/minus signs) but the question remains the same.






linear-algebra matrices inverse







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Your solution is correct. It doesn't matter that the determinant of $A$ is $0$. What's important is that the determinant of $A+I$ is non-zero.
    $endgroup$
    – John Douma
    Nov 6 '18 at 23:40












  • $begingroup$
    Hint: what's $x^{n} + y^{n}$?
    $endgroup$
    – Matija Sreckovic
    Nov 6 '18 at 23:43










  • $begingroup$
    Your idea is right, but you need to argue that the product in the other order is also $I$.
    $endgroup$
    – darij grinberg
    Nov 7 '18 at 0:50


















2












$begingroup$


The question goes $A^k = 0$, for some $k>0$ ($A$ is square). Prove that $A+I$ is invertible.



I did $A^k = 0 implies A^k+I=I$. So, $(A+I)(A^{k-1} +.... + I) = I$. So it's invertible with the inverse given above.



I feel like this is wrong. I'm not very confident that I'm allowed to factorise it like that since $det{A^k}=(det{A})^k=0.$ And doing $A^{k-1}$ during factorisation doesn't feel right since you are multiplying with $A^{-1}$.



Can you please direct me to the correct direction and tell me why the thing I did is incorrect?



Thanks in advance.



Edit: I see that my factorisation is incorrect (plus/minus signs) but the question remains the same.






linear-algebra matrices inverse







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Your solution is correct. It doesn't matter that the determinant of $A$ is $0$. What's important is that the determinant of $A+I$ is non-zero.
    $endgroup$
    – John Douma
    Nov 6 '18 at 23:40












  • $begingroup$
    Hint: what's $x^{n} + y^{n}$?
    $endgroup$
    – Matija Sreckovic
    Nov 6 '18 at 23:43










  • $begingroup$
    Your idea is right, but you need to argue that the product in the other order is also $I$.
    $endgroup$
    – darij grinberg
    Nov 7 '18 at 0:50
















2












2








2


1



$begingroup$


The question goes $A^k = 0$, for some $k>0$ ($A$ is square). Prove that $A+I$ is invertible.



I did $A^k = 0 implies A^k+I=I$. So, $(A+I)(A^{k-1} +.... + I) = I$. So it's invertible with the inverse given above.



I feel like this is wrong. I'm not very confident that I'm allowed to factorise it like that since $det{A^k}=(det{A})^k=0.$ And doing $A^{k-1}$ during factorisation doesn't feel right since you are multiplying with $A^{-1}$.



Can you please direct me to the correct direction and tell me why the thing I did is incorrect?



Thanks in advance.



Edit: I see that my factorisation is incorrect (plus/minus signs) but the question remains the same.






linear-algebra matrices inverse







share|cite|improve this question











$endgroup$




The question goes $A^k = 0$, for some $k>0$ ($A$ is square). Prove that $A+I$ is invertible.



I did $A^k = 0 implies A^k+I=I$. So, $(A+I)(A^{k-1} +.... + I) = I$. So it's invertible with the inverse given above.



I feel like this is wrong. I'm not very confident that I'm allowed to factorise it like that since $det{A^k}=(det{A})^k=0.$ And doing $A^{k-1}$ during factorisation doesn't feel right since you are multiplying with $A^{-1}$.



Can you please direct me to the correct direction and tell me why the thing I did is incorrect?



Thanks in advance.



Edit: I see that my factorisation is incorrect (plus/minus signs) but the question remains the same.






linear-algebra matrices inverse




linear-algebra matrices inverse






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 7 '18 at 9:03









José Carlos Santos

160k22127232




160k22127232










asked Nov 6 '18 at 23:34









Nutan NepalNutan Nepal

505




505








  • 2




    $begingroup$
    Your solution is correct. It doesn't matter that the determinant of $A$ is $0$. What's important is that the determinant of $A+I$ is non-zero.
    $endgroup$
    – John Douma
    Nov 6 '18 at 23:40












  • $begingroup$
    Hint: what's $x^{n} + y^{n}$?
    $endgroup$
    – Matija Sreckovic
    Nov 6 '18 at 23:43










  • $begingroup$
    Your idea is right, but you need to argue that the product in the other order is also $I$.
    $endgroup$
    – darij grinberg
    Nov 7 '18 at 0:50
















  • 2




    $begingroup$
    Your solution is correct. It doesn't matter that the determinant of $A$ is $0$. What's important is that the determinant of $A+I$ is non-zero.
    $endgroup$
    – John Douma
    Nov 6 '18 at 23:40












  • $begingroup$
    Hint: what's $x^{n} + y^{n}$?
    $endgroup$
    – Matija Sreckovic
    Nov 6 '18 at 23:43










  • $begingroup$
    Your idea is right, but you need to argue that the product in the other order is also $I$.
    $endgroup$
    – darij grinberg
    Nov 7 '18 at 0:50










2




2




$begingroup$
Your solution is correct. It doesn't matter that the determinant of $A$ is $0$. What's important is that the determinant of $A+I$ is non-zero.
$endgroup$
– John Douma
Nov 6 '18 at 23:40






$begingroup$
Your solution is correct. It doesn't matter that the determinant of $A$ is $0$. What's important is that the determinant of $A+I$ is non-zero.
$endgroup$
– John Douma
Nov 6 '18 at 23:40














$begingroup$
Hint: what's $x^{n} + y^{n}$?
$endgroup$
– Matija Sreckovic
Nov 6 '18 at 23:43




$begingroup$
Hint: what's $x^{n} + y^{n}$?
$endgroup$
– Matija Sreckovic
Nov 6 '18 at 23:43












$begingroup$
Your idea is right, but you need to argue that the product in the other order is also $I$.
$endgroup$
– darij grinberg
Nov 7 '18 at 0:50






$begingroup$
Your idea is right, but you need to argue that the product in the other order is also $I$.
$endgroup$
– darij grinberg
Nov 7 '18 at 0:50












2 Answers
2






active

oldest

votes


















10












$begingroup$

What you did is almost correct. In fact:$$(A+operatorname{Id})bigl((-1)^{k-1}A^{k-1}+(-1)^{k-2}A^{k-2}+cdots-A+operatorname{Id}bigr)=operatorname{Id},$$and so $(-1)^{k-1}A^{k-1}+(-1)^{k-2}A^{k-2}+cdots-A+operatorname{Id}$ is the inverse of $A+operatorname{Id}$. What does it matter that $det(A^k)=(det A)^k=0$? This has nothing to do with determinants.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I thought that since the determinant of A is 0, factorising it the same as doing $A^kA^{-1}$. I failed to see that $k>0 $ so factorising it like that is correct.
    $endgroup$
    – Nutan Nepal
    Nov 6 '18 at 23:47





















8












$begingroup$

The identity:
$;x^n+y^n=(x+y)(x^{n-1}-x^{n-2}y+dots-xy^{n-2}+y^{n-1})$ $;(ntext{ odd})$ is valid in any ring provided $x$ and $y$ commute.



Here $I$ and $A$ commute, and we may, without loss of generality, suppose $k$ is odd (if $A^k=0$, then $A^{k+1}=0$ as well). So your proof is essentially correct.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Oh wow, thanks! I just learned something.
    $endgroup$
    – Nutan Nepal
    Nov 6 '18 at 23:50











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%2f2987859%2fartins-algebra-exercise-1-1-16%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









10












$begingroup$

What you did is almost correct. In fact:$$(A+operatorname{Id})bigl((-1)^{k-1}A^{k-1}+(-1)^{k-2}A^{k-2}+cdots-A+operatorname{Id}bigr)=operatorname{Id},$$and so $(-1)^{k-1}A^{k-1}+(-1)^{k-2}A^{k-2}+cdots-A+operatorname{Id}$ is the inverse of $A+operatorname{Id}$. What does it matter that $det(A^k)=(det A)^k=0$? This has nothing to do with determinants.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I thought that since the determinant of A is 0, factorising it the same as doing $A^kA^{-1}$. I failed to see that $k>0 $ so factorising it like that is correct.
    $endgroup$
    – Nutan Nepal
    Nov 6 '18 at 23:47


















10












$begingroup$

What you did is almost correct. In fact:$$(A+operatorname{Id})bigl((-1)^{k-1}A^{k-1}+(-1)^{k-2}A^{k-2}+cdots-A+operatorname{Id}bigr)=operatorname{Id},$$and so $(-1)^{k-1}A^{k-1}+(-1)^{k-2}A^{k-2}+cdots-A+operatorname{Id}$ is the inverse of $A+operatorname{Id}$. What does it matter that $det(A^k)=(det A)^k=0$? This has nothing to do with determinants.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I thought that since the determinant of A is 0, factorising it the same as doing $A^kA^{-1}$. I failed to see that $k>0 $ so factorising it like that is correct.
    $endgroup$
    – Nutan Nepal
    Nov 6 '18 at 23:47
















10












10








10





$begingroup$

What you did is almost correct. In fact:$$(A+operatorname{Id})bigl((-1)^{k-1}A^{k-1}+(-1)^{k-2}A^{k-2}+cdots-A+operatorname{Id}bigr)=operatorname{Id},$$and so $(-1)^{k-1}A^{k-1}+(-1)^{k-2}A^{k-2}+cdots-A+operatorname{Id}$ is the inverse of $A+operatorname{Id}$. What does it matter that $det(A^k)=(det A)^k=0$? This has nothing to do with determinants.






share|cite|improve this answer











$endgroup$



What you did is almost correct. In fact:$$(A+operatorname{Id})bigl((-1)^{k-1}A^{k-1}+(-1)^{k-2}A^{k-2}+cdots-A+operatorname{Id}bigr)=operatorname{Id},$$and so $(-1)^{k-1}A^{k-1}+(-1)^{k-2}A^{k-2}+cdots-A+operatorname{Id}$ is the inverse of $A+operatorname{Id}$. What does it matter that $det(A^k)=(det A)^k=0$? This has nothing to do with determinants.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 14 '18 at 13:15

























answered Nov 6 '18 at 23:42









José Carlos SantosJosé Carlos Santos

160k22127232




160k22127232












  • $begingroup$
    I thought that since the determinant of A is 0, factorising it the same as doing $A^kA^{-1}$. I failed to see that $k>0 $ so factorising it like that is correct.
    $endgroup$
    – Nutan Nepal
    Nov 6 '18 at 23:47




















  • $begingroup$
    I thought that since the determinant of A is 0, factorising it the same as doing $A^kA^{-1}$. I failed to see that $k>0 $ so factorising it like that is correct.
    $endgroup$
    – Nutan Nepal
    Nov 6 '18 at 23:47


















$begingroup$
I thought that since the determinant of A is 0, factorising it the same as doing $A^kA^{-1}$. I failed to see that $k>0 $ so factorising it like that is correct.
$endgroup$
– Nutan Nepal
Nov 6 '18 at 23:47






$begingroup$
I thought that since the determinant of A is 0, factorising it the same as doing $A^kA^{-1}$. I failed to see that $k>0 $ so factorising it like that is correct.
$endgroup$
– Nutan Nepal
Nov 6 '18 at 23:47













8












$begingroup$

The identity:
$;x^n+y^n=(x+y)(x^{n-1}-x^{n-2}y+dots-xy^{n-2}+y^{n-1})$ $;(ntext{ odd})$ is valid in any ring provided $x$ and $y$ commute.



Here $I$ and $A$ commute, and we may, without loss of generality, suppose $k$ is odd (if $A^k=0$, then $A^{k+1}=0$ as well). So your proof is essentially correct.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Oh wow, thanks! I just learned something.
    $endgroup$
    – Nutan Nepal
    Nov 6 '18 at 23:50
















8












$begingroup$

The identity:
$;x^n+y^n=(x+y)(x^{n-1}-x^{n-2}y+dots-xy^{n-2}+y^{n-1})$ $;(ntext{ odd})$ is valid in any ring provided $x$ and $y$ commute.



Here $I$ and $A$ commute, and we may, without loss of generality, suppose $k$ is odd (if $A^k=0$, then $A^{k+1}=0$ as well). So your proof is essentially correct.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Oh wow, thanks! I just learned something.
    $endgroup$
    – Nutan Nepal
    Nov 6 '18 at 23:50














8












8








8





$begingroup$

The identity:
$;x^n+y^n=(x+y)(x^{n-1}-x^{n-2}y+dots-xy^{n-2}+y^{n-1})$ $;(ntext{ odd})$ is valid in any ring provided $x$ and $y$ commute.



Here $I$ and $A$ commute, and we may, without loss of generality, suppose $k$ is odd (if $A^k=0$, then $A^{k+1}=0$ as well). So your proof is essentially correct.






share|cite|improve this answer











$endgroup$



The identity:
$;x^n+y^n=(x+y)(x^{n-1}-x^{n-2}y+dots-xy^{n-2}+y^{n-1})$ $;(ntext{ odd})$ is valid in any ring provided $x$ and $y$ commute.



Here $I$ and $A$ commute, and we may, without loss of generality, suppose $k$ is odd (if $A^k=0$, then $A^{k+1}=0$ as well). So your proof is essentially correct.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 7 '18 at 9:59

























answered Nov 6 '18 at 23:49









BernardBernard

121k740116




121k740116












  • $begingroup$
    Oh wow, thanks! I just learned something.
    $endgroup$
    – Nutan Nepal
    Nov 6 '18 at 23:50


















  • $begingroup$
    Oh wow, thanks! I just learned something.
    $endgroup$
    – Nutan Nepal
    Nov 6 '18 at 23:50
















$begingroup$
Oh wow, thanks! I just learned something.
$endgroup$
– Nutan Nepal
Nov 6 '18 at 23:50




$begingroup$
Oh wow, thanks! I just learned something.
$endgroup$
– Nutan Nepal
Nov 6 '18 at 23:50


















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%2f2987859%2fartins-algebra-exercise-1-1-16%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

Plaza Victoria

Image
Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
Read more

Puebla de Zaragoza

Image
Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
Read more

Musa

Image
Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
Read more