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











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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


















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