Artin's Algebra Exercise 1.1.16

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the question goes $A^k = 0$, for some $k>0$. 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 with inverse given above.



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



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 inverse







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  • 1




    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.
    – John Douma
    3 hours ago











  • Hint: what's $x^n + y^n$?
    – Matija Sreckovic
    3 hours ago










  • Your idea is right, but you need to argue that the product in the other order is also $I$.
    – darij grinberg
    2 hours ago















up vote
1
down vote

favorite












the question goes $A^k = 0$, for some $k>0$. 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 with inverse given above.



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



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 inverse







share|cite|improve this question



















  • 1




    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.
    – John Douma
    3 hours ago











  • Hint: what's $x^n + y^n$?
    – Matija Sreckovic
    3 hours ago










  • Your idea is right, but you need to argue that the product in the other order is also $I$.
    – darij grinberg
    2 hours ago













up vote
1
down vote

favorite









up vote
1
down vote

favorite











the question goes $A^k = 0$, for some $k>0$. 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 with inverse given above.



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



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 inverse







share|cite|improve this question















the question goes $A^k = 0$, for some $k>0$. 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 with inverse given above.



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



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 inverse




linear-algebra inverse






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 3 hours ago









José Carlos Santos

135k17108198




135k17108198










asked 3 hours ago









Nutan Nepal

184




184







  • 1




    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.
    – John Douma
    3 hours ago











  • Hint: what's $x^n + y^n$?
    – Matija Sreckovic
    3 hours ago










  • Your idea is right, but you need to argue that the product in the other order is also $I$.
    – darij grinberg
    2 hours ago













  • 1




    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.
    – John Douma
    3 hours ago











  • Hint: what's $x^n + y^n$?
    – Matija Sreckovic
    3 hours ago










  • Your idea is right, but you need to argue that the product in the other order is also $I$.
    – darij grinberg
    2 hours ago








1




1




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.
– John Douma
3 hours ago





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.
– John Douma
3 hours ago













Hint: what's $x^n + y^n$?
– Matija Sreckovic
3 hours ago




Hint: what's $x^n + y^n$?
– Matija Sreckovic
3 hours ago












Your idea is right, but you need to argue that the product in the other order is also $I$.
– darij grinberg
2 hours ago





Your idea is right, but you need to argue that the product in the other order is also $I$.
– darij grinberg
2 hours ago











2 Answers
2






active

oldest

votes

















up vote
4
down vote



accepted










What you did is almost right. In fact:$$(A+operatornameId)bigl((-1)^k-1A^k-1+(-1)^k-2A^k-2+cdots-A+operatornameIdbigr)=operatornameId,$$and therefore $(-1)^k-1A^k-1+(-1)^k-2A^k-2+cdots-A+operatornameId$ is the inverse of $A+operatornameId$. What does it matter that $det(A^k)=(det A)^k=0$? This has nothing to do with determinants.






share|cite|improve this answer




















  • 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.
    – Nutan Nepal
    3 hours ago











  • I'm glad I could help.
    – José Carlos Santos
    3 hours ago

















up vote
2
down vote













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



Here $I$ and $A$ commute, and we may suppose, 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




















  • Oh wow, thanks! I just learned something.
    – Nutan Nepal
    3 hours ago










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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
4
down vote



accepted










What you did is almost right. In fact:$$(A+operatornameId)bigl((-1)^k-1A^k-1+(-1)^k-2A^k-2+cdots-A+operatornameIdbigr)=operatornameId,$$and therefore $(-1)^k-1A^k-1+(-1)^k-2A^k-2+cdots-A+operatornameId$ is the inverse of $A+operatornameId$. What does it matter that $det(A^k)=(det A)^k=0$? This has nothing to do with determinants.






share|cite|improve this answer




















  • 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.
    – Nutan Nepal
    3 hours ago











  • I'm glad I could help.
    – José Carlos Santos
    3 hours ago














up vote
4
down vote



accepted










What you did is almost right. In fact:$$(A+operatornameId)bigl((-1)^k-1A^k-1+(-1)^k-2A^k-2+cdots-A+operatornameIdbigr)=operatornameId,$$and therefore $(-1)^k-1A^k-1+(-1)^k-2A^k-2+cdots-A+operatornameId$ is the inverse of $A+operatornameId$. What does it matter that $det(A^k)=(det A)^k=0$? This has nothing to do with determinants.






share|cite|improve this answer




















  • 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.
    – Nutan Nepal
    3 hours ago











  • I'm glad I could help.
    – José Carlos Santos
    3 hours ago












up vote
4
down vote



accepted







up vote
4
down vote



accepted






What you did is almost right. In fact:$$(A+operatornameId)bigl((-1)^k-1A^k-1+(-1)^k-2A^k-2+cdots-A+operatornameIdbigr)=operatornameId,$$and therefore $(-1)^k-1A^k-1+(-1)^k-2A^k-2+cdots-A+operatornameId$ is the inverse of $A+operatornameId$. What does it matter that $det(A^k)=(det A)^k=0$? This has nothing to do with determinants.






share|cite|improve this answer












What you did is almost right. In fact:$$(A+operatornameId)bigl((-1)^k-1A^k-1+(-1)^k-2A^k-2+cdots-A+operatornameIdbigr)=operatornameId,$$and therefore $(-1)^k-1A^k-1+(-1)^k-2A^k-2+cdots-A+operatornameId$ is the inverse of $A+operatornameId$. 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










answered 3 hours ago









José Carlos Santos

135k17108198




135k17108198











  • 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.
    – Nutan Nepal
    3 hours ago











  • I'm glad I could help.
    – José Carlos Santos
    3 hours ago
















  • 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.
    – Nutan Nepal
    3 hours ago











  • I'm glad I could help.
    – José Carlos Santos
    3 hours ago















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.
– Nutan Nepal
3 hours ago





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.
– Nutan Nepal
3 hours ago













I'm glad I could help.
– José Carlos Santos
3 hours ago




I'm glad I could help.
– José Carlos Santos
3 hours ago










up vote
2
down vote













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



Here $I$ and $A$ commute, and we may suppose, 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




















  • Oh wow, thanks! I just learned something.
    – Nutan Nepal
    3 hours ago














up vote
2
down vote













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



Here $I$ and $A$ commute, and we may suppose, 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




















  • Oh wow, thanks! I just learned something.
    – Nutan Nepal
    3 hours ago












up vote
2
down vote










up vote
2
down vote









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



Here $I$ and $A$ commute, and we may suppose, 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












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



Here $I$ and $A$ commute, and we may suppose, 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










answered 3 hours ago









Bernard

115k637107




115k637107











  • Oh wow, thanks! I just learned something.
    – Nutan Nepal
    3 hours ago
















  • Oh wow, thanks! I just learned something.
    – Nutan Nepal
    3 hours ago















Oh wow, thanks! I just learned something.
– Nutan Nepal
3 hours ago




Oh wow, thanks! I just learned something.
– Nutan Nepal
3 hours ago

















 

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