Artin's Algebra Exercise 1.1.16
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3
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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 inverse
asked Nov 6 at 23:34
Nutan Nepal
315
add a comment |
up vote
3
down vote
favorite
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 inverse
asked Nov 6 at 23:34
Nutan Nepal
315
2
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
Nov 6 at 23:40
Hint: what's $x^{n} + y^{n}$?
– Matija Sreckovic
Nov 6 at 23:43
Your idea is right, but you need to argue that the product in the other order is also $I$.
– darij grinberg
Nov 7 at 0:50
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
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 inverse
asked Nov 6 at 23:34
Nutan Nepal
315
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 inverse
linear-algebra inverse
asked Nov 6 at 23:34
Nutan Nepal
315
asked Nov 6 at 23:34
Nutan Nepal
315
edited Nov 7 at 16:31
asked Nov 6 at 23:34
Nutan Nepal
315
asked Nov 6 at 23:34
Nutan Nepal
315
asked Nov 6 at 23:34
Nutan Nepal
315
315
2
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
Nov 6 at 23:40
Hint: what's $x^{n} + y^{n}$?
– Matija Sreckovic
Nov 6 at 23:43
Your idea is right, but you need to argue that the product in the other order is also $I$.
– darij grinberg
Nov 7 at 0:50
add a comment |
2
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
Nov 6 at 23:40
Hint: what's $x^{n} + y^{n}$?
– Matija Sreckovic
Nov 6 at 23:43
Your idea is right, but you need to argue that the product in the other order is also $I$.
– darij grinberg
Nov 7 at 0:50
2
2
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
Nov 6 at 23:40
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
Nov 6 at 23:40
Hint: what's $x^{n} + y^{n}$?
– Matija Sreckovic
Nov 6 at 23:43
Hint: what's $x^{n} + y^{n}$?
– Matija Sreckovic
Nov 6 at 23:43
Your idea is right, but you need to argue that the product in the other order is also $I$.
– darij grinberg
Nov 7 at 0:50
Your idea is right, but you need to argue that the product in the other order is also $I$.
– darij grinberg
Nov 7 at 0:50
add a comment |
2 Answers
2
active
oldest
votes
up vote
9
down vote
accepted
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.
answered Nov 6 at 23:42
José Carlos Santos
139k18111203
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
Nov 6 at 23:47
add a comment |
up vote
7
down vote
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.
answered Nov 6 at 23:49
Bernard
115k637108
Oh wow, thanks! I just learned something.
– Nutan Nepal
Nov 6 at 23:50
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
9
down vote
accepted
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.
answered Nov 6 at 23:42
José Carlos Santos
139k18111203
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
Nov 6 at 23:47
add a comment |
up vote
9
down vote
accepted
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.
answered Nov 6 at 23:42
José Carlos Santos
139k18111203
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
Nov 6 at 23:47
add a comment |
up vote
9
down vote
accepted
up vote
9
down vote
accepted
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.
answered Nov 6 at 23:42
José Carlos Santos
139k18111203
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.
answered Nov 6 at 23:42
José Carlos Santos
139k18111203
edited Nov 14 at 13:15
answered Nov 6 at 23:42
José Carlos Santos
139k18111203
answered Nov 6 at 23:42
José Carlos Santos
139k18111203
answered Nov 6 at 23:42
José Carlos Santos
139k18111203
139k18111203
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
Nov 6 at 23:47
add a comment |
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
Nov 6 at 23:47
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
Nov 6 at 23:47
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
Nov 6 at 23:47
add a comment |
up vote
7
down vote
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.
answered Nov 6 at 23:49
Bernard
115k637108
Oh wow, thanks! I just learned something.
– Nutan Nepal
Nov 6 at 23:50
add a comment |
up vote
7
down vote
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.
answered Nov 6 at 23:49
Bernard
115k637108
Oh wow, thanks! I just learned something.
– Nutan Nepal
Nov 6 at 23:50
add a comment |
up vote
7
down vote
up vote
7
down vote
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.
answered Nov 6 at 23:49
Bernard
115k637108
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.
answered Nov 6 at 23:49
Bernard
115k637108
edited Nov 7 at 9:59
answered Nov 6 at 23:49
Bernard
115k637108
answered Nov 6 at 23:49
Bernard
115k637108
answered Nov 6 at 23:49
Bernard
115k637108
115k637108
Oh wow, thanks! I just learned something.
– Nutan Nepal
Nov 6 at 23:50
add a comment |
Oh wow, thanks! I just learned something.
– Nutan Nepal
Nov 6 at 23:50
Oh wow, thanks! I just learned something.
– Nutan Nepal
Nov 6 at 23:50
Oh wow, thanks! I just learned something.
– Nutan Nepal
Nov 6 at 23:50
add a comment |
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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
Nov 6 at 23:40
Hint: what's $x^{n} + y^{n}$?
– Matija Sreckovic
Nov 6 at 23:43
Your idea is right, but you need to argue that the product in the other order is also $I$.
– darij grinberg
Nov 7 at 0:50