Question about matrices and eigenvectors
$begingroup$
For two $ntimes n$ square matrices $A$ and $B$ satisfying $AB=A+B,;$
suppose $A$ has an eigenvector $u.$
How to show that $u$ is also and eigenvector of $B?$
linear-algebra
$endgroup$
add a comment |
$begingroup$
For two $ntimes n$ square matrices $A$ and $B$ satisfying $AB=A+B,;$
suppose $A$ has an eigenvector $u.$
How to show that $u$ is also and eigenvector of $B?$
linear-algebra
$endgroup$
$begingroup$
Show us what you tried please.
$endgroup$
– user376343
Dec 13 '18 at 12:14
4
$begingroup$
I tried from definition $Au=lambda u$ and multiply both side by u i.e. ABu=Au+Bu, then Bu=ABu-Au=(AB-A)u. But I don’t know how to continue onward.
$endgroup$
– Real_Galois
Dec 13 '18 at 12:19
2
$begingroup$
But AB doesnt have to equal BA
$endgroup$
– Real_Galois
Dec 13 '18 at 12:23
1
$begingroup$
Also, it seems that proof by contradiction is not working (or I don’t know the way)
$endgroup$
– Real_Galois
Dec 13 '18 at 12:27
$begingroup$
Well! Could you please enclose your comments above into your question?
$endgroup$
– user376343
Dec 13 '18 at 13:29
add a comment |
$begingroup$
For two $ntimes n$ square matrices $A$ and $B$ satisfying $AB=A+B,;$
suppose $A$ has an eigenvector $u.$
How to show that $u$ is also and eigenvector of $B?$
linear-algebra
$endgroup$
For two $ntimes n$ square matrices $A$ and $B$ satisfying $AB=A+B,;$
suppose $A$ has an eigenvector $u.$
How to show that $u$ is also and eigenvector of $B?$
linear-algebra
linear-algebra
edited Dec 13 '18 at 14:44
user376343
3,6783827
3,6783827
asked Dec 13 '18 at 12:12
Real_GaloisReal_Galois
255
255
$begingroup$
Show us what you tried please.
$endgroup$
– user376343
Dec 13 '18 at 12:14
4
$begingroup$
I tried from definition $Au=lambda u$ and multiply both side by u i.e. ABu=Au+Bu, then Bu=ABu-Au=(AB-A)u. But I don’t know how to continue onward.
$endgroup$
– Real_Galois
Dec 13 '18 at 12:19
2
$begingroup$
But AB doesnt have to equal BA
$endgroup$
– Real_Galois
Dec 13 '18 at 12:23
1
$begingroup$
Also, it seems that proof by contradiction is not working (or I don’t know the way)
$endgroup$
– Real_Galois
Dec 13 '18 at 12:27
$begingroup$
Well! Could you please enclose your comments above into your question?
$endgroup$
– user376343
Dec 13 '18 at 13:29
add a comment |
$begingroup$
Show us what you tried please.
$endgroup$
– user376343
Dec 13 '18 at 12:14
4
$begingroup$
I tried from definition $Au=lambda u$ and multiply both side by u i.e. ABu=Au+Bu, then Bu=ABu-Au=(AB-A)u. But I don’t know how to continue onward.
$endgroup$
– Real_Galois
Dec 13 '18 at 12:19
2
$begingroup$
But AB doesnt have to equal BA
$endgroup$
– Real_Galois
Dec 13 '18 at 12:23
1
$begingroup$
Also, it seems that proof by contradiction is not working (or I don’t know the way)
$endgroup$
– Real_Galois
Dec 13 '18 at 12:27
$begingroup$
Well! Could you please enclose your comments above into your question?
$endgroup$
– user376343
Dec 13 '18 at 13:29
$begingroup$
Show us what you tried please.
$endgroup$
– user376343
Dec 13 '18 at 12:14
$begingroup$
Show us what you tried please.
$endgroup$
– user376343
Dec 13 '18 at 12:14
4
4
$begingroup$
I tried from definition $Au=lambda u$ and multiply both side by u i.e. ABu=Au+Bu, then Bu=ABu-Au=(AB-A)u. But I don’t know how to continue onward.
$endgroup$
– Real_Galois
Dec 13 '18 at 12:19
$begingroup$
I tried from definition $Au=lambda u$ and multiply both side by u i.e. ABu=Au+Bu, then Bu=ABu-Au=(AB-A)u. But I don’t know how to continue onward.
$endgroup$
– Real_Galois
Dec 13 '18 at 12:19
2
2
$begingroup$
But AB doesnt have to equal BA
$endgroup$
– Real_Galois
Dec 13 '18 at 12:23
$begingroup$
But AB doesnt have to equal BA
$endgroup$
– Real_Galois
Dec 13 '18 at 12:23
1
1
$begingroup$
Also, it seems that proof by contradiction is not working (or I don’t know the way)
$endgroup$
– Real_Galois
Dec 13 '18 at 12:27
$begingroup$
Also, it seems that proof by contradiction is not working (or I don’t know the way)
$endgroup$
– Real_Galois
Dec 13 '18 at 12:27
$begingroup$
Well! Could you please enclose your comments above into your question?
$endgroup$
– user376343
Dec 13 '18 at 13:29
$begingroup$
Well! Could you please enclose your comments above into your question?
$endgroup$
– user376343
Dec 13 '18 at 13:29
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint: from the given condition show that $A-I$ and $B-I$ are inverse of each of each other. Then from the given equation solve for $B$.
It is easy to check that $(A-I)(B-I)=I$. Now $$(A-I)B=A. $$ Multiply bothsides by $B-I$ to get $$B= (B-I)A=BA-A.$$
You should be able to continue from here since now $AB= BA$.
$endgroup$
1
$begingroup$
Show $B=(A-I)^{-1}A=(B-I) A$.
$endgroup$
– user9077
Dec 13 '18 at 13:49
2
$begingroup$
If you have show that $(A−I)(B−I)=I$ it's enough. $A−I$ has the same eigenvectors as $A$ , $B−I$ as inverse the same as $A−I$ and $B$ the same as $B−I$
$endgroup$
– Widawensen
Dec 13 '18 at 16:31
1
$begingroup$
Nice observations, thanks!
$endgroup$
– user9077
Dec 13 '18 at 16:48
add a comment |
$begingroup$
With the hint of @user9077, and without need of commutativity of $A,B,$ assume $(lambda,v);$ is an eigenpair of $A.$
Then $$Bv=(B-I)Av=(B-I)lambda v=lambda B v- lambda v$$ hence $$(1-lambda)Bv=-lambda v.$$ This means that $(frac{lambda}{lambda-1},v)$ is an eigenpair of $B.$
$endgroup$
add a comment |
Your Answer
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2 Answers
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active
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2 Answers
2
active
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active
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oldest
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$begingroup$
Hint: from the given condition show that $A-I$ and $B-I$ are inverse of each of each other. Then from the given equation solve for $B$.
It is easy to check that $(A-I)(B-I)=I$. Now $$(A-I)B=A. $$ Multiply bothsides by $B-I$ to get $$B= (B-I)A=BA-A.$$
You should be able to continue from here since now $AB= BA$.
$endgroup$
1
$begingroup$
Show $B=(A-I)^{-1}A=(B-I) A$.
$endgroup$
– user9077
Dec 13 '18 at 13:49
2
$begingroup$
If you have show that $(A−I)(B−I)=I$ it's enough. $A−I$ has the same eigenvectors as $A$ , $B−I$ as inverse the same as $A−I$ and $B$ the same as $B−I$
$endgroup$
– Widawensen
Dec 13 '18 at 16:31
1
$begingroup$
Nice observations, thanks!
$endgroup$
– user9077
Dec 13 '18 at 16:48
add a comment |
$begingroup$
Hint: from the given condition show that $A-I$ and $B-I$ are inverse of each of each other. Then from the given equation solve for $B$.
It is easy to check that $(A-I)(B-I)=I$. Now $$(A-I)B=A. $$ Multiply bothsides by $B-I$ to get $$B= (B-I)A=BA-A.$$
You should be able to continue from here since now $AB= BA$.
$endgroup$
1
$begingroup$
Show $B=(A-I)^{-1}A=(B-I) A$.
$endgroup$
– user9077
Dec 13 '18 at 13:49
2
$begingroup$
If you have show that $(A−I)(B−I)=I$ it's enough. $A−I$ has the same eigenvectors as $A$ , $B−I$ as inverse the same as $A−I$ and $B$ the same as $B−I$
$endgroup$
– Widawensen
Dec 13 '18 at 16:31
1
$begingroup$
Nice observations, thanks!
$endgroup$
– user9077
Dec 13 '18 at 16:48
add a comment |
$begingroup$
Hint: from the given condition show that $A-I$ and $B-I$ are inverse of each of each other. Then from the given equation solve for $B$.
It is easy to check that $(A-I)(B-I)=I$. Now $$(A-I)B=A. $$ Multiply bothsides by $B-I$ to get $$B= (B-I)A=BA-A.$$
You should be able to continue from here since now $AB= BA$.
$endgroup$
Hint: from the given condition show that $A-I$ and $B-I$ are inverse of each of each other. Then from the given equation solve for $B$.
It is easy to check that $(A-I)(B-I)=I$. Now $$(A-I)B=A. $$ Multiply bothsides by $B-I$ to get $$B= (B-I)A=BA-A.$$
You should be able to continue from here since now $AB= BA$.
edited Dec 13 '18 at 13:58
answered Dec 13 '18 at 13:25
user9077user9077
1,239612
1,239612
1
$begingroup$
Show $B=(A-I)^{-1}A=(B-I) A$.
$endgroup$
– user9077
Dec 13 '18 at 13:49
2
$begingroup$
If you have show that $(A−I)(B−I)=I$ it's enough. $A−I$ has the same eigenvectors as $A$ , $B−I$ as inverse the same as $A−I$ and $B$ the same as $B−I$
$endgroup$
– Widawensen
Dec 13 '18 at 16:31
1
$begingroup$
Nice observations, thanks!
$endgroup$
– user9077
Dec 13 '18 at 16:48
add a comment |
1
$begingroup$
Show $B=(A-I)^{-1}A=(B-I) A$.
$endgroup$
– user9077
Dec 13 '18 at 13:49
2
$begingroup$
If you have show that $(A−I)(B−I)=I$ it's enough. $A−I$ has the same eigenvectors as $A$ , $B−I$ as inverse the same as $A−I$ and $B$ the same as $B−I$
$endgroup$
– Widawensen
Dec 13 '18 at 16:31
1
$begingroup$
Nice observations, thanks!
$endgroup$
– user9077
Dec 13 '18 at 16:48
1
1
$begingroup$
Show $B=(A-I)^{-1}A=(B-I) A$.
$endgroup$
– user9077
Dec 13 '18 at 13:49
$begingroup$
Show $B=(A-I)^{-1}A=(B-I) A$.
$endgroup$
– user9077
Dec 13 '18 at 13:49
2
2
$begingroup$
If you have show that $(A−I)(B−I)=I$ it's enough. $A−I$ has the same eigenvectors as $A$ , $B−I$ as inverse the same as $A−I$ and $B$ the same as $B−I$
$endgroup$
– Widawensen
Dec 13 '18 at 16:31
$begingroup$
If you have show that $(A−I)(B−I)=I$ it's enough. $A−I$ has the same eigenvectors as $A$ , $B−I$ as inverse the same as $A−I$ and $B$ the same as $B−I$
$endgroup$
– Widawensen
Dec 13 '18 at 16:31
1
1
$begingroup$
Nice observations, thanks!
$endgroup$
– user9077
Dec 13 '18 at 16:48
$begingroup$
Nice observations, thanks!
$endgroup$
– user9077
Dec 13 '18 at 16:48
add a comment |
$begingroup$
With the hint of @user9077, and without need of commutativity of $A,B,$ assume $(lambda,v);$ is an eigenpair of $A.$
Then $$Bv=(B-I)Av=(B-I)lambda v=lambda B v- lambda v$$ hence $$(1-lambda)Bv=-lambda v.$$ This means that $(frac{lambda}{lambda-1},v)$ is an eigenpair of $B.$
$endgroup$
add a comment |
$begingroup$
With the hint of @user9077, and without need of commutativity of $A,B,$ assume $(lambda,v);$ is an eigenpair of $A.$
Then $$Bv=(B-I)Av=(B-I)lambda v=lambda B v- lambda v$$ hence $$(1-lambda)Bv=-lambda v.$$ This means that $(frac{lambda}{lambda-1},v)$ is an eigenpair of $B.$
$endgroup$
add a comment |
$begingroup$
With the hint of @user9077, and without need of commutativity of $A,B,$ assume $(lambda,v);$ is an eigenpair of $A.$
Then $$Bv=(B-I)Av=(B-I)lambda v=lambda B v- lambda v$$ hence $$(1-lambda)Bv=-lambda v.$$ This means that $(frac{lambda}{lambda-1},v)$ is an eigenpair of $B.$
$endgroup$
With the hint of @user9077, and without need of commutativity of $A,B,$ assume $(lambda,v);$ is an eigenpair of $A.$
Then $$Bv=(B-I)Av=(B-I)lambda v=lambda B v- lambda v$$ hence $$(1-lambda)Bv=-lambda v.$$ This means that $(frac{lambda}{lambda-1},v)$ is an eigenpair of $B.$
answered Dec 13 '18 at 14:41
user376343user376343
3,6783827
3,6783827
add a comment |
add a comment |
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$begingroup$
Show us what you tried please.
$endgroup$
– user376343
Dec 13 '18 at 12:14
4
$begingroup$
I tried from definition $Au=lambda u$ and multiply both side by u i.e. ABu=Au+Bu, then Bu=ABu-Au=(AB-A)u. But I don’t know how to continue onward.
$endgroup$
– Real_Galois
Dec 13 '18 at 12:19
2
$begingroup$
But AB doesnt have to equal BA
$endgroup$
– Real_Galois
Dec 13 '18 at 12:23
1
$begingroup$
Also, it seems that proof by contradiction is not working (or I don’t know the way)
$endgroup$
– Real_Galois
Dec 13 '18 at 12:27
$begingroup$
Well! Could you please enclose your comments above into your question?
$endgroup$
– user376343
Dec 13 '18 at 13:29