Mixing
If $I-AB$ is invertible, then is $I-BA$ invertible? [duplicate]
$begingroup$
This question already has an answer here:
$I_m - AB$ is invertible if and only if $I_n - BA$ is invertible.
4 answers
If $A$, $B$ are square matrices and $I-AB$ is invertible how do I prove that $I-BA$ is invertible?
This is exercise 8 of section 6.2 in Linear Algebra by Hoffman and Kunze.
My thoughts.
If $A$ and $B$ are invertible then $AB$ and $BA$ are similar, so we can use that to show that $I-AB$ and $I-BA$ are similar, and hence if $I-AB$ is invertible then so is $I-BA$.
However, $A$ and $B$ are not given to be invertible, so I am not able to apply this idea to show that $I-AB$ and $I-BA$ will be similar in general. Can anyone give me a hint to prove this in the general case?
EDIT: As pointed out in the comments below, it is not true in general that $I-AB$ and $I-BA$ are similar. @JulianRosen gave the example
$$
A = begin{bmatrix}
0 & 1 \
0 & 0
end{bmatrix}, quad
B = begin{bmatrix}
0 & 0 \
0 & 1
end{bmatrix}.
$$
So, my original approach was completely incorrect. However, it is still true that $I-AB$ invertible implies $I-BA$ invertible.
linear-algebra matrices linear-transformations inverse
edited Jan 3 at 18:40
Brahadeesh
6,19742361
asked Nov 10 '14 at 17:05
SSHSSH
395
$endgroup$
marked as duplicate by Brahadeesh, Adrian Keister, Leucippus, max_zorn, metamorphy Jan 4 at 2:47
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
|
show 7 more comments
$begingroup$
This question already has an answer here:
$I_m - AB$ is invertible if and only if $I_n - BA$ is invertible.
4 answers
If $A$, $B$ are square matrices and $I-AB$ is invertible how do I prove that $I-BA$ is invertible?
This is exercise 8 of section 6.2 in Linear Algebra by Hoffman and Kunze.
My thoughts.
If $A$ and $B$ are invertible then $AB$ and $BA$ are similar, so we can use that to show that $I-AB$ and $I-BA$ are similar, and hence if $I-AB$ is invertible then so is $I-BA$.
However, $A$ and $B$ are not given to be invertible, so I am not able to apply this idea to show that $I-AB$ and $I-BA$ will be similar in general. Can anyone give me a hint to prove this in the general case?
EDIT: As pointed out in the comments below, it is not true in general that $I-AB$ and $I-BA$ are similar. @JulianRosen gave the example
$$
A = begin{bmatrix}
0 & 1 \
0 & 0
end{bmatrix}, quad
B = begin{bmatrix}
0 & 0 \
0 & 1
end{bmatrix}.
$$
So, my original approach was completely incorrect. However, it is still true that $I-AB$ invertible implies $I-BA$ invertible.
linear-algebra matrices linear-transformations inverse
edited Jan 3 at 18:40
Brahadeesh
6,19742361
asked Nov 10 '14 at 17:05
SSHSSH
395
$endgroup$
marked as duplicate by Brahadeesh, Adrian Keister, Leucippus, max_zorn, metamorphy Jan 4 at 2:47
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
4
$begingroup$
$A=left[begin{array}{cc}0&1\0&0end{array}right]$, $B=left[begin{array}{cc}0&0\0&1end{array}right]$ is a counterexample
$endgroup$
– Julian Rosen
Nov 10 '14 at 17:45
$begingroup$
we assume that I-AB is invertible
$endgroup$
– SSH
Nov 10 '14 at 18:06
1
$begingroup$
In this example $I-AB=left[begin{array}{cc}1&-1\0&1end{array}right]$ is invertible
$endgroup$
– Julian Rosen
Nov 10 '14 at 18:07
1
$begingroup$
Do you mean that either $A$ or $B$ is invertible? Because then the statement would be correct
$endgroup$
– Omnomnomnom
Nov 10 '14 at 18:12
1
$begingroup$
@Salem then Julian's counterexample applies, and there's nothing we can do. If the question is correct, then there must be some part of the question that you're not sharing with us.
$endgroup$
– Omnomnomnom
Nov 10 '14 at 18:17
|
show 7 more comments
$begingroup$
This question already has an answer here:
$I_m - AB$ is invertible if and only if $I_n - BA$ is invertible.
4 answers
If $A$, $B$ are square matrices and $I-AB$ is invertible how do I prove that $I-BA$ is invertible?
This is exercise 8 of section 6.2 in Linear Algebra by Hoffman and Kunze.
My thoughts.
If $A$ and $B$ are invertible then $AB$ and $BA$ are similar, so we can use that to show that $I-AB$ and $I-BA$ are similar, and hence if $I-AB$ is invertible then so is $I-BA$.
However, $A$ and $B$ are not given to be invertible, so I am not able to apply this idea to show that $I-AB$ and $I-BA$ will be similar in general. Can anyone give me a hint to prove this in the general case?
EDIT: As pointed out in the comments below, it is not true in general that $I-AB$ and $I-BA$ are similar. @JulianRosen gave the example
$$
A = begin{bmatrix}
0 & 1 \
0 & 0
end{bmatrix}, quad
B = begin{bmatrix}
0 & 0 \
0 & 1
end{bmatrix}.
$$
So, my original approach was completely incorrect. However, it is still true that $I-AB$ invertible implies $I-BA$ invertible.
linear-algebra matrices linear-transformations inverse
edited Jan 3 at 18:40
Brahadeesh
6,19742361
asked Nov 10 '14 at 17:05
SSHSSH
395
$endgroup$
This question already has an answer here:
$I_m - AB$ is invertible if and only if $I_n - BA$ is invertible.
4 answers
If $A$, $B$ are square matrices and $I-AB$ is invertible how do I prove that $I-BA$ is invertible?
This is exercise 8 of section 6.2 in Linear Algebra by Hoffman and Kunze.
My thoughts.
If $A$ and $B$ are invertible then $AB$ and $BA$ are similar, so we can use that to show that $I-AB$ and $I-BA$ are similar, and hence if $I-AB$ is invertible then so is $I-BA$.
However, $A$ and $B$ are not given to be invertible, so I am not able to apply this idea to show that $I-AB$ and $I-BA$ will be similar in general. Can anyone give me a hint to prove this in the general case?
EDIT: As pointed out in the comments below, it is not true in general that $I-AB$ and $I-BA$ are similar. @JulianRosen gave the example
$$
A = begin{bmatrix}
0 & 1 \
0 & 0
end{bmatrix}, quad
B = begin{bmatrix}
0 & 0 \
0 & 1
end{bmatrix}.
$$
So, my original approach was completely incorrect. However, it is still true that $I-AB$ invertible implies $I-BA$ invertible.
This question already has an answer here:
$I_m - AB$ is invertible if and only if $I_n - BA$ is invertible.
4 answers
linear-algebra matrices linear-transformations inverse
linear-algebra matrices linear-transformations inverse
edited Jan 3 at 18:40
Brahadeesh
6,19742361
asked Nov 10 '14 at 17:05
SSHSSH
395
edited Jan 3 at 18:40
Brahadeesh
6,19742361
asked Nov 10 '14 at 17:05
SSHSSH
395
edited Jan 3 at 18:40
Brahadeesh
6,19742361
edited Jan 3 at 18:40
Brahadeesh
6,19742361
edited Jan 3 at 18:40
Brahadeesh
6,19742361
6,19742361
asked Nov 10 '14 at 17:05
SSHSSH
395
asked Nov 10 '14 at 17:05
SSHSSH
395
asked Nov 10 '14 at 17:05
SSHSSH
395
395
marked as duplicate by Brahadeesh, Adrian Keister, Leucippus, max_zorn, metamorphy Jan 4 at 2:47
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Brahadeesh, Adrian Keister, Leucippus, max_zorn, metamorphy Jan 4 at 2:47
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
4
$begingroup$
$A=left[begin{array}{cc}0&1\0&0end{array}right]$, $B=left[begin{array}{cc}0&0\0&1end{array}right]$ is a counterexample
$endgroup$
– Julian Rosen
Nov 10 '14 at 17:45
$begingroup$
we assume that I-AB is invertible
$endgroup$
– SSH
Nov 10 '14 at 18:06
1
$begingroup$
In this example $I-AB=left[begin{array}{cc}1&-1\0&1end{array}right]$ is invertible
$endgroup$
– Julian Rosen
Nov 10 '14 at 18:07
1
$begingroup$
Do you mean that either $A$ or $B$ is invertible? Because then the statement would be correct
$endgroup$
– Omnomnomnom
Nov 10 '14 at 18:12
1
$begingroup$
@Salem then Julian's counterexample applies, and there's nothing we can do. If the question is correct, then there must be some part of the question that you're not sharing with us.
$endgroup$
– Omnomnomnom
Nov 10 '14 at 18:17
|
show 7 more comments
4
$begingroup$
$A=left[begin{array}{cc}0&1\0&0end{array}right]$, $B=left[begin{array}{cc}0&0\0&1end{array}right]$ is a counterexample
$endgroup$
– Julian Rosen
Nov 10 '14 at 17:45
$begingroup$
we assume that I-AB is invertible
$endgroup$
– SSH
Nov 10 '14 at 18:06
1
$begingroup$
In this example $I-AB=left[begin{array}{cc}1&-1\0&1end{array}right]$ is invertible
$endgroup$
– Julian Rosen
Nov 10 '14 at 18:07
1
$begingroup$
Do you mean that either $A$ or $B$ is invertible? Because then the statement would be correct
$endgroup$
– Omnomnomnom
Nov 10 '14 at 18:12
1
$begingroup$
@Salem then Julian's counterexample applies, and there's nothing we can do. If the question is correct, then there must be some part of the question that you're not sharing with us.
$endgroup$
– Omnomnomnom
Nov 10 '14 at 18:17
4
4
$begingroup$
$A=left[begin{array}{cc}0&1\0&0end{array}right]$, $B=left[begin{array}{cc}0&0\0&1end{array}right]$ is a counterexample
$endgroup$
– Julian Rosen
Nov 10 '14 at 17:45
$begingroup$
$A=left[begin{array}{cc}0&1\0&0end{array}right]$, $B=left[begin{array}{cc}0&0\0&1end{array}right]$ is a counterexample
$endgroup$
– Julian Rosen
Nov 10 '14 at 17:45
$begingroup$
we assume that I-AB is invertible
$endgroup$
– SSH
Nov 10 '14 at 18:06
$begingroup$
we assume that I-AB is invertible
$endgroup$
– SSH
Nov 10 '14 at 18:06
1
1
$begingroup$
In this example $I-AB=left[begin{array}{cc}1&-1\0&1end{array}right]$ is invertible
$endgroup$
– Julian Rosen
Nov 10 '14 at 18:07
$begingroup$
In this example $I-AB=left[begin{array}{cc}1&-1\0&1end{array}right]$ is invertible
$endgroup$
– Julian Rosen
Nov 10 '14 at 18:07
1
1
$begingroup$
Do you mean that either $A$ or $B$ is invertible? Because then the statement would be correct
$endgroup$
– Omnomnomnom
Nov 10 '14 at 18:12
$begingroup$
Do you mean that either $A$ or $B$ is invertible? Because then the statement would be correct
$endgroup$
– Omnomnomnom
Nov 10 '14 at 18:12
1
1
$begingroup$
@Salem then Julian's counterexample applies, and there's nothing we can do. If the question is correct, then there must be some part of the question that you're not sharing with us.
$endgroup$
– Omnomnomnom
Nov 10 '14 at 18:17
$begingroup$
@Salem then Julian's counterexample applies, and there's nothing we can do. If the question is correct, then there must be some part of the question that you're not sharing with us.
$endgroup$
– Omnomnomnom
Nov 10 '14 at 18:17
|
show 7 more comments
3 Answers
3
active
oldest
votes
$begingroup$
Hint: Note that the two matrices have the same determinant.
There are several proofs of this, such as those given here.
Note that if either of $A$ and $B$ are invertible, we may further state that the two matrices are similar.
edited Apr 13 '17 at 12:21
Community♦
1
answered Nov 10 '14 at 18:29
OmnomnomnomOmnomnomnom
127k790178
$endgroup$
add a comment |
$begingroup$
Write $X = (I - AB)^{-1}$ and set $Y = I + BXA$. It is easy to check that $Y$ is the inverse of $I - BA$, but of course this solution will make you wonder how you come up with this in the first place.. see the mathoverflow question here.
edited Apr 13 '17 at 12:58
Community♦
1
answered Nov 10 '14 at 18:49
spinspin
3,82512257
$endgroup$
add a comment |
$begingroup$
A linear operator on a finite-dimensional vector space is invertible if and only if it is injective, and thus if and only if its kernel is trivial. An $n times n$ matrix over a field $F$ defines a linear operator on $F^n$.
Suppose $X in F^n$ such that $(I-BA)X = 0$. We want to show that $X = 0$. Left-multiplying by $A$, we get $AX - AB(AX) = 0$. If we let $Y = AX$, then this just says that $(I - AB)Y = 0$. Since $I-AB$ is given to be invertible, this implies that $Y = 0$, that is, $AX = 0$. Now, from $(I-BA)X = 0$ we get $X = B(AX) = B0 = 0$. Hence proved.
answered Jan 3 at 18:30
BrahadeeshBrahadeesh
6,19742361
$endgroup$
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: Note that the two matrices have the same determinant.
There are several proofs of this, such as those given here.
Note that if either of $A$ and $B$ are invertible, we may further state that the two matrices are similar.
edited Apr 13 '17 at 12:21
Community♦
1
answered Nov 10 '14 at 18:29
OmnomnomnomOmnomnomnom
127k790178
$endgroup$
add a comment |
$begingroup$
Hint: Note that the two matrices have the same determinant.
There are several proofs of this, such as those given here.
Note that if either of $A$ and $B$ are invertible, we may further state that the two matrices are similar.
edited Apr 13 '17 at 12:21
Community♦
1
answered Nov 10 '14 at 18:29
OmnomnomnomOmnomnomnom
127k790178
$endgroup$
add a comment |
$begingroup$
Hint: Note that the two matrices have the same determinant.
There are several proofs of this, such as those given here.
Note that if either of $A$ and $B$ are invertible, we may further state that the two matrices are similar.
edited Apr 13 '17 at 12:21
Community♦
1
answered Nov 10 '14 at 18:29
OmnomnomnomOmnomnomnom
127k790178
$endgroup$
Hint: Note that the two matrices have the same determinant.
There are several proofs of this, such as those given here.
Note that if either of $A$ and $B$ are invertible, we may further state that the two matrices are similar.
edited Apr 13 '17 at 12:21
Community♦
1
answered Nov 10 '14 at 18:29
OmnomnomnomOmnomnomnom
127k790178
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
1
answered Nov 10 '14 at 18:29
OmnomnomnomOmnomnomnom
127k790178
answered Nov 10 '14 at 18:29
OmnomnomnomOmnomnomnom
127k790178
answered Nov 10 '14 at 18:29
OmnomnomnomOmnomnomnom
127k790178
127k790178
add a comment |
add a comment |
$begingroup$
Write $X = (I - AB)^{-1}$ and set $Y = I + BXA$. It is easy to check that $Y$ is the inverse of $I - BA$, but of course this solution will make you wonder how you come up with this in the first place.. see the mathoverflow question here.
edited Apr 13 '17 at 12:58
Community♦
1
answered Nov 10 '14 at 18:49
spinspin
3,82512257
$endgroup$
add a comment |
$begingroup$
Write $X = (I - AB)^{-1}$ and set $Y = I + BXA$. It is easy to check that $Y$ is the inverse of $I - BA$, but of course this solution will make you wonder how you come up with this in the first place.. see the mathoverflow question here.
edited Apr 13 '17 at 12:58
Community♦
1
answered Nov 10 '14 at 18:49
spinspin
3,82512257
$endgroup$
add a comment |
$begingroup$
Write $X = (I - AB)^{-1}$ and set $Y = I + BXA$. It is easy to check that $Y$ is the inverse of $I - BA$, but of course this solution will make you wonder how you come up with this in the first place.. see the mathoverflow question here.
edited Apr 13 '17 at 12:58
Community♦
1
answered Nov 10 '14 at 18:49
spinspin
3,82512257
$endgroup$
Write $X = (I - AB)^{-1}$ and set $Y = I + BXA$. It is easy to check that $Y$ is the inverse of $I - BA$, but of course this solution will make you wonder how you come up with this in the first place.. see the mathoverflow question here.
edited Apr 13 '17 at 12:58
Community♦
1
answered Nov 10 '14 at 18:49
spinspin
3,82512257
edited Apr 13 '17 at 12:58
Community♦
1
edited Apr 13 '17 at 12:58
Community♦
1
edited Apr 13 '17 at 12:58
Community♦
1
1
answered Nov 10 '14 at 18:49
spinspin
3,82512257
answered Nov 10 '14 at 18:49
spinspin
3,82512257
answered Nov 10 '14 at 18:49
spinspin
3,82512257
3,82512257
add a comment |
add a comment |
$begingroup$
A linear operator on a finite-dimensional vector space is invertible if and only if it is injective, and thus if and only if its kernel is trivial. An $n times n$ matrix over a field $F$ defines a linear operator on $F^n$.
Suppose $X in F^n$ such that $(I-BA)X = 0$. We want to show that $X = 0$. Left-multiplying by $A$, we get $AX - AB(AX) = 0$. If we let $Y = AX$, then this just says that $(I - AB)Y = 0$. Since $I-AB$ is given to be invertible, this implies that $Y = 0$, that is, $AX = 0$. Now, from $(I-BA)X = 0$ we get $X = B(AX) = B0 = 0$. Hence proved.
answered Jan 3 at 18:30
BrahadeeshBrahadeesh
6,19742361
$endgroup$
add a comment |
$begingroup$
A linear operator on a finite-dimensional vector space is invertible if and only if it is injective, and thus if and only if its kernel is trivial. An $n times n$ matrix over a field $F$ defines a linear operator on $F^n$.
Suppose $X in F^n$ such that $(I-BA)X = 0$. We want to show that $X = 0$. Left-multiplying by $A$, we get $AX - AB(AX) = 0$. If we let $Y = AX$, then this just says that $(I - AB)Y = 0$. Since $I-AB$ is given to be invertible, this implies that $Y = 0$, that is, $AX = 0$. Now, from $(I-BA)X = 0$ we get $X = B(AX) = B0 = 0$. Hence proved.
answered Jan 3 at 18:30
BrahadeeshBrahadeesh
6,19742361
$endgroup$
add a comment |
$begingroup$
A linear operator on a finite-dimensional vector space is invertible if and only if it is injective, and thus if and only if its kernel is trivial. An $n times n$ matrix over a field $F$ defines a linear operator on $F^n$.
Suppose $X in F^n$ such that $(I-BA)X = 0$. We want to show that $X = 0$. Left-multiplying by $A$, we get $AX - AB(AX) = 0$. If we let $Y = AX$, then this just says that $(I - AB)Y = 0$. Since $I-AB$ is given to be invertible, this implies that $Y = 0$, that is, $AX = 0$. Now, from $(I-BA)X = 0$ we get $X = B(AX) = B0 = 0$. Hence proved.
answered Jan 3 at 18:30
BrahadeeshBrahadeesh
6,19742361
$endgroup$
A linear operator on a finite-dimensional vector space is invertible if and only if it is injective, and thus if and only if its kernel is trivial. An $n times n$ matrix over a field $F$ defines a linear operator on $F^n$.
Suppose $X in F^n$ such that $(I-BA)X = 0$. We want to show that $X = 0$. Left-multiplying by $A$, we get $AX - AB(AX) = 0$. If we let $Y = AX$, then this just says that $(I - AB)Y = 0$. Since $I-AB$ is given to be invertible, this implies that $Y = 0$, that is, $AX = 0$. Now, from $(I-BA)X = 0$ we get $X = B(AX) = B0 = 0$. Hence proved.
answered Jan 3 at 18:30
BrahadeeshBrahadeesh
6,19742361
answered Jan 3 at 18:30
BrahadeeshBrahadeesh
6,19742361
answered Jan 3 at 18:30
BrahadeeshBrahadeesh
6,19742361
answered Jan 3 at 18:30
BrahadeeshBrahadeesh
6,19742361
6,19742361
add a comment |
add a comment |
4
$begingroup$
$A=left[begin{array}{cc}0&1\0&0end{array}right]$, $B=left[begin{array}{cc}0&0\0&1end{array}right]$ is a counterexample
$endgroup$
– Julian Rosen
Nov 10 '14 at 17:45
$begingroup$
we assume that I-AB is invertible
$endgroup$
– SSH
Nov 10 '14 at 18:06
1
$begingroup$
In this example $I-AB=left[begin{array}{cc}1&-1\0&1end{array}right]$ is invertible
$endgroup$
– Julian Rosen
Nov 10 '14 at 18:07
1
$begingroup$
Do you mean that either $A$ or $B$ is invertible? Because then the statement would be correct
$endgroup$
– Omnomnomnom
Nov 10 '14 at 18:12
1
$begingroup$
@Salem then Julian's counterexample applies, and there's nothing we can do. If the question is correct, then there must be some part of the question that you're not sharing with us.
$endgroup$
– Omnomnomnom
Nov 10 '14 at 18:17