Mixing
$A$ and $B$ are $3times 3$ real matrices such that $operatorname{rank}(AB)=1$, then $operatorname{rank}(BA$)...
$begingroup$
I was thinking about the problem that says:
If $A$ and $B$ are $3times 3$ real matrices such that $operatorname{rank}(AB)=1$, then $operatorname{rank}(BA)$ can not be which of the following?
(a) $0$
(b) $1$
(c) $2$
(d) $3$.
My attempt: I have chosen suitable $3 times 3 $ matrices for $A$ and $B$ keeping in mind that $operatorname{rank}(AB)=1$. Say for example if I take $A$ and $B$ to be
$$A = begin{pmatrix}
1 &2 &0 \
0 & 0 &0 \
0 & 0 &0
end{pmatrix}$$
and
$$B = begin{pmatrix}
-2 &1 &0 \
1 & 0 &0 \
0 & 0 &0
end{pmatrix}$$
respectively, then I see $operatorname{rank}(AB) = operatorname{rank}(BA) = 1$. So, option (b) can not be correct. Do I have to keep choosing the matrices and then observe which of the option holds good. Is this kind of approach right to tackle the problem? I am looking for a direct way (e.g. application to some theorem) which can give me the result. I have also noticed that $AB$ and $BA$ are similar matrices as we see that $A^{-1}(AB)A=BA$. Is this observation going to help me in any way? Thanks in advance for your time.
linear-algebra matrices
edited Dec 31 '12 at 15:25
TMM
9,12032848
asked Dec 31 '12 at 13:41
learnerlearner
3,39532268
$endgroup$
add a comment |
$begingroup$
I was thinking about the problem that says:
If $A$ and $B$ are $3times 3$ real matrices such that $operatorname{rank}(AB)=1$, then $operatorname{rank}(BA)$ can not be which of the following?
(a) $0$
(b) $1$
(c) $2$
(d) $3$.
My attempt: I have chosen suitable $3 times 3 $ matrices for $A$ and $B$ keeping in mind that $operatorname{rank}(AB)=1$. Say for example if I take $A$ and $B$ to be
$$A = begin{pmatrix}
1 &2 &0 \
0 & 0 &0 \
0 & 0 &0
end{pmatrix}$$
and
$$B = begin{pmatrix}
-2 &1 &0 \
1 & 0 &0 \
0 & 0 &0
end{pmatrix}$$
respectively, then I see $operatorname{rank}(AB) = operatorname{rank}(BA) = 1$. So, option (b) can not be correct. Do I have to keep choosing the matrices and then observe which of the option holds good. Is this kind of approach right to tackle the problem? I am looking for a direct way (e.g. application to some theorem) which can give me the result. I have also noticed that $AB$ and $BA$ are similar matrices as we see that $A^{-1}(AB)A=BA$. Is this observation going to help me in any way? Thanks in advance for your time.
linear-algebra matrices
edited Dec 31 '12 at 15:25
TMM
9,12032848
asked Dec 31 '12 at 13:41
learnerlearner
3,39532268
$endgroup$
$begingroup$
How do you know det$Aneq0$?
$endgroup$
– Sugata Adhya
Dec 31 '12 at 15:55
$begingroup$
Thanks for pointing out..Yes ,indeed det($A$) can be zero.
$endgroup$
– learner
Dec 31 '12 at 15:59
add a comment |
$begingroup$
I was thinking about the problem that says:
If $A$ and $B$ are $3times 3$ real matrices such that $operatorname{rank}(AB)=1$, then $operatorname{rank}(BA)$ can not be which of the following?
(a) $0$
(b) $1$
(c) $2$
(d) $3$.
My attempt: I have chosen suitable $3 times 3 $ matrices for $A$ and $B$ keeping in mind that $operatorname{rank}(AB)=1$. Say for example if I take $A$ and $B$ to be
$$A = begin{pmatrix}
1 &2 &0 \
0 & 0 &0 \
0 & 0 &0
end{pmatrix}$$
and
$$B = begin{pmatrix}
-2 &1 &0 \
1 & 0 &0 \
0 & 0 &0
end{pmatrix}$$
respectively, then I see $operatorname{rank}(AB) = operatorname{rank}(BA) = 1$. So, option (b) can not be correct. Do I have to keep choosing the matrices and then observe which of the option holds good. Is this kind of approach right to tackle the problem? I am looking for a direct way (e.g. application to some theorem) which can give me the result. I have also noticed that $AB$ and $BA$ are similar matrices as we see that $A^{-1}(AB)A=BA$. Is this observation going to help me in any way? Thanks in advance for your time.
linear-algebra matrices
edited Dec 31 '12 at 15:25
TMM
9,12032848
asked Dec 31 '12 at 13:41
learnerlearner
3,39532268
$endgroup$
I was thinking about the problem that says:
If $A$ and $B$ are $3times 3$ real matrices such that $operatorname{rank}(AB)=1$, then $operatorname{rank}(BA)$ can not be which of the following?
(a) $0$
(b) $1$
(c) $2$
(d) $3$.
My attempt: I have chosen suitable $3 times 3 $ matrices for $A$ and $B$ keeping in mind that $operatorname{rank}(AB)=1$. Say for example if I take $A$ and $B$ to be
$$A = begin{pmatrix}
1 &2 &0 \
0 & 0 &0 \
0 & 0 &0
end{pmatrix}$$
and
$$B = begin{pmatrix}
-2 &1 &0 \
1 & 0 &0 \
0 & 0 &0
end{pmatrix}$$
respectively, then I see $operatorname{rank}(AB) = operatorname{rank}(BA) = 1$. So, option (b) can not be correct. Do I have to keep choosing the matrices and then observe which of the option holds good. Is this kind of approach right to tackle the problem? I am looking for a direct way (e.g. application to some theorem) which can give me the result. I have also noticed that $AB$ and $BA$ are similar matrices as we see that $A^{-1}(AB)A=BA$. Is this observation going to help me in any way? Thanks in advance for your time.
linear-algebra matrices
linear-algebra matrices
edited Dec 31 '12 at 15:25
TMM
9,12032848
asked Dec 31 '12 at 13:41
learnerlearner
3,39532268
edited Dec 31 '12 at 15:25
TMM
9,12032848
asked Dec 31 '12 at 13:41
learnerlearner
3,39532268
edited Dec 31 '12 at 15:25
TMM
9,12032848
edited Dec 31 '12 at 15:25
TMM
9,12032848
edited Dec 31 '12 at 15:25
TMM
9,12032848
9,12032848
asked Dec 31 '12 at 13:41
learnerlearner
3,39532268
asked Dec 31 '12 at 13:41
learnerlearner
3,39532268
asked Dec 31 '12 at 13:41
learnerlearner
3,39532268
3,39532268
$begingroup$
How do you know det$Aneq0$?
$endgroup$
– Sugata Adhya
Dec 31 '12 at 15:55
$begingroup$
Thanks for pointing out..Yes ,indeed det($A$) can be zero.
$endgroup$
– learner
Dec 31 '12 at 15:59
add a comment |
$begingroup$
How do you know det$Aneq0$?
$endgroup$
– Sugata Adhya
Dec 31 '12 at 15:55
$begingroup$
Thanks for pointing out..Yes ,indeed det($A$) can be zero.
$endgroup$
– learner
Dec 31 '12 at 15:59
$begingroup$
How do you know det$Aneq0$?
$endgroup$
– Sugata Adhya
Dec 31 '12 at 15:55
$begingroup$
How do you know det$Aneq0$?
$endgroup$
– Sugata Adhya
Dec 31 '12 at 15:55
$begingroup$
Thanks for pointing out..Yes ,indeed det($A$) can be zero.
$endgroup$
– learner
Dec 31 '12 at 15:59
$begingroup$
Thanks for pointing out..Yes ,indeed det($A$) can be zero.
$endgroup$
– learner
Dec 31 '12 at 15:59
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
According to these properties of ranks listed on Wikiepdia :
http://en.wikipedia.org/wiki/Rank_(linear_algebra)
If $A$ is $m times n$
- $operatorname{rank}(AB) leq min(operatorname{rank}(A), operatorname{rank}(B))$ if $B$ is $n times k$
- $operatorname{rank}(AB) = operatorname{rank}(A)$ if $B$ is $n times k$ of rank $n$
Note that in your situation $m$, $n$ and $k$ is $3$
By property (1) the only way to have $operatorname{rank}(BA) = 3$ is to have both $operatorname{rank}(A) = 3$ and $operatorname{rank}(B) = 3$. Since $min(operatorname{rank}(A), operatorname{rank}(B)) = min(operatorname{rank}(B), operatorname{rank}(A))$.
But in that situation, we can apply property (2): $B$ is $3 times 3$ of rank $3$ and so $operatorname{rank}(AB) = operatorname{rank}(A) = 3$.
So you can't have at the same time $operatorname{rank}(BA) = 3$ and $operatorname{rank}(AB) neq 3$ using $3 times 3$ matrix only.
Thus answer (d) is correct.
edited Dec 31 '12 at 14:49
TMM
9,12032848
answered Dec 31 '12 at 13:52
almathiealmathie
1663
$endgroup$
$begingroup$
thanks a lot.I have got it.
$endgroup$
– learner
Dec 31 '12 at 14:31
add a comment |
$begingroup$
If you only want to solve the problem, then as the others have pointed out, (d) is impossible and hence the answer is (d). It is helpful, however, to constuct some examples to convince yourself that (a)-(c) are really possible (this is also a sanity check against misprints). For this purpose, you may consider the matrices
$$
A=begin{pmatrix}1\&1\&&0end{pmatrix}, B=begin{pmatrix}b&0&1\0&0&0\0&c&0end{pmatrix}.
$$
Then $operatorname{rank}(AB)=1$. Now you may try to choose some appropriate values of $b$ and $c$ to make $operatorname{rank}(BA)$ equal to $0,1$ or $2$.
answered Dec 31 '12 at 14:45
user1551user1551
72.5k566127
$endgroup$
add a comment |
$begingroup$
The answer is (d), i.e. if $rank(AB)=1$, then $rank(BA)$ cannot be $3$. Let's prove it by contradiction. Suppose $A, B$ are $3times 3$ matrices such that $rank(AB)=3$, then $AB$ is invertible. This implies that $A$ and $B$ are invertible (otherwise, if $A$ is not invertible, then $A$ could not be full rank, i.e. $rank(A)<3$. This would imply that $rank(AB)leq rank(A)<3$, which contradicts $rank(AB)=3$. Similarly you can prove that $B$ is invertible). Then $BA$ is invertible since product of invertible matrices is invertible. This implies that $rank(BA)=3$, which contradicts to the assumption that $rank(AB)=1$.
To see that (a), (b), (c) are possible, we can construct some examples of $A$ and $B$. As you have shown that (b) is possible (In fact, we can just take $A$ to be any rank 1 matrix, i.e. $rank(A)=1$ and $B$ to be the identity. Then $AB=BA=A$ has rank $1$).
I will leave other cases to you.
answered Dec 31 '12 at 14:13
PaulPaul
15.9k33767
$endgroup$
1
$begingroup$
i have one question.If i want to prove it by contradiction,then should not i keep the condition that $rank(AB)=1$ intact and then take the supposition that if possible let us assume that $rank(BA)=3$ and then finally see what happens? with regards..
$endgroup$
– learner
Dec 31 '12 at 14:26
$begingroup$
@Paul: Is (a) possible? For rank$(BA)=0implies BA=0implies (AB)^2=0implies$ rank$(AB)^2=0implies$ min{rank($AB$), rank($AB$)} $ =0implies$rank $(AB)=0.$
$endgroup$
– Sugata Adhya
Dec 31 '12 at 15:40
add a comment |
$begingroup$
This question has lots of good answers.But I just want to share a very short answer.It may already be discussed in the previous answers,so really sorry to answer again.
$AB$ is a $3times 3$ matrix,rank$(AB)=1ne3Rightarrow AB$ is not of full rank$Rightarrow operatorname{det}(AB)=0Rightarrow operatorname{det}(BA)=0Rightarrow BA$ is not of full rank$Rightarrow$rank$(BA)ne3$
answered Jun 12 '14 at 6:36
usermathusermath
2,5561227
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add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
According to these properties of ranks listed on Wikiepdia :
http://en.wikipedia.org/wiki/Rank_(linear_algebra)
If $A$ is $m times n$
- $operatorname{rank}(AB) leq min(operatorname{rank}(A), operatorname{rank}(B))$ if $B$ is $n times k$
- $operatorname{rank}(AB) = operatorname{rank}(A)$ if $B$ is $n times k$ of rank $n$
Note that in your situation $m$, $n$ and $k$ is $3$
By property (1) the only way to have $operatorname{rank}(BA) = 3$ is to have both $operatorname{rank}(A) = 3$ and $operatorname{rank}(B) = 3$. Since $min(operatorname{rank}(A), operatorname{rank}(B)) = min(operatorname{rank}(B), operatorname{rank}(A))$.
But in that situation, we can apply property (2): $B$ is $3 times 3$ of rank $3$ and so $operatorname{rank}(AB) = operatorname{rank}(A) = 3$.
So you can't have at the same time $operatorname{rank}(BA) = 3$ and $operatorname{rank}(AB) neq 3$ using $3 times 3$ matrix only.
Thus answer (d) is correct.
edited Dec 31 '12 at 14:49
TMM
9,12032848
answered Dec 31 '12 at 13:52
almathiealmathie
1663
$endgroup$
$begingroup$
thanks a lot.I have got it.
$endgroup$
– learner
Dec 31 '12 at 14:31
add a comment |
$begingroup$
According to these properties of ranks listed on Wikiepdia :
http://en.wikipedia.org/wiki/Rank_(linear_algebra)
If $A$ is $m times n$
- $operatorname{rank}(AB) leq min(operatorname{rank}(A), operatorname{rank}(B))$ if $B$ is $n times k$
- $operatorname{rank}(AB) = operatorname{rank}(A)$ if $B$ is $n times k$ of rank $n$
Note that in your situation $m$, $n$ and $k$ is $3$
By property (1) the only way to have $operatorname{rank}(BA) = 3$ is to have both $operatorname{rank}(A) = 3$ and $operatorname{rank}(B) = 3$. Since $min(operatorname{rank}(A), operatorname{rank}(B)) = min(operatorname{rank}(B), operatorname{rank}(A))$.
But in that situation, we can apply property (2): $B$ is $3 times 3$ of rank $3$ and so $operatorname{rank}(AB) = operatorname{rank}(A) = 3$.
So you can't have at the same time $operatorname{rank}(BA) = 3$ and $operatorname{rank}(AB) neq 3$ using $3 times 3$ matrix only.
Thus answer (d) is correct.
edited Dec 31 '12 at 14:49
TMM
9,12032848
answered Dec 31 '12 at 13:52
almathiealmathie
1663
$endgroup$
$begingroup$
thanks a lot.I have got it.
$endgroup$
– learner
Dec 31 '12 at 14:31
add a comment |
$begingroup$
According to these properties of ranks listed on Wikiepdia :
http://en.wikipedia.org/wiki/Rank_(linear_algebra)
If $A$ is $m times n$
- $operatorname{rank}(AB) leq min(operatorname{rank}(A), operatorname{rank}(B))$ if $B$ is $n times k$
- $operatorname{rank}(AB) = operatorname{rank}(A)$ if $B$ is $n times k$ of rank $n$
Note that in your situation $m$, $n$ and $k$ is $3$
By property (1) the only way to have $operatorname{rank}(BA) = 3$ is to have both $operatorname{rank}(A) = 3$ and $operatorname{rank}(B) = 3$. Since $min(operatorname{rank}(A), operatorname{rank}(B)) = min(operatorname{rank}(B), operatorname{rank}(A))$.
But in that situation, we can apply property (2): $B$ is $3 times 3$ of rank $3$ and so $operatorname{rank}(AB) = operatorname{rank}(A) = 3$.
So you can't have at the same time $operatorname{rank}(BA) = 3$ and $operatorname{rank}(AB) neq 3$ using $3 times 3$ matrix only.
Thus answer (d) is correct.
edited Dec 31 '12 at 14:49
TMM
9,12032848
answered Dec 31 '12 at 13:52
almathiealmathie
1663
$endgroup$
According to these properties of ranks listed on Wikiepdia :
http://en.wikipedia.org/wiki/Rank_(linear_algebra)
If $A$ is $m times n$
- $operatorname{rank}(AB) leq min(operatorname{rank}(A), operatorname{rank}(B))$ if $B$ is $n times k$
- $operatorname{rank}(AB) = operatorname{rank}(A)$ if $B$ is $n times k$ of rank $n$
Note that in your situation $m$, $n$ and $k$ is $3$
By property (1) the only way to have $operatorname{rank}(BA) = 3$ is to have both $operatorname{rank}(A) = 3$ and $operatorname{rank}(B) = 3$. Since $min(operatorname{rank}(A), operatorname{rank}(B)) = min(operatorname{rank}(B), operatorname{rank}(A))$.
But in that situation, we can apply property (2): $B$ is $3 times 3$ of rank $3$ and so $operatorname{rank}(AB) = operatorname{rank}(A) = 3$.
So you can't have at the same time $operatorname{rank}(BA) = 3$ and $operatorname{rank}(AB) neq 3$ using $3 times 3$ matrix only.
Thus answer (d) is correct.
edited Dec 31 '12 at 14:49
TMM
9,12032848
answered Dec 31 '12 at 13:52
almathiealmathie
1663
edited Dec 31 '12 at 14:49
TMM
9,12032848
edited Dec 31 '12 at 14:49
TMM
9,12032848
edited Dec 31 '12 at 14:49
TMM
9,12032848
9,12032848
answered Dec 31 '12 at 13:52
almathiealmathie
1663
answered Dec 31 '12 at 13:52
almathiealmathie
1663
answered Dec 31 '12 at 13:52
almathiealmathie
1663
1663
$begingroup$
thanks a lot.I have got it.
$endgroup$
– learner
Dec 31 '12 at 14:31
add a comment |
$begingroup$
thanks a lot.I have got it.
$endgroup$
– learner
Dec 31 '12 at 14:31
$begingroup$
thanks a lot.I have got it.
$endgroup$
– learner
Dec 31 '12 at 14:31
$begingroup$
thanks a lot.I have got it.
$endgroup$
– learner
Dec 31 '12 at 14:31
add a comment |
$begingroup$
If you only want to solve the problem, then as the others have pointed out, (d) is impossible and hence the answer is (d). It is helpful, however, to constuct some examples to convince yourself that (a)-(c) are really possible (this is also a sanity check against misprints). For this purpose, you may consider the matrices
$$
A=begin{pmatrix}1\&1\&&0end{pmatrix}, B=begin{pmatrix}b&0&1\0&0&0\0&c&0end{pmatrix}.
$$
Then $operatorname{rank}(AB)=1$. Now you may try to choose some appropriate values of $b$ and $c$ to make $operatorname{rank}(BA)$ equal to $0,1$ or $2$.
answered Dec 31 '12 at 14:45
user1551user1551
72.5k566127
$endgroup$
add a comment |
$begingroup$
If you only want to solve the problem, then as the others have pointed out, (d) is impossible and hence the answer is (d). It is helpful, however, to constuct some examples to convince yourself that (a)-(c) are really possible (this is also a sanity check against misprints). For this purpose, you may consider the matrices
$$
A=begin{pmatrix}1\&1\&&0end{pmatrix}, B=begin{pmatrix}b&0&1\0&0&0\0&c&0end{pmatrix}.
$$
Then $operatorname{rank}(AB)=1$. Now you may try to choose some appropriate values of $b$ and $c$ to make $operatorname{rank}(BA)$ equal to $0,1$ or $2$.
answered Dec 31 '12 at 14:45
user1551user1551
72.5k566127
$endgroup$
add a comment |
$begingroup$
If you only want to solve the problem, then as the others have pointed out, (d) is impossible and hence the answer is (d). It is helpful, however, to constuct some examples to convince yourself that (a)-(c) are really possible (this is also a sanity check against misprints). For this purpose, you may consider the matrices
$$
A=begin{pmatrix}1\&1\&&0end{pmatrix}, B=begin{pmatrix}b&0&1\0&0&0\0&c&0end{pmatrix}.
$$
Then $operatorname{rank}(AB)=1$. Now you may try to choose some appropriate values of $b$ and $c$ to make $operatorname{rank}(BA)$ equal to $0,1$ or $2$.
answered Dec 31 '12 at 14:45
user1551user1551
72.5k566127
$endgroup$
If you only want to solve the problem, then as the others have pointed out, (d) is impossible and hence the answer is (d). It is helpful, however, to constuct some examples to convince yourself that (a)-(c) are really possible (this is also a sanity check against misprints). For this purpose, you may consider the matrices
$$
A=begin{pmatrix}1\&1\&&0end{pmatrix}, B=begin{pmatrix}b&0&1\0&0&0\0&c&0end{pmatrix}.
$$
Then $operatorname{rank}(AB)=1$. Now you may try to choose some appropriate values of $b$ and $c$ to make $operatorname{rank}(BA)$ equal to $0,1$ or $2$.
answered Dec 31 '12 at 14:45
user1551user1551
72.5k566127
answered Dec 31 '12 at 14:45
user1551user1551
72.5k566127
answered Dec 31 '12 at 14:45
user1551user1551
72.5k566127
answered Dec 31 '12 at 14:45
user1551user1551
72.5k566127
72.5k566127
add a comment |
add a comment |
$begingroup$
The answer is (d), i.e. if $rank(AB)=1$, then $rank(BA)$ cannot be $3$. Let's prove it by contradiction. Suppose $A, B$ are $3times 3$ matrices such that $rank(AB)=3$, then $AB$ is invertible. This implies that $A$ and $B$ are invertible (otherwise, if $A$ is not invertible, then $A$ could not be full rank, i.e. $rank(A)<3$. This would imply that $rank(AB)leq rank(A)<3$, which contradicts $rank(AB)=3$. Similarly you can prove that $B$ is invertible). Then $BA$ is invertible since product of invertible matrices is invertible. This implies that $rank(BA)=3$, which contradicts to the assumption that $rank(AB)=1$.
To see that (a), (b), (c) are possible, we can construct some examples of $A$ and $B$. As you have shown that (b) is possible (In fact, we can just take $A$ to be any rank 1 matrix, i.e. $rank(A)=1$ and $B$ to be the identity. Then $AB=BA=A$ has rank $1$).
I will leave other cases to you.
answered Dec 31 '12 at 14:13
PaulPaul
15.9k33767
$endgroup$
1
$begingroup$
i have one question.If i want to prove it by contradiction,then should not i keep the condition that $rank(AB)=1$ intact and then take the supposition that if possible let us assume that $rank(BA)=3$ and then finally see what happens? with regards..
$endgroup$
– learner
Dec 31 '12 at 14:26
$begingroup$
@Paul: Is (a) possible? For rank$(BA)=0implies BA=0implies (AB)^2=0implies$ rank$(AB)^2=0implies$ min{rank($AB$), rank($AB$)} $ =0implies$rank $(AB)=0.$
$endgroup$
– Sugata Adhya
Dec 31 '12 at 15:40
add a comment |
$begingroup$
The answer is (d), i.e. if $rank(AB)=1$, then $rank(BA)$ cannot be $3$. Let's prove it by contradiction. Suppose $A, B$ are $3times 3$ matrices such that $rank(AB)=3$, then $AB$ is invertible. This implies that $A$ and $B$ are invertible (otherwise, if $A$ is not invertible, then $A$ could not be full rank, i.e. $rank(A)<3$. This would imply that $rank(AB)leq rank(A)<3$, which contradicts $rank(AB)=3$. Similarly you can prove that $B$ is invertible). Then $BA$ is invertible since product of invertible matrices is invertible. This implies that $rank(BA)=3$, which contradicts to the assumption that $rank(AB)=1$.
To see that (a), (b), (c) are possible, we can construct some examples of $A$ and $B$. As you have shown that (b) is possible (In fact, we can just take $A$ to be any rank 1 matrix, i.e. $rank(A)=1$ and $B$ to be the identity. Then $AB=BA=A$ has rank $1$).
I will leave other cases to you.
answered Dec 31 '12 at 14:13
PaulPaul
15.9k33767
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1
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i have one question.If i want to prove it by contradiction,then should not i keep the condition that $rank(AB)=1$ intact and then take the supposition that if possible let us assume that $rank(BA)=3$ and then finally see what happens? with regards..
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– learner
Dec 31 '12 at 14:26
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@Paul: Is (a) possible? For rank$(BA)=0implies BA=0implies (AB)^2=0implies$ rank$(AB)^2=0implies$ min{rank($AB$), rank($AB$)} $ =0implies$rank $(AB)=0.$
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– Sugata Adhya
Dec 31 '12 at 15:40
add a comment |
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The answer is (d), i.e. if $rank(AB)=1$, then $rank(BA)$ cannot be $3$. Let's prove it by contradiction. Suppose $A, B$ are $3times 3$ matrices such that $rank(AB)=3$, then $AB$ is invertible. This implies that $A$ and $B$ are invertible (otherwise, if $A$ is not invertible, then $A$ could not be full rank, i.e. $rank(A)<3$. This would imply that $rank(AB)leq rank(A)<3$, which contradicts $rank(AB)=3$. Similarly you can prove that $B$ is invertible). Then $BA$ is invertible since product of invertible matrices is invertible. This implies that $rank(BA)=3$, which contradicts to the assumption that $rank(AB)=1$.
To see that (a), (b), (c) are possible, we can construct some examples of $A$ and $B$. As you have shown that (b) is possible (In fact, we can just take $A$ to be any rank 1 matrix, i.e. $rank(A)=1$ and $B$ to be the identity. Then $AB=BA=A$ has rank $1$).
I will leave other cases to you.
answered Dec 31 '12 at 14:13
PaulPaul
15.9k33767
$endgroup$
The answer is (d), i.e. if $rank(AB)=1$, then $rank(BA)$ cannot be $3$. Let's prove it by contradiction. Suppose $A, B$ are $3times 3$ matrices such that $rank(AB)=3$, then $AB$ is invertible. This implies that $A$ and $B$ are invertible (otherwise, if $A$ is not invertible, then $A$ could not be full rank, i.e. $rank(A)<3$. This would imply that $rank(AB)leq rank(A)<3$, which contradicts $rank(AB)=3$. Similarly you can prove that $B$ is invertible). Then $BA$ is invertible since product of invertible matrices is invertible. This implies that $rank(BA)=3$, which contradicts to the assumption that $rank(AB)=1$.
To see that (a), (b), (c) are possible, we can construct some examples of $A$ and $B$. As you have shown that (b) is possible (In fact, we can just take $A$ to be any rank 1 matrix, i.e. $rank(A)=1$ and $B$ to be the identity. Then $AB=BA=A$ has rank $1$).
I will leave other cases to you.
answered Dec 31 '12 at 14:13
PaulPaul
15.9k33767
answered Dec 31 '12 at 14:13
PaulPaul
15.9k33767
answered Dec 31 '12 at 14:13
PaulPaul
15.9k33767
answered Dec 31 '12 at 14:13
PaulPaul
15.9k33767
15.9k33767
1
$begingroup$
i have one question.If i want to prove it by contradiction,then should not i keep the condition that $rank(AB)=1$ intact and then take the supposition that if possible let us assume that $rank(BA)=3$ and then finally see what happens? with regards..
$endgroup$
– learner
Dec 31 '12 at 14:26
$begingroup$
@Paul: Is (a) possible? For rank$(BA)=0implies BA=0implies (AB)^2=0implies$ rank$(AB)^2=0implies$ min{rank($AB$), rank($AB$)} $ =0implies$rank $(AB)=0.$
$endgroup$
– Sugata Adhya
Dec 31 '12 at 15:40
add a comment |
1
$begingroup$
i have one question.If i want to prove it by contradiction,then should not i keep the condition that $rank(AB)=1$ intact and then take the supposition that if possible let us assume that $rank(BA)=3$ and then finally see what happens? with regards..
$endgroup$
– learner
Dec 31 '12 at 14:26
$begingroup$
@Paul: Is (a) possible? For rank$(BA)=0implies BA=0implies (AB)^2=0implies$ rank$(AB)^2=0implies$ min{rank($AB$), rank($AB$)} $ =0implies$rank $(AB)=0.$
$endgroup$
– Sugata Adhya
Dec 31 '12 at 15:40
1
1
$begingroup$
i have one question.If i want to prove it by contradiction,then should not i keep the condition that $rank(AB)=1$ intact and then take the supposition that if possible let us assume that $rank(BA)=3$ and then finally see what happens? with regards..
$endgroup$
– learner
Dec 31 '12 at 14:26
$begingroup$
i have one question.If i want to prove it by contradiction,then should not i keep the condition that $rank(AB)=1$ intact and then take the supposition that if possible let us assume that $rank(BA)=3$ and then finally see what happens? with regards..
$endgroup$
– learner
Dec 31 '12 at 14:26
$begingroup$
@Paul: Is (a) possible? For rank$(BA)=0implies BA=0implies (AB)^2=0implies$ rank$(AB)^2=0implies$ min{rank($AB$), rank($AB$)} $ =0implies$rank $(AB)=0.$
$endgroup$
– Sugata Adhya
Dec 31 '12 at 15:40
$begingroup$
@Paul: Is (a) possible? For rank$(BA)=0implies BA=0implies (AB)^2=0implies$ rank$(AB)^2=0implies$ min{rank($AB$), rank($AB$)} $ =0implies$rank $(AB)=0.$
$endgroup$
– Sugata Adhya
Dec 31 '12 at 15:40
add a comment |
$begingroup$
This question has lots of good answers.But I just want to share a very short answer.It may already be discussed in the previous answers,so really sorry to answer again.
$AB$ is a $3times 3$ matrix,rank$(AB)=1ne3Rightarrow AB$ is not of full rank$Rightarrow operatorname{det}(AB)=0Rightarrow operatorname{det}(BA)=0Rightarrow BA$ is not of full rank$Rightarrow$rank$(BA)ne3$
answered Jun 12 '14 at 6:36
usermathusermath
2,5561227
$endgroup$
add a comment |
$begingroup$
This question has lots of good answers.But I just want to share a very short answer.It may already be discussed in the previous answers,so really sorry to answer again.
$AB$ is a $3times 3$ matrix,rank$(AB)=1ne3Rightarrow AB$ is not of full rank$Rightarrow operatorname{det}(AB)=0Rightarrow operatorname{det}(BA)=0Rightarrow BA$ is not of full rank$Rightarrow$rank$(BA)ne3$
answered Jun 12 '14 at 6:36
usermathusermath
2,5561227
$endgroup$
add a comment |
$begingroup$
This question has lots of good answers.But I just want to share a very short answer.It may already be discussed in the previous answers,so really sorry to answer again.
$AB$ is a $3times 3$ matrix,rank$(AB)=1ne3Rightarrow AB$ is not of full rank$Rightarrow operatorname{det}(AB)=0Rightarrow operatorname{det}(BA)=0Rightarrow BA$ is not of full rank$Rightarrow$rank$(BA)ne3$
answered Jun 12 '14 at 6:36
usermathusermath
2,5561227
$endgroup$
This question has lots of good answers.But I just want to share a very short answer.It may already be discussed in the previous answers,so really sorry to answer again.
$AB$ is a $3times 3$ matrix,rank$(AB)=1ne3Rightarrow AB$ is not of full rank$Rightarrow operatorname{det}(AB)=0Rightarrow operatorname{det}(BA)=0Rightarrow BA$ is not of full rank$Rightarrow$rank$(BA)ne3$
answered Jun 12 '14 at 6:36
usermathusermath
2,5561227
answered Jun 12 '14 at 6:36
usermathusermath
2,5561227
answered Jun 12 '14 at 6:36
usermathusermath
2,5561227
answered Jun 12 '14 at 6:36
usermathusermath
2,5561227
2,5561227
add a comment |
add a comment |
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$begingroup$
How do you know det$Aneq0$?
$endgroup$
– Sugata Adhya
Dec 31 '12 at 15:55
$begingroup$
Thanks for pointing out..Yes ,indeed det($A$) can be zero.
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– learner
Dec 31 '12 at 15:59