The existence of a matrix?
up vote
2
down vote
favorite
Suppose that the matrix $Ain{mathbb{R}}^{ntimes r}$, $textrm{rank}(A)=r$, and $I_{n}$ is the identity matrix. Is there a matrix $Bin{mathbb{R}}^{rtimes n}$, such that
$$AB=I_{n}?$$
What is the requirement for the matrix $A$? Thank you!
linear-algebra matrices
edited Dec 4 at 8:13
Alex Silva
2,70331332
asked Dec 4 at 7:55
Wei Jiang
243
add a comment |
up vote
2
down vote
favorite
Suppose that the matrix $Ain{mathbb{R}}^{ntimes r}$, $textrm{rank}(A)=r$, and $I_{n}$ is the identity matrix. Is there a matrix $Bin{mathbb{R}}^{rtimes n}$, such that
$$AB=I_{n}?$$
What is the requirement for the matrix $A$? Thank you!
linear-algebra matrices
edited Dec 4 at 8:13
Alex Silva
2,70331332
asked Dec 4 at 7:55
Wei Jiang
243
1
You mean $rank(A)$?
– Patricio
Dec 4 at 8:02
2
@Yasmin, $A$ needs not be square, I think
– Patricio
Dec 4 at 8:03
1
@Yasmin, that's my point, $A$ is $n times r$ and $n$ and $r$ need not be equal to each other.
– Patricio
Dec 4 at 8:10
2
@Yasmin In case $n=r$, the question states that $rank(A)=n$, so it is automatically invertible. I guess the meat of the question is what happens if $n neq r$.
– lisyarus
Dec 4 at 10:41
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Suppose that the matrix $Ain{mathbb{R}}^{ntimes r}$, $textrm{rank}(A)=r$, and $I_{n}$ is the identity matrix. Is there a matrix $Bin{mathbb{R}}^{rtimes n}$, such that
$$AB=I_{n}?$$
What is the requirement for the matrix $A$? Thank you!
linear-algebra matrices
edited Dec 4 at 8:13
Alex Silva
2,70331332
asked Dec 4 at 7:55
Wei Jiang
243
Suppose that the matrix $Ain{mathbb{R}}^{ntimes r}$, $textrm{rank}(A)=r$, and $I_{n}$ is the identity matrix. Is there a matrix $Bin{mathbb{R}}^{rtimes n}$, such that
$$AB=I_{n}?$$
What is the requirement for the matrix $A$? Thank you!
linear-algebra matrices
linear-algebra matrices
edited Dec 4 at 8:13
Alex Silva
2,70331332
asked Dec 4 at 7:55
Wei Jiang
243
edited Dec 4 at 8:13
Alex Silva
2,70331332
asked Dec 4 at 7:55
Wei Jiang
243
edited Dec 4 at 8:13
Alex Silva
2,70331332
edited Dec 4 at 8:13
Alex Silva
2,70331332
edited Dec 4 at 8:13
Alex Silva
2,70331332
2,70331332
asked Dec 4 at 7:55
Wei Jiang
243
asked Dec 4 at 7:55
Wei Jiang
243
asked Dec 4 at 7:55
Wei Jiang
243
243
1
You mean $rank(A)$?
– Patricio
Dec 4 at 8:02
2
@Yasmin, $A$ needs not be square, I think
– Patricio
Dec 4 at 8:03
1
@Yasmin, that's my point, $A$ is $n times r$ and $n$ and $r$ need not be equal to each other.
– Patricio
Dec 4 at 8:10
2
@Yasmin In case $n=r$, the question states that $rank(A)=n$, so it is automatically invertible. I guess the meat of the question is what happens if $n neq r$.
– lisyarus
Dec 4 at 10:41
add a comment |
1
You mean $rank(A)$?
– Patricio
Dec 4 at 8:02
2
@Yasmin, $A$ needs not be square, I think
– Patricio
Dec 4 at 8:03
1
@Yasmin, that's my point, $A$ is $n times r$ and $n$ and $r$ need not be equal to each other.
– Patricio
Dec 4 at 8:10
2
@Yasmin In case $n=r$, the question states that $rank(A)=n$, so it is automatically invertible. I guess the meat of the question is what happens if $n neq r$.
– lisyarus
Dec 4 at 10:41
1
1
You mean $rank(A)$?
– Patricio
Dec 4 at 8:02
You mean $rank(A)$?
– Patricio
Dec 4 at 8:02
2
2
@Yasmin, $A$ needs not be square, I think
– Patricio
Dec 4 at 8:03
@Yasmin, $A$ needs not be square, I think
– Patricio
Dec 4 at 8:03
1
1
@Yasmin, that's my point, $A$ is $n times r$ and $n$ and $r$ need not be equal to each other.
– Patricio
Dec 4 at 8:10
@Yasmin, that's my point, $A$ is $n times r$ and $n$ and $r$ need not be equal to each other.
– Patricio
Dec 4 at 8:10
2
2
@Yasmin In case $n=r$, the question states that $rank(A)=n$, so it is automatically invertible. I guess the meat of the question is what happens if $n neq r$.
– lisyarus
Dec 4 at 10:41
@Yasmin In case $n=r$, the question states that $rank(A)=n$, so it is automatically invertible. I guess the meat of the question is what happens if $n neq r$.
– lisyarus
Dec 4 at 10:41
add a comment |
4 Answers
4
active
oldest
votes
up vote
6
down vote
From $n=rank (AB)leqmin{rank (A),rank (B)}$ and also $rank(A)leqmin{n,r}$ you can deduce something about the size of $A$ and its rank.
answered Dec 4 at 8:12
AnyAD
2,098812
add a comment |
up vote
6
down vote
Consider
$$
A=begin{bmatrix}1\0end{bmatrix}
$$
then
$$
AB=begin{bmatrix}1\0end{bmatrix}begin{bmatrix}b_1 & b_2end{bmatrix}=begin{bmatrix}b_1 & b_2\0 & 0end{bmatrix}.
$$
Do you see the trouble?
answered Dec 4 at 8:23
A.Γ.
21.7k22455
add a comment |
up vote
4
down vote
One of the solution for $B$ is the Moore-Penrose inverse $A^+$ of $A$.
Solution(s) exist as soon as $n le r$ and that $rank(A)=n$.
If $n = r$, $B$ is the inverse of $A$.
If $n < r$, the set of solutions corresponds to the Moore-Penrose inverse plus an element ($C$ such that $AC = 0$) owing to a set of dimension equal to $r-n$.
Look at the corresponding Wikipendia entry for example.
answered Dec 4 at 8:43
Damien
3974
The Moore-Penrose inverse is just one of many other possibilities for $B$.
– A.Γ.
Dec 4 at 8:46
1
@A.Γ. Of course. I was editing the answer when you commented it.
– Damien
Dec 4 at 8:50
OP specified that $operatorname{rank}(A)=r$, so this disqualifies cases with $r>n$.
– Federico Poloni
Dec 4 at 17:49
@FedericoPoloni The problem can only have a solution if $n le r$. I don't think that OP ignores this basic point. Therefore, I assumed it was a typo when OP writes $rank(A)=r$. It is also why in this answer, I insisted on this point in the second line, to avoid confusion.
– Damien
Dec 4 at 18:17
I think OP is asking precisely an exercise about this basic point.
– Federico Poloni
Dec 4 at 18:37
add a comment |
up vote
1
down vote
Yet another sufficient way of doing this. Since $rank(A) = r$, we have $rleq n$. Thinking in terms of columns of $A$ will help you see this. You picked $r$ vectors in a $n$-dimensional space. Since $rank(A)=r$, they are all linearly independent. Since you can have only $n$ independent vectors at most in a $n$-dimensional space, $rleq n$. In case $r=n$, then $B=A^{-1}$. In case $r<n$, $B$ doesn't exist.To see this, observe that determinant of RHS is $1$ ($neq 0$). Now determinant of LHS (i.e. AB) is zero as it is not a full-rank matrix. To see that, you can think of columns of $AB$ as a linear combination of columns $A$. Since you took $r$ independent vectors and constructed $n(>r)$ vectors out of it, they have to be dependent and hence $AB$ should have determinant zero.
answered Dec 4 at 13:00
dineshdileep
5,85711735
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
up vote
6
down vote
From $n=rank (AB)leqmin{rank (A),rank (B)}$ and also $rank(A)leqmin{n,r}$ you can deduce something about the size of $A$ and its rank.
answered Dec 4 at 8:12
AnyAD
2,098812
add a comment |
up vote
6
down vote
From $n=rank (AB)leqmin{rank (A),rank (B)}$ and also $rank(A)leqmin{n,r}$ you can deduce something about the size of $A$ and its rank.
answered Dec 4 at 8:12
AnyAD
2,098812
add a comment |
up vote
6
down vote
up vote
6
down vote
From $n=rank (AB)leqmin{rank (A),rank (B)}$ and also $rank(A)leqmin{n,r}$ you can deduce something about the size of $A$ and its rank.
answered Dec 4 at 8:12
AnyAD
2,098812
From $n=rank (AB)leqmin{rank (A),rank (B)}$ and also $rank(A)leqmin{n,r}$ you can deduce something about the size of $A$ and its rank.
answered Dec 4 at 8:12
AnyAD
2,098812
answered Dec 4 at 8:12
AnyAD
2,098812
answered Dec 4 at 8:12
AnyAD
2,098812
answered Dec 4 at 8:12
AnyAD
2,098812
2,098812
add a comment |
add a comment |
up vote
6
down vote
Consider
$$
A=begin{bmatrix}1\0end{bmatrix}
$$
then
$$
AB=begin{bmatrix}1\0end{bmatrix}begin{bmatrix}b_1 & b_2end{bmatrix}=begin{bmatrix}b_1 & b_2\0 & 0end{bmatrix}.
$$
Do you see the trouble?
answered Dec 4 at 8:23
A.Γ.
21.7k22455
add a comment |
up vote
6
down vote
Consider
$$
A=begin{bmatrix}1\0end{bmatrix}
$$
then
$$
AB=begin{bmatrix}1\0end{bmatrix}begin{bmatrix}b_1 & b_2end{bmatrix}=begin{bmatrix}b_1 & b_2\0 & 0end{bmatrix}.
$$
Do you see the trouble?
answered Dec 4 at 8:23
A.Γ.
21.7k22455
add a comment |
up vote
6
down vote
up vote
6
down vote
Consider
$$
A=begin{bmatrix}1\0end{bmatrix}
$$
then
$$
AB=begin{bmatrix}1\0end{bmatrix}begin{bmatrix}b_1 & b_2end{bmatrix}=begin{bmatrix}b_1 & b_2\0 & 0end{bmatrix}.
$$
Do you see the trouble?
answered Dec 4 at 8:23
A.Γ.
21.7k22455
Consider
$$
A=begin{bmatrix}1\0end{bmatrix}
$$
then
$$
AB=begin{bmatrix}1\0end{bmatrix}begin{bmatrix}b_1 & b_2end{bmatrix}=begin{bmatrix}b_1 & b_2\0 & 0end{bmatrix}.
$$
Do you see the trouble?
answered Dec 4 at 8:23
A.Γ.
21.7k22455
answered Dec 4 at 8:23
A.Γ.
21.7k22455
answered Dec 4 at 8:23
A.Γ.
21.7k22455
answered Dec 4 at 8:23
A.Γ.
21.7k22455
21.7k22455
add a comment |
add a comment |
up vote
4
down vote
One of the solution for $B$ is the Moore-Penrose inverse $A^+$ of $A$.
Solution(s) exist as soon as $n le r$ and that $rank(A)=n$.
If $n = r$, $B$ is the inverse of $A$.
If $n < r$, the set of solutions corresponds to the Moore-Penrose inverse plus an element ($C$ such that $AC = 0$) owing to a set of dimension equal to $r-n$.
Look at the corresponding Wikipendia entry for example.
answered Dec 4 at 8:43
Damien
3974
The Moore-Penrose inverse is just one of many other possibilities for $B$.
– A.Γ.
Dec 4 at 8:46
1
@A.Γ. Of course. I was editing the answer when you commented it.
– Damien
Dec 4 at 8:50
OP specified that $operatorname{rank}(A)=r$, so this disqualifies cases with $r>n$.
– Federico Poloni
Dec 4 at 17:49
@FedericoPoloni The problem can only have a solution if $n le r$. I don't think that OP ignores this basic point. Therefore, I assumed it was a typo when OP writes $rank(A)=r$. It is also why in this answer, I insisted on this point in the second line, to avoid confusion.
– Damien
Dec 4 at 18:17
I think OP is asking precisely an exercise about this basic point.
– Federico Poloni
Dec 4 at 18:37
add a comment |
up vote
4
down vote
One of the solution for $B$ is the Moore-Penrose inverse $A^+$ of $A$.
Solution(s) exist as soon as $n le r$ and that $rank(A)=n$.
If $n = r$, $B$ is the inverse of $A$.
If $n < r$, the set of solutions corresponds to the Moore-Penrose inverse plus an element ($C$ such that $AC = 0$) owing to a set of dimension equal to $r-n$.
Look at the corresponding Wikipendia entry for example.
answered Dec 4 at 8:43
Damien
3974
The Moore-Penrose inverse is just one of many other possibilities for $B$.
– A.Γ.
Dec 4 at 8:46
1
@A.Γ. Of course. I was editing the answer when you commented it.
– Damien
Dec 4 at 8:50
OP specified that $operatorname{rank}(A)=r$, so this disqualifies cases with $r>n$.
– Federico Poloni
Dec 4 at 17:49
@FedericoPoloni The problem can only have a solution if $n le r$. I don't think that OP ignores this basic point. Therefore, I assumed it was a typo when OP writes $rank(A)=r$. It is also why in this answer, I insisted on this point in the second line, to avoid confusion.
– Damien
Dec 4 at 18:17
I think OP is asking precisely an exercise about this basic point.
– Federico Poloni
Dec 4 at 18:37
add a comment |
up vote
4
down vote
up vote
4
down vote
One of the solution for $B$ is the Moore-Penrose inverse $A^+$ of $A$.
Solution(s) exist as soon as $n le r$ and that $rank(A)=n$.
If $n = r$, $B$ is the inverse of $A$.
If $n < r$, the set of solutions corresponds to the Moore-Penrose inverse plus an element ($C$ such that $AC = 0$) owing to a set of dimension equal to $r-n$.
Look at the corresponding Wikipendia entry for example.
answered Dec 4 at 8:43
Damien
3974
One of the solution for $B$ is the Moore-Penrose inverse $A^+$ of $A$.
Solution(s) exist as soon as $n le r$ and that $rank(A)=n$.
If $n = r$, $B$ is the inverse of $A$.
If $n < r$, the set of solutions corresponds to the Moore-Penrose inverse plus an element ($C$ such that $AC = 0$) owing to a set of dimension equal to $r-n$.
Look at the corresponding Wikipendia entry for example.
answered Dec 4 at 8:43
Damien
3974
edited Dec 4 at 9:02
answered Dec 4 at 8:43
Damien
3974
answered Dec 4 at 8:43
Damien
3974
answered Dec 4 at 8:43
Damien
3974
3974
The Moore-Penrose inverse is just one of many other possibilities for $B$.
– A.Γ.
Dec 4 at 8:46
1
@A.Γ. Of course. I was editing the answer when you commented it.
– Damien
Dec 4 at 8:50
OP specified that $operatorname{rank}(A)=r$, so this disqualifies cases with $r>n$.
– Federico Poloni
Dec 4 at 17:49
@FedericoPoloni The problem can only have a solution if $n le r$. I don't think that OP ignores this basic point. Therefore, I assumed it was a typo when OP writes $rank(A)=r$. It is also why in this answer, I insisted on this point in the second line, to avoid confusion.
– Damien
Dec 4 at 18:17
I think OP is asking precisely an exercise about this basic point.
– Federico Poloni
Dec 4 at 18:37
add a comment |
The Moore-Penrose inverse is just one of many other possibilities for $B$.
– A.Γ.
Dec 4 at 8:46
1
@A.Γ. Of course. I was editing the answer when you commented it.
– Damien
Dec 4 at 8:50
OP specified that $operatorname{rank}(A)=r$, so this disqualifies cases with $r>n$.
– Federico Poloni
Dec 4 at 17:49
@FedericoPoloni The problem can only have a solution if $n le r$. I don't think that OP ignores this basic point. Therefore, I assumed it was a typo when OP writes $rank(A)=r$. It is also why in this answer, I insisted on this point in the second line, to avoid confusion.
– Damien
Dec 4 at 18:17
I think OP is asking precisely an exercise about this basic point.
– Federico Poloni
Dec 4 at 18:37
The Moore-Penrose inverse is just one of many other possibilities for $B$.
– A.Γ.
Dec 4 at 8:46
The Moore-Penrose inverse is just one of many other possibilities for $B$.
– A.Γ.
Dec 4 at 8:46
1
1
@A.Γ. Of course. I was editing the answer when you commented it.
– Damien
Dec 4 at 8:50
@A.Γ. Of course. I was editing the answer when you commented it.
– Damien
Dec 4 at 8:50
OP specified that $operatorname{rank}(A)=r$, so this disqualifies cases with $r>n$.
– Federico Poloni
Dec 4 at 17:49
OP specified that $operatorname{rank}(A)=r$, so this disqualifies cases with $r>n$.
– Federico Poloni
Dec 4 at 17:49
@FedericoPoloni The problem can only have a solution if $n le r$. I don't think that OP ignores this basic point. Therefore, I assumed it was a typo when OP writes $rank(A)=r$. It is also why in this answer, I insisted on this point in the second line, to avoid confusion.
– Damien
Dec 4 at 18:17
@FedericoPoloni The problem can only have a solution if $n le r$. I don't think that OP ignores this basic point. Therefore, I assumed it was a typo when OP writes $rank(A)=r$. It is also why in this answer, I insisted on this point in the second line, to avoid confusion.
– Damien
Dec 4 at 18:17
I think OP is asking precisely an exercise about this basic point.
– Federico Poloni
Dec 4 at 18:37
I think OP is asking precisely an exercise about this basic point.
– Federico Poloni
Dec 4 at 18:37
add a comment |
up vote
1
down vote
Yet another sufficient way of doing this. Since $rank(A) = r$, we have $rleq n$. Thinking in terms of columns of $A$ will help you see this. You picked $r$ vectors in a $n$-dimensional space. Since $rank(A)=r$, they are all linearly independent. Since you can have only $n$ independent vectors at most in a $n$-dimensional space, $rleq n$. In case $r=n$, then $B=A^{-1}$. In case $r<n$, $B$ doesn't exist.To see this, observe that determinant of RHS is $1$ ($neq 0$). Now determinant of LHS (i.e. AB) is zero as it is not a full-rank matrix. To see that, you can think of columns of $AB$ as a linear combination of columns $A$. Since you took $r$ independent vectors and constructed $n(>r)$ vectors out of it, they have to be dependent and hence $AB$ should have determinant zero.
answered Dec 4 at 13:00
dineshdileep
5,85711735
add a comment |
up vote
1
down vote
Yet another sufficient way of doing this. Since $rank(A) = r$, we have $rleq n$. Thinking in terms of columns of $A$ will help you see this. You picked $r$ vectors in a $n$-dimensional space. Since $rank(A)=r$, they are all linearly independent. Since you can have only $n$ independent vectors at most in a $n$-dimensional space, $rleq n$. In case $r=n$, then $B=A^{-1}$. In case $r<n$, $B$ doesn't exist.To see this, observe that determinant of RHS is $1$ ($neq 0$). Now determinant of LHS (i.e. AB) is zero as it is not a full-rank matrix. To see that, you can think of columns of $AB$ as a linear combination of columns $A$. Since you took $r$ independent vectors and constructed $n(>r)$ vectors out of it, they have to be dependent and hence $AB$ should have determinant zero.
answered Dec 4 at 13:00
dineshdileep
5,85711735
add a comment |
up vote
1
down vote
up vote
1
down vote
Yet another sufficient way of doing this. Since $rank(A) = r$, we have $rleq n$. Thinking in terms of columns of $A$ will help you see this. You picked $r$ vectors in a $n$-dimensional space. Since $rank(A)=r$, they are all linearly independent. Since you can have only $n$ independent vectors at most in a $n$-dimensional space, $rleq n$. In case $r=n$, then $B=A^{-1}$. In case $r<n$, $B$ doesn't exist.To see this, observe that determinant of RHS is $1$ ($neq 0$). Now determinant of LHS (i.e. AB) is zero as it is not a full-rank matrix. To see that, you can think of columns of $AB$ as a linear combination of columns $A$. Since you took $r$ independent vectors and constructed $n(>r)$ vectors out of it, they have to be dependent and hence $AB$ should have determinant zero.
answered Dec 4 at 13:00
dineshdileep
5,85711735
Yet another sufficient way of doing this. Since $rank(A) = r$, we have $rleq n$. Thinking in terms of columns of $A$ will help you see this. You picked $r$ vectors in a $n$-dimensional space. Since $rank(A)=r$, they are all linearly independent. Since you can have only $n$ independent vectors at most in a $n$-dimensional space, $rleq n$. In case $r=n$, then $B=A^{-1}$. In case $r<n$, $B$ doesn't exist.To see this, observe that determinant of RHS is $1$ ($neq 0$). Now determinant of LHS (i.e. AB) is zero as it is not a full-rank matrix. To see that, you can think of columns of $AB$ as a linear combination of columns $A$. Since you took $r$ independent vectors and constructed $n(>r)$ vectors out of it, they have to be dependent and hence $AB$ should have determinant zero.
answered Dec 4 at 13:00
dineshdileep
5,85711735
edited Dec 4 at 13:08
answered Dec 4 at 13:00
dineshdileep
5,85711735
answered Dec 4 at 13:00
dineshdileep
5,85711735
answered Dec 4 at 13:00
dineshdileep
5,85711735
5,85711735
add a comment |
add a comment |
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1
You mean $rank(A)$?
– Patricio
Dec 4 at 8:02
2
@Yasmin, $A$ needs not be square, I think
– Patricio
Dec 4 at 8:03
1
@Yasmin, that's my point, $A$ is $n times r$ and $n$ and $r$ need not be equal to each other.
– Patricio
Dec 4 at 8:10
2
@Yasmin In case $n=r$, the question states that $rank(A)=n$, so it is automatically invertible. I guess the meat of the question is what happens if $n neq r$.
– lisyarus
Dec 4 at 10:41