how to deduce two matrices $P$ and $Q$ are square if $PQ=I_n$ and $QP=I_m$ The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Which of the following matrices are diagonalizable over $mathbbC$polar decomposition for non square matricesUnderstanding dimension of matrixInteger matrices whose $m$-th power are identity matrixAre $10times 10$ matrices spanned by powers of a single matrix?If $A$ and $B$ are real matrices and $X,Y$ are are non-singular square matrices such that $XA=BY$Are these two matrices necessarily entrywise-nonnegative?Tensors and matrices multiplicationProve that a square matrix with integer entries can be decomposed into a product of square matrices with integer matrixesTwo invertible matrices
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how to deduce two matrices $P$ and $Q$ are square if $PQ=I_n$ and $QP=I_m$
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Which of the following matrices are diagonalizable over $mathbbC$polar decomposition for non square matricesUnderstanding dimension of matrixInteger matrices whose $m$-th power are identity matrixAre $10times 10$ matrices spanned by powers of a single matrix?If $A$ and $B$ are real matrices and $X,Y$ are are non-singular square matrices such that $XA=BY$Are these two matrices necessarily entrywise-nonnegative?Tensors and matrices multiplicationProve that a square matrix with integer entries can be decomposed into a product of square matrices with integer matrixesTwo invertible matrices
$begingroup$
If you had two matrices $P$ and $Q$ where $P$ is an $ntimes m$ matrix and $Q$ is an $m times n$ matrix both with integer entries satisfying:
$$PQ = I_n text and QP = I_m$$
How would you prove that $n=m$? I read somewhere that because $mathbbZ subset mathbbQ $ you can think of the matrices with entries over $mathbbQ$ so that makes $P=Q^-1$ but I don't understand how that conclusion has been made.
linear-algebra matrices free-abelian-group
edited Apr 8 at 14:49
TonyK
44.1k358137
asked Apr 8 at 13:09
BigWigBigWig
12511
$endgroup$
add a comment |
$begingroup$
If you had two matrices $P$ and $Q$ where $P$ is an $ntimes m$ matrix and $Q$ is an $m times n$ matrix both with integer entries satisfying:
$$PQ = I_n text and QP = I_m$$
How would you prove that $n=m$? I read somewhere that because $mathbbZ subset mathbbQ $ you can think of the matrices with entries over $mathbbQ$ so that makes $P=Q^-1$ but I don't understand how that conclusion has been made.
linear-algebra matrices free-abelian-group
edited Apr 8 at 14:49
TonyK
44.1k358137
asked Apr 8 at 13:09
BigWigBigWig
12511
$endgroup$
1
$begingroup$
The rank of $PQ$, $QP$ is at most max of the ranks for $P,Q$.
$endgroup$
– dan_fulea
Apr 8 at 13:15
1
$begingroup$
This has absolutely nothing to do with the fact that $Bbb ZsubsetBbb Q$.
$endgroup$
– TonyK
Apr 8 at 14:50
add a comment |
$begingroup$
If you had two matrices $P$ and $Q$ where $P$ is an $ntimes m$ matrix and $Q$ is an $m times n$ matrix both with integer entries satisfying:
$$PQ = I_n text and QP = I_m$$
How would you prove that $n=m$? I read somewhere that because $mathbbZ subset mathbbQ $ you can think of the matrices with entries over $mathbbQ$ so that makes $P=Q^-1$ but I don't understand how that conclusion has been made.
linear-algebra matrices free-abelian-group
edited Apr 8 at 14:49
TonyK
44.1k358137
asked Apr 8 at 13:09
BigWigBigWig
12511
$endgroup$
If you had two matrices $P$ and $Q$ where $P$ is an $ntimes m$ matrix and $Q$ is an $m times n$ matrix both with integer entries satisfying:
$$PQ = I_n text and QP = I_m$$
How would you prove that $n=m$? I read somewhere that because $mathbbZ subset mathbbQ $ you can think of the matrices with entries over $mathbbQ$ so that makes $P=Q^-1$ but I don't understand how that conclusion has been made.
linear-algebra matrices free-abelian-group
linear-algebra matrices free-abelian-group
edited Apr 8 at 14:49
TonyK
44.1k358137
asked Apr 8 at 13:09
BigWigBigWig
12511
edited Apr 8 at 14:49
TonyK
44.1k358137
asked Apr 8 at 13:09
BigWigBigWig
12511
edited Apr 8 at 14:49
TonyK
44.1k358137
edited Apr 8 at 14:49
TonyK
44.1k358137
edited Apr 8 at 14:49
TonyK
44.1k358137
44.1k358137
asked Apr 8 at 13:09
BigWigBigWig
12511
asked Apr 8 at 13:09
BigWigBigWig
12511
asked Apr 8 at 13:09
BigWigBigWig
12511
12511
1
$begingroup$
The rank of $PQ$, $QP$ is at most max of the ranks for $P,Q$.
$endgroup$
– dan_fulea
Apr 8 at 13:15
1
$begingroup$
This has absolutely nothing to do with the fact that $Bbb ZsubsetBbb Q$.
$endgroup$
– TonyK
Apr 8 at 14:50
add a comment |
1
$begingroup$
The rank of $PQ$, $QP$ is at most max of the ranks for $P,Q$.
$endgroup$
– dan_fulea
Apr 8 at 13:15
1
$begingroup$
This has absolutely nothing to do with the fact that $Bbb ZsubsetBbb Q$.
$endgroup$
– TonyK
Apr 8 at 14:50
1
1
$begingroup$
The rank of $PQ$, $QP$ is at most max of the ranks for $P,Q$.
$endgroup$
– dan_fulea
Apr 8 at 13:15
$begingroup$
The rank of $PQ$, $QP$ is at most max of the ranks for $P,Q$.
$endgroup$
– dan_fulea
Apr 8 at 13:15
1
1
$begingroup$
This has absolutely nothing to do with the fact that $Bbb ZsubsetBbb Q$.
$endgroup$
– TonyK
Apr 8 at 14:50
$begingroup$
This has absolutely nothing to do with the fact that $Bbb ZsubsetBbb Q$.
$endgroup$
– TonyK
Apr 8 at 14:50
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
One way is to see it as a rank thing: an operator cannot increase rank. So $PQ=I_n$ tells you that the ranks of $P$ and $Q$ are at least $n$; thus $mgeq n$. The other equality gives you $mleq n$.
Another, direct proof, can be obtained by taking the trace. The equality $operatornameTr(AB)=operatornameTr(BA)$ holds whenever $AB$ and $BA $ make sense.
Thus
$$
n=operatornameTr(I_n)=operatornameTr(PQ)=operatornameTr(QP)=operatornameTr(I_m)=m.
$$
answered Apr 8 at 17:57
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
$begingroup$
That's a fun one! I never thought that characteristic $0$ would help that much in this problem :)
$endgroup$
– darij grinberg
Apr 8 at 20:18
add a comment |
$begingroup$
One way to see this is by thinking of an $n times m$ matrix as a linear function $mathbbR^m to mathbbR^n$. If you let $S : mathbbR^m to mathbbR^n$ be the linear function whose matrix with respect to the standard basis is $P$ and $T : mathbbR^n to mathbbR^m$ be the linear function with matrix $Q$, then we see that $T circ S = textId_m$ and $S circ T = textId_n$ means that $T$ has a two-sided inverse. Thus, $S$ must be a bijection, and therefore a vector space isomorphism. So, we see that $mathbbR^m$ is isomorphic to $mathbbR^n$. Thus, it must be the case that $n = m$, so $S : mathbbR^m to mathbbR^m$ and its matrix form, $P$, must be an $m times m$ matrix. Therefore, $P$ (and likewise $Q$) are square matrices.
answered Apr 8 at 13:20
PamelloesPamelloes
1212
New contributor
Pamelloes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
All well and good. But I fear that this won't help the OP very much.
$endgroup$
– TonyK
Apr 8 at 14:53
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
One way is to see it as a rank thing: an operator cannot increase rank. So $PQ=I_n$ tells you that the ranks of $P$ and $Q$ are at least $n$; thus $mgeq n$. The other equality gives you $mleq n$.
Another, direct proof, can be obtained by taking the trace. The equality $operatornameTr(AB)=operatornameTr(BA)$ holds whenever $AB$ and $BA $ make sense.
Thus
$$
n=operatornameTr(I_n)=operatornameTr(PQ)=operatornameTr(QP)=operatornameTr(I_m)=m.
$$
answered Apr 8 at 17:57
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
$begingroup$
That's a fun one! I never thought that characteristic $0$ would help that much in this problem :)
$endgroup$
– darij grinberg
Apr 8 at 20:18
add a comment |
$begingroup$
One way is to see it as a rank thing: an operator cannot increase rank. So $PQ=I_n$ tells you that the ranks of $P$ and $Q$ are at least $n$; thus $mgeq n$. The other equality gives you $mleq n$.
Another, direct proof, can be obtained by taking the trace. The equality $operatornameTr(AB)=operatornameTr(BA)$ holds whenever $AB$ and $BA $ make sense.
Thus
$$
n=operatornameTr(I_n)=operatornameTr(PQ)=operatornameTr(QP)=operatornameTr(I_m)=m.
$$
answered Apr 8 at 17:57
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
$begingroup$
That's a fun one! I never thought that characteristic $0$ would help that much in this problem :)
$endgroup$
– darij grinberg
Apr 8 at 20:18
add a comment |
$begingroup$
One way is to see it as a rank thing: an operator cannot increase rank. So $PQ=I_n$ tells you that the ranks of $P$ and $Q$ are at least $n$; thus $mgeq n$. The other equality gives you $mleq n$.
Another, direct proof, can be obtained by taking the trace. The equality $operatornameTr(AB)=operatornameTr(BA)$ holds whenever $AB$ and $BA $ make sense.
Thus
$$
n=operatornameTr(I_n)=operatornameTr(PQ)=operatornameTr(QP)=operatornameTr(I_m)=m.
$$
answered Apr 8 at 17:57
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
One way is to see it as a rank thing: an operator cannot increase rank. So $PQ=I_n$ tells you that the ranks of $P$ and $Q$ are at least $n$; thus $mgeq n$. The other equality gives you $mleq n$.
Another, direct proof, can be obtained by taking the trace. The equality $operatornameTr(AB)=operatornameTr(BA)$ holds whenever $AB$ and $BA $ make sense.
Thus
$$
n=operatornameTr(I_n)=operatornameTr(PQ)=operatornameTr(QP)=operatornameTr(I_m)=m.
$$
answered Apr 8 at 17:57
Martin ArgeramiMartin Argerami
129k1184185
answered Apr 8 at 17:57
Martin ArgeramiMartin Argerami
129k1184185
answered Apr 8 at 17:57
Martin ArgeramiMartin Argerami
129k1184185
answered Apr 8 at 17:57
Martin ArgeramiMartin Argerami
129k1184185
129k1184185
$begingroup$
That's a fun one! I never thought that characteristic $0$ would help that much in this problem :)
$endgroup$
– darij grinberg
Apr 8 at 20:18
add a comment |
$begingroup$
That's a fun one! I never thought that characteristic $0$ would help that much in this problem :)
$endgroup$
– darij grinberg
Apr 8 at 20:18
$begingroup$
That's a fun one! I never thought that characteristic $0$ would help that much in this problem :)
$endgroup$
– darij grinberg
Apr 8 at 20:18
$begingroup$
That's a fun one! I never thought that characteristic $0$ would help that much in this problem :)
$endgroup$
– darij grinberg
Apr 8 at 20:18
add a comment |
$begingroup$
One way to see this is by thinking of an $n times m$ matrix as a linear function $mathbbR^m to mathbbR^n$. If you let $S : mathbbR^m to mathbbR^n$ be the linear function whose matrix with respect to the standard basis is $P$ and $T : mathbbR^n to mathbbR^m$ be the linear function with matrix $Q$, then we see that $T circ S = textId_m$ and $S circ T = textId_n$ means that $T$ has a two-sided inverse. Thus, $S$ must be a bijection, and therefore a vector space isomorphism. So, we see that $mathbbR^m$ is isomorphic to $mathbbR^n$. Thus, it must be the case that $n = m$, so $S : mathbbR^m to mathbbR^m$ and its matrix form, $P$, must be an $m times m$ matrix. Therefore, $P$ (and likewise $Q$) are square matrices.
answered Apr 8 at 13:20
PamelloesPamelloes
1212
New contributor
Pamelloes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
All well and good. But I fear that this won't help the OP very much.
$endgroup$
– TonyK
Apr 8 at 14:53
add a comment |
$begingroup$
One way to see this is by thinking of an $n times m$ matrix as a linear function $mathbbR^m to mathbbR^n$. If you let $S : mathbbR^m to mathbbR^n$ be the linear function whose matrix with respect to the standard basis is $P$ and $T : mathbbR^n to mathbbR^m$ be the linear function with matrix $Q$, then we see that $T circ S = textId_m$ and $S circ T = textId_n$ means that $T$ has a two-sided inverse. Thus, $S$ must be a bijection, and therefore a vector space isomorphism. So, we see that $mathbbR^m$ is isomorphic to $mathbbR^n$. Thus, it must be the case that $n = m$, so $S : mathbbR^m to mathbbR^m$ and its matrix form, $P$, must be an $m times m$ matrix. Therefore, $P$ (and likewise $Q$) are square matrices.
answered Apr 8 at 13:20
PamelloesPamelloes
1212
New contributor
Pamelloes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
All well and good. But I fear that this won't help the OP very much.
$endgroup$
– TonyK
Apr 8 at 14:53
add a comment |
$begingroup$
One way to see this is by thinking of an $n times m$ matrix as a linear function $mathbbR^m to mathbbR^n$. If you let $S : mathbbR^m to mathbbR^n$ be the linear function whose matrix with respect to the standard basis is $P$ and $T : mathbbR^n to mathbbR^m$ be the linear function with matrix $Q$, then we see that $T circ S = textId_m$ and $S circ T = textId_n$ means that $T$ has a two-sided inverse. Thus, $S$ must be a bijection, and therefore a vector space isomorphism. So, we see that $mathbbR^m$ is isomorphic to $mathbbR^n$. Thus, it must be the case that $n = m$, so $S : mathbbR^m to mathbbR^m$ and its matrix form, $P$, must be an $m times m$ matrix. Therefore, $P$ (and likewise $Q$) are square matrices.
answered Apr 8 at 13:20
PamelloesPamelloes
1212
New contributor
Pamelloes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
One way to see this is by thinking of an $n times m$ matrix as a linear function $mathbbR^m to mathbbR^n$. If you let $S : mathbbR^m to mathbbR^n$ be the linear function whose matrix with respect to the standard basis is $P$ and $T : mathbbR^n to mathbbR^m$ be the linear function with matrix $Q$, then we see that $T circ S = textId_m$ and $S circ T = textId_n$ means that $T$ has a two-sided inverse. Thus, $S$ must be a bijection, and therefore a vector space isomorphism. So, we see that $mathbbR^m$ is isomorphic to $mathbbR^n$. Thus, it must be the case that $n = m$, so $S : mathbbR^m to mathbbR^m$ and its matrix form, $P$, must be an $m times m$ matrix. Therefore, $P$ (and likewise $Q$) are square matrices.
answered Apr 8 at 13:20
PamelloesPamelloes
1212
New contributor
Pamelloes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Apr 8 at 13:20
PamelloesPamelloes
1212
New contributor
Pamelloes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Apr 8 at 13:20
PamelloesPamelloes
1212
answered Apr 8 at 13:20
PamelloesPamelloes
1212
1212
New contributor
Pamelloes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Pamelloes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Pamelloes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
All well and good. But I fear that this won't help the OP very much.
$endgroup$
– TonyK
Apr 8 at 14:53
add a comment |
$begingroup$
All well and good. But I fear that this won't help the OP very much.
$endgroup$
– TonyK
Apr 8 at 14:53
$begingroup$
All well and good. But I fear that this won't help the OP very much.
$endgroup$
– TonyK
Apr 8 at 14:53
$begingroup$
All well and good. But I fear that this won't help the OP very much.
$endgroup$
– TonyK
Apr 8 at 14:53
add a comment |
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$begingroup$
The rank of $PQ$, $QP$ is at most max of the ranks for $P,Q$.
$endgroup$
– dan_fulea
Apr 8 at 13:15
1
$begingroup$
This has absolutely nothing to do with the fact that $Bbb ZsubsetBbb Q$.
$endgroup$
– TonyK
Apr 8 at 14:50