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Linear algebra question with rank-1 matrices
Clash Royale CLAN TAG#URR8PPP
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Suppose I have a column vector $v in mathbbR^n$ and I want to find solutions $X in mathbbR^ntimes n$ such that
$$
X^T v v^T X = v v^T
$$
The solutions I found are
$$
X = fracv v^Tv^T v, qquad X = pm I
$$
However, suppose I have two column vectors $u,v in mathbbR^n$ which may be assumed to be orthogonal ($u^T v = 0$). Then do there exist solutions $X$ to
$$
X^T left( u u^T + v v^T right) X = u u^T + v v^T
$$
Do any solutions exist besides the trivial $X = pm I$?
linear-algebra linear-transformations matrix-rank
asked Jul 30 at 21:21
vibe
1388
 |Â
show 8 more comments
up vote
1
down vote
favorite
Suppose I have a column vector $v in mathbbR^n$ and I want to find solutions $X in mathbbR^ntimes n$ such that
$$
X^T v v^T X = v v^T
$$
The solutions I found are
$$
X = fracv v^Tv^T v, qquad X = pm I
$$
However, suppose I have two column vectors $u,v in mathbbR^n$ which may be assumed to be orthogonal ($u^T v = 0$). Then do there exist solutions $X$ to
$$
X^T left( u u^T + v v^T right) X = u u^T + v v^T
$$
Do any solutions exist besides the trivial $X = pm I$?
linear-algebra linear-transformations matrix-rank
asked Jul 30 at 21:21
vibe
1388
I don't understand how you obtain $$X^T v v^T X = v v^Timplies X=pm I$$
– gimusi
Jul 30 at 21:25
@gimusi The OP meant that $X=pm I$ are solutions to $X^top v v^top X=v v^top$.
– Batominovski
Jul 30 at 21:25
@Batominovski But it is not the only solution I guess, we can take also $X^T=$ a projection matrix onto $v$.
– gimusi
Jul 30 at 21:27
@gimusi The OP never claimed that whatever he found were the only solutions. Plus, if $X$ is a solution, then $X+N^top$ is a solution, where $N$ is a matrix with $vinker(N)$.
– Batominovski
Jul 30 at 21:29
@Batominovski: Since $u u^T + v v^T$ is rank 2 and therefore not invertible for $n > 2$, does this mean I can't write down a closed form expression for its pseudo inverse?
– vibe
Jul 30 at 21:29
 |Â
show 8 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Suppose I have a column vector $v in mathbbR^n$ and I want to find solutions $X in mathbbR^ntimes n$ such that
$$
X^T v v^T X = v v^T
$$
The solutions I found are
$$
X = fracv v^Tv^T v, qquad X = pm I
$$
However, suppose I have two column vectors $u,v in mathbbR^n$ which may be assumed to be orthogonal ($u^T v = 0$). Then do there exist solutions $X$ to
$$
X^T left( u u^T + v v^T right) X = u u^T + v v^T
$$
Do any solutions exist besides the trivial $X = pm I$?
linear-algebra linear-transformations matrix-rank
asked Jul 30 at 21:21
vibe
1388
Suppose I have a column vector $v in mathbbR^n$ and I want to find solutions $X in mathbbR^ntimes n$ such that
$$
X^T v v^T X = v v^T
$$
The solutions I found are
$$
X = fracv v^Tv^T v, qquad X = pm I
$$
However, suppose I have two column vectors $u,v in mathbbR^n$ which may be assumed to be orthogonal ($u^T v = 0$). Then do there exist solutions $X$ to
$$
X^T left( u u^T + v v^T right) X = u u^T + v v^T
$$
Do any solutions exist besides the trivial $X = pm I$?
linear-algebra linear-transformations matrix-rank
asked Jul 30 at 21:21
vibe
1388
asked Jul 30 at 21:21
vibe
1388
asked Jul 30 at 21:21
vibe
1388
asked Jul 30 at 21:21
vibe
1388
1388
I don't understand how you obtain $$X^T v v^T X = v v^Timplies X=pm I$$
– gimusi
Jul 30 at 21:25
@gimusi The OP meant that $X=pm I$ are solutions to $X^top v v^top X=v v^top$.
– Batominovski
Jul 30 at 21:25
@Batominovski But it is not the only solution I guess, we can take also $X^T=$ a projection matrix onto $v$.
– gimusi
Jul 30 at 21:27
@gimusi The OP never claimed that whatever he found were the only solutions. Plus, if $X$ is a solution, then $X+N^top$ is a solution, where $N$ is a matrix with $vinker(N)$.
– Batominovski
Jul 30 at 21:29
@Batominovski: Since $u u^T + v v^T$ is rank 2 and therefore not invertible for $n > 2$, does this mean I can't write down a closed form expression for its pseudo inverse?
– vibe
Jul 30 at 21:29
 |Â
show 8 more comments
I don't understand how you obtain $$X^T v v^T X = v v^Timplies X=pm I$$
– gimusi
Jul 30 at 21:25
@gimusi The OP meant that $X=pm I$ are solutions to $X^top v v^top X=v v^top$.
– Batominovski
Jul 30 at 21:25
@Batominovski But it is not the only solution I guess, we can take also $X^T=$ a projection matrix onto $v$.
– gimusi
Jul 30 at 21:27
@gimusi The OP never claimed that whatever he found were the only solutions. Plus, if $X$ is a solution, then $X+N^top$ is a solution, where $N$ is a matrix with $vinker(N)$.
– Batominovski
Jul 30 at 21:29
@Batominovski: Since $u u^T + v v^T$ is rank 2 and therefore not invertible for $n > 2$, does this mean I can't write down a closed form expression for its pseudo inverse?
– vibe
Jul 30 at 21:29
I don't understand how you obtain $$X^T v v^T X = v v^Timplies X=pm I$$
– gimusi
Jul 30 at 21:25
I don't understand how you obtain $$X^T v v^T X = v v^Timplies X=pm I$$
– gimusi
Jul 30 at 21:25
@gimusi The OP meant that $X=pm I$ are solutions to $X^top v v^top X=v v^top$.
– Batominovski
Jul 30 at 21:25
@gimusi The OP meant that $X=pm I$ are solutions to $X^top v v^top X=v v^top$.
– Batominovski
Jul 30 at 21:25
@Batominovski But it is not the only solution I guess, we can take also $X^T=$ a projection matrix onto $v$.
– gimusi
Jul 30 at 21:27
@Batominovski But it is not the only solution I guess, we can take also $X^T=$ a projection matrix onto $v$.
– gimusi
Jul 30 at 21:27
@gimusi The OP never claimed that whatever he found were the only solutions. Plus, if $X$ is a solution, then $X+N^top$ is a solution, where $N$ is a matrix with $vinker(N)$.
– Batominovski
Jul 30 at 21:29
@gimusi The OP never claimed that whatever he found were the only solutions. Plus, if $X$ is a solution, then $X+N^top$ is a solution, where $N$ is a matrix with $vinker(N)$.
– Batominovski
Jul 30 at 21:29
@Batominovski: Since $u u^T + v v^T$ is rank 2 and therefore not invertible for $n > 2$, does this mean I can't write down a closed form expression for its pseudo inverse?
– vibe
Jul 30 at 21:29
@Batominovski: Since $u u^T + v v^T$ is rank 2 and therefore not invertible for $n > 2$, does this mean I can't write down a closed form expression for its pseudo inverse?
– vibe
Jul 30 at 21:29
 |Â
show 8 more comments
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
Set
$$A:=fracu,u^topu^top u+fracv,v^topv^top v,.$$
Because $u^top ,v=0$ and $v^top,u=0$, as $u$ and $v$ are orthogonal, we conclude that $X=A$ is a solution to
$$X^top,left(u,u^top+v,v^topright),X=u,u^top+v,v^top,.$$
If $N$ is an $n$-by-$n$ matrix such that $u$ and $v$ are in $ker(N)$, then $X=A+N^top$ is also a solution. For $ngeq 3$, there are infinitely many such $N$. Furthermore, $X=I+N^top$ is also a nontrivial solution.
For a fixed positive integer $mleq n$, let $u_1,u_2,ldots,u_m$ be (not necessarily orthogonal) linearly independent elements of $mathbbR^n$. Write $$J:=sum_j=1^m,u_j,u_j^top,.$$
I shall find all solutions $XintextMat_ntimes n(mathbbR)$ such that
$$X^top , J ,X=J,.$$
Let $e_1,e_2,ldots,e_n$ be the standard basis vectors of $mathbbR^n$. Choose an invertible matrix $VintextGL_n(mathbbR)$ such that $V,u_j=e_j$ for $j=1,2,ldots,m$. Define the matrix $K$ to be $$K:=sum_j=1^m,e_j,e_j^top=V,left(sum_j=1^m,u_j,u_j^topright),V^top=V,J,V^top,.$$
Setting $Y:=V^-1,X,V^top$, we see that $X^top,J,X=J$ if and only if $$Y^top,K,Y=K,.$$
Write
$$Y=beginbmatrixP&Q\R&Sendbmatrix,,$$
where $PintextMat_mtimes m(mathbbR)$, $QintextMat_mtimes (n-m)(mathbbR)$, $RintextMat_(n-m)times m(mathbbR)$, and $SintextMat_(n-m)times(n-m)(mathbbR)$. Also, $$K=beginbmatrixI_mtimes m&0_mtimes (n-m)\0_(n-m)times m&0_mtimes mendbmatrix,,$$
where $I_ktimes k$ is the $k$-by-$k$ identity matrix and $0_rtimes s$ is the $r$-by-$s$ zero matrix.
Thus, the condition $Y^top,K,Y=K$ means that
$$beginalign
beginbmatrixI_mtimes m&0_mtimes (n-m)\0_(n-m)times m&0_mtimes mendbmatrix
&=K=Y^top,K,Y
\
&=
beginbmatrix P^top& R^top\Q^top &S^topendbmatrixbeginbmatrixI_mtimes m&0_mtimes (n-m)\0_(n-m)times m&0_mtimes mendbmatrixbeginbmatrixP&Q\ R&Sendbmatrix
\
&=beginbmatrix P^top& R^top\Q^top &S^topendbmatrixbeginbmatrix P&Q\ 0_(n-m)times m&0_mtimes mendbmatrix=beginbmatrixP^top, P&P^top, Q\ Q^top, P &Q^top ,Qendbmatrix
endalign,.$$
That is, $Q^top, Q=0_(n-m)times (n-m)$. This shows that $Q=0_(n-m)times (n-m)$. Furthermore, we have
$$P^top,P =I_mtimes mtext or Pin O_m(mathbbR),.$$
There are no other conditions on $R$ and $S$.
Since $X=V,Y,left(V^-1right)^top$, we conclude that all solutions $X$ to $X^top,J,X=J$ take the form
$$V,beginbmatrixP&0_mtimes(n-m)\R&Sendbmatrix,left(V^-1right)^top,,$$
where $Pin O_m(mathbbR)$ is a real orthogonal matrix, $RintextMat_(n-m)times n(mathbbR)$, and $SintextMat_(n-m)times (n-m)(mathbbR)$.
The solution set is therefore a smooth real manifold, which is diffeomorphic to
$$O_m(mathbbR)times mathbbR^m(n-m)+(n-m)^2,.$$
Hence, the solution set is of dimension $$fracm(m-1)2+m(n-m)+(n-m)^2=n(n-m)+fracm(m-1)2geq fracn(n-1)2,.$$
There are therefore infinitely many nontrivial solutions when $ngeq2$.
answered Jul 30 at 21:43
Batominovski
22.8k22776
Then can we conclude that all the solution are the projection matrix onto span(u,v)?
– gimusi
Jul 30 at 21:50
This is great, and would generalize nicely to any number of rank-1 matrices. I am also curious what other solutions could exist.
– vibe
Jul 30 at 21:59
Interesting solution. Since in this case the $u_j$ are orthogonal, I guess you can choose $V = I$ to simplify the solution a little.
– vibe
Jul 31 at 17:26
Also in the case $m = 1$, we know that $X = u u^T / (u^T u)$ is a solution, but this would not necessarily have $0$ in the upper right block of the matrix?
– vibe
Jul 31 at 17:29
You can't always choose $V=I$ even when the $u_j$'s are orthogonal. Work on the small cases to see that $V=I$ doesn't usually work.
– Batominovski
Jul 31 at 17:32
add a comment |Â
up vote
0
down vote
With reference to
$$X^T v v^T X = v v^T$$
assuming $X^T=fracvv^T$ that is a projection matrix onto $operatornamespan(v)$ we have
$$X^T v v^T X = fracvv^Tv v^Tfracvv^T=vv^T$$
For
$$X^T left( u u^T + v v^T right) X = u u^T + v v^T$$
we can assume $X^T$ as a projection matrix onto $operatornamespan(v,u)$.
answered Jul 30 at 21:36
gimusi
64.1k73480
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Set
$$A:=fracu,u^topu^top u+fracv,v^topv^top v,.$$
Because $u^top ,v=0$ and $v^top,u=0$, as $u$ and $v$ are orthogonal, we conclude that $X=A$ is a solution to
$$X^top,left(u,u^top+v,v^topright),X=u,u^top+v,v^top,.$$
If $N$ is an $n$-by-$n$ matrix such that $u$ and $v$ are in $ker(N)$, then $X=A+N^top$ is also a solution. For $ngeq 3$, there are infinitely many such $N$. Furthermore, $X=I+N^top$ is also a nontrivial solution.
For a fixed positive integer $mleq n$, let $u_1,u_2,ldots,u_m$ be (not necessarily orthogonal) linearly independent elements of $mathbbR^n$. Write $$J:=sum_j=1^m,u_j,u_j^top,.$$
I shall find all solutions $XintextMat_ntimes n(mathbbR)$ such that
$$X^top , J ,X=J,.$$
Let $e_1,e_2,ldots,e_n$ be the standard basis vectors of $mathbbR^n$. Choose an invertible matrix $VintextGL_n(mathbbR)$ such that $V,u_j=e_j$ for $j=1,2,ldots,m$. Define the matrix $K$ to be $$K:=sum_j=1^m,e_j,e_j^top=V,left(sum_j=1^m,u_j,u_j^topright),V^top=V,J,V^top,.$$
Setting $Y:=V^-1,X,V^top$, we see that $X^top,J,X=J$ if and only if $$Y^top,K,Y=K,.$$
Write
$$Y=beginbmatrixP&Q\R&Sendbmatrix,,$$
where $PintextMat_mtimes m(mathbbR)$, $QintextMat_mtimes (n-m)(mathbbR)$, $RintextMat_(n-m)times m(mathbbR)$, and $SintextMat_(n-m)times(n-m)(mathbbR)$. Also, $$K=beginbmatrixI_mtimes m&0_mtimes (n-m)\0_(n-m)times m&0_mtimes mendbmatrix,,$$
where $I_ktimes k$ is the $k$-by-$k$ identity matrix and $0_rtimes s$ is the $r$-by-$s$ zero matrix.
Thus, the condition $Y^top,K,Y=K$ means that
$$beginalign
beginbmatrixI_mtimes m&0_mtimes (n-m)\0_(n-m)times m&0_mtimes mendbmatrix
&=K=Y^top,K,Y
\
&=
beginbmatrix P^top& R^top\Q^top &S^topendbmatrixbeginbmatrixI_mtimes m&0_mtimes (n-m)\0_(n-m)times m&0_mtimes mendbmatrixbeginbmatrixP&Q\ R&Sendbmatrix
\
&=beginbmatrix P^top& R^top\Q^top &S^topendbmatrixbeginbmatrix P&Q\ 0_(n-m)times m&0_mtimes mendbmatrix=beginbmatrixP^top, P&P^top, Q\ Q^top, P &Q^top ,Qendbmatrix
endalign,.$$
That is, $Q^top, Q=0_(n-m)times (n-m)$. This shows that $Q=0_(n-m)times (n-m)$. Furthermore, we have
$$P^top,P =I_mtimes mtext or Pin O_m(mathbbR),.$$
There are no other conditions on $R$ and $S$.
Since $X=V,Y,left(V^-1right)^top$, we conclude that all solutions $X$ to $X^top,J,X=J$ take the form
$$V,beginbmatrixP&0_mtimes(n-m)\R&Sendbmatrix,left(V^-1right)^top,,$$
where $Pin O_m(mathbbR)$ is a real orthogonal matrix, $RintextMat_(n-m)times n(mathbbR)$, and $SintextMat_(n-m)times (n-m)(mathbbR)$.
The solution set is therefore a smooth real manifold, which is diffeomorphic to
$$O_m(mathbbR)times mathbbR^m(n-m)+(n-m)^2,.$$
Hence, the solution set is of dimension $$fracm(m-1)2+m(n-m)+(n-m)^2=n(n-m)+fracm(m-1)2geq fracn(n-1)2,.$$
There are therefore infinitely many nontrivial solutions when $ngeq2$.
answered Jul 30 at 21:43
Batominovski
22.8k22776
Then can we conclude that all the solution are the projection matrix onto span(u,v)?
– gimusi
Jul 30 at 21:50
This is great, and would generalize nicely to any number of rank-1 matrices. I am also curious what other solutions could exist.
– vibe
Jul 30 at 21:59
Interesting solution. Since in this case the $u_j$ are orthogonal, I guess you can choose $V = I$ to simplify the solution a little.
– vibe
Jul 31 at 17:26
Also in the case $m = 1$, we know that $X = u u^T / (u^T u)$ is a solution, but this would not necessarily have $0$ in the upper right block of the matrix?
– vibe
Jul 31 at 17:29
You can't always choose $V=I$ even when the $u_j$'s are orthogonal. Work on the small cases to see that $V=I$ doesn't usually work.
– Batominovski
Jul 31 at 17:32
add a comment |Â
up vote
1
down vote
accepted
Set
$$A:=fracu,u^topu^top u+fracv,v^topv^top v,.$$
Because $u^top ,v=0$ and $v^top,u=0$, as $u$ and $v$ are orthogonal, we conclude that $X=A$ is a solution to
$$X^top,left(u,u^top+v,v^topright),X=u,u^top+v,v^top,.$$
If $N$ is an $n$-by-$n$ matrix such that $u$ and $v$ are in $ker(N)$, then $X=A+N^top$ is also a solution. For $ngeq 3$, there are infinitely many such $N$. Furthermore, $X=I+N^top$ is also a nontrivial solution.
For a fixed positive integer $mleq n$, let $u_1,u_2,ldots,u_m$ be (not necessarily orthogonal) linearly independent elements of $mathbbR^n$. Write $$J:=sum_j=1^m,u_j,u_j^top,.$$
I shall find all solutions $XintextMat_ntimes n(mathbbR)$ such that
$$X^top , J ,X=J,.$$
Let $e_1,e_2,ldots,e_n$ be the standard basis vectors of $mathbbR^n$. Choose an invertible matrix $VintextGL_n(mathbbR)$ such that $V,u_j=e_j$ for $j=1,2,ldots,m$. Define the matrix $K$ to be $$K:=sum_j=1^m,e_j,e_j^top=V,left(sum_j=1^m,u_j,u_j^topright),V^top=V,J,V^top,.$$
Setting $Y:=V^-1,X,V^top$, we see that $X^top,J,X=J$ if and only if $$Y^top,K,Y=K,.$$
Write
$$Y=beginbmatrixP&Q\R&Sendbmatrix,,$$
where $PintextMat_mtimes m(mathbbR)$, $QintextMat_mtimes (n-m)(mathbbR)$, $RintextMat_(n-m)times m(mathbbR)$, and $SintextMat_(n-m)times(n-m)(mathbbR)$. Also, $$K=beginbmatrixI_mtimes m&0_mtimes (n-m)\0_(n-m)times m&0_mtimes mendbmatrix,,$$
where $I_ktimes k$ is the $k$-by-$k$ identity matrix and $0_rtimes s$ is the $r$-by-$s$ zero matrix.
Thus, the condition $Y^top,K,Y=K$ means that
$$beginalign
beginbmatrixI_mtimes m&0_mtimes (n-m)\0_(n-m)times m&0_mtimes mendbmatrix
&=K=Y^top,K,Y
\
&=
beginbmatrix P^top& R^top\Q^top &S^topendbmatrixbeginbmatrixI_mtimes m&0_mtimes (n-m)\0_(n-m)times m&0_mtimes mendbmatrixbeginbmatrixP&Q\ R&Sendbmatrix
\
&=beginbmatrix P^top& R^top\Q^top &S^topendbmatrixbeginbmatrix P&Q\ 0_(n-m)times m&0_mtimes mendbmatrix=beginbmatrixP^top, P&P^top, Q\ Q^top, P &Q^top ,Qendbmatrix
endalign,.$$
That is, $Q^top, Q=0_(n-m)times (n-m)$. This shows that $Q=0_(n-m)times (n-m)$. Furthermore, we have
$$P^top,P =I_mtimes mtext or Pin O_m(mathbbR),.$$
There are no other conditions on $R$ and $S$.
Since $X=V,Y,left(V^-1right)^top$, we conclude that all solutions $X$ to $X^top,J,X=J$ take the form
$$V,beginbmatrixP&0_mtimes(n-m)\R&Sendbmatrix,left(V^-1right)^top,,$$
where $Pin O_m(mathbbR)$ is a real orthogonal matrix, $RintextMat_(n-m)times n(mathbbR)$, and $SintextMat_(n-m)times (n-m)(mathbbR)$.
The solution set is therefore a smooth real manifold, which is diffeomorphic to
$$O_m(mathbbR)times mathbbR^m(n-m)+(n-m)^2,.$$
Hence, the solution set is of dimension $$fracm(m-1)2+m(n-m)+(n-m)^2=n(n-m)+fracm(m-1)2geq fracn(n-1)2,.$$
There are therefore infinitely many nontrivial solutions when $ngeq2$.
answered Jul 30 at 21:43
Batominovski
22.8k22776
Then can we conclude that all the solution are the projection matrix onto span(u,v)?
– gimusi
Jul 30 at 21:50
This is great, and would generalize nicely to any number of rank-1 matrices. I am also curious what other solutions could exist.
– vibe
Jul 30 at 21:59
Interesting solution. Since in this case the $u_j$ are orthogonal, I guess you can choose $V = I$ to simplify the solution a little.
– vibe
Jul 31 at 17:26
Also in the case $m = 1$, we know that $X = u u^T / (u^T u)$ is a solution, but this would not necessarily have $0$ in the upper right block of the matrix?
– vibe
Jul 31 at 17:29
You can't always choose $V=I$ even when the $u_j$'s are orthogonal. Work on the small cases to see that $V=I$ doesn't usually work.
– Batominovski
Jul 31 at 17:32
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Set
$$A:=fracu,u^topu^top u+fracv,v^topv^top v,.$$
Because $u^top ,v=0$ and $v^top,u=0$, as $u$ and $v$ are orthogonal, we conclude that $X=A$ is a solution to
$$X^top,left(u,u^top+v,v^topright),X=u,u^top+v,v^top,.$$
If $N$ is an $n$-by-$n$ matrix such that $u$ and $v$ are in $ker(N)$, then $X=A+N^top$ is also a solution. For $ngeq 3$, there are infinitely many such $N$. Furthermore, $X=I+N^top$ is also a nontrivial solution.
For a fixed positive integer $mleq n$, let $u_1,u_2,ldots,u_m$ be (not necessarily orthogonal) linearly independent elements of $mathbbR^n$. Write $$J:=sum_j=1^m,u_j,u_j^top,.$$
I shall find all solutions $XintextMat_ntimes n(mathbbR)$ such that
$$X^top , J ,X=J,.$$
Let $e_1,e_2,ldots,e_n$ be the standard basis vectors of $mathbbR^n$. Choose an invertible matrix $VintextGL_n(mathbbR)$ such that $V,u_j=e_j$ for $j=1,2,ldots,m$. Define the matrix $K$ to be $$K:=sum_j=1^m,e_j,e_j^top=V,left(sum_j=1^m,u_j,u_j^topright),V^top=V,J,V^top,.$$
Setting $Y:=V^-1,X,V^top$, we see that $X^top,J,X=J$ if and only if $$Y^top,K,Y=K,.$$
Write
$$Y=beginbmatrixP&Q\R&Sendbmatrix,,$$
where $PintextMat_mtimes m(mathbbR)$, $QintextMat_mtimes (n-m)(mathbbR)$, $RintextMat_(n-m)times m(mathbbR)$, and $SintextMat_(n-m)times(n-m)(mathbbR)$. Also, $$K=beginbmatrixI_mtimes m&0_mtimes (n-m)\0_(n-m)times m&0_mtimes mendbmatrix,,$$
where $I_ktimes k$ is the $k$-by-$k$ identity matrix and $0_rtimes s$ is the $r$-by-$s$ zero matrix.
Thus, the condition $Y^top,K,Y=K$ means that
$$beginalign
beginbmatrixI_mtimes m&0_mtimes (n-m)\0_(n-m)times m&0_mtimes mendbmatrix
&=K=Y^top,K,Y
\
&=
beginbmatrix P^top& R^top\Q^top &S^topendbmatrixbeginbmatrixI_mtimes m&0_mtimes (n-m)\0_(n-m)times m&0_mtimes mendbmatrixbeginbmatrixP&Q\ R&Sendbmatrix
\
&=beginbmatrix P^top& R^top\Q^top &S^topendbmatrixbeginbmatrix P&Q\ 0_(n-m)times m&0_mtimes mendbmatrix=beginbmatrixP^top, P&P^top, Q\ Q^top, P &Q^top ,Qendbmatrix
endalign,.$$
That is, $Q^top, Q=0_(n-m)times (n-m)$. This shows that $Q=0_(n-m)times (n-m)$. Furthermore, we have
$$P^top,P =I_mtimes mtext or Pin O_m(mathbbR),.$$
There are no other conditions on $R$ and $S$.
Since $X=V,Y,left(V^-1right)^top$, we conclude that all solutions $X$ to $X^top,J,X=J$ take the form
$$V,beginbmatrixP&0_mtimes(n-m)\R&Sendbmatrix,left(V^-1right)^top,,$$
where $Pin O_m(mathbbR)$ is a real orthogonal matrix, $RintextMat_(n-m)times n(mathbbR)$, and $SintextMat_(n-m)times (n-m)(mathbbR)$.
The solution set is therefore a smooth real manifold, which is diffeomorphic to
$$O_m(mathbbR)times mathbbR^m(n-m)+(n-m)^2,.$$
Hence, the solution set is of dimension $$fracm(m-1)2+m(n-m)+(n-m)^2=n(n-m)+fracm(m-1)2geq fracn(n-1)2,.$$
There are therefore infinitely many nontrivial solutions when $ngeq2$.
answered Jul 30 at 21:43
Batominovski
22.8k22776
Set
$$A:=fracu,u^topu^top u+fracv,v^topv^top v,.$$
Because $u^top ,v=0$ and $v^top,u=0$, as $u$ and $v$ are orthogonal, we conclude that $X=A$ is a solution to
$$X^top,left(u,u^top+v,v^topright),X=u,u^top+v,v^top,.$$
If $N$ is an $n$-by-$n$ matrix such that $u$ and $v$ are in $ker(N)$, then $X=A+N^top$ is also a solution. For $ngeq 3$, there are infinitely many such $N$. Furthermore, $X=I+N^top$ is also a nontrivial solution.
For a fixed positive integer $mleq n$, let $u_1,u_2,ldots,u_m$ be (not necessarily orthogonal) linearly independent elements of $mathbbR^n$. Write $$J:=sum_j=1^m,u_j,u_j^top,.$$
I shall find all solutions $XintextMat_ntimes n(mathbbR)$ such that
$$X^top , J ,X=J,.$$
Let $e_1,e_2,ldots,e_n$ be the standard basis vectors of $mathbbR^n$. Choose an invertible matrix $VintextGL_n(mathbbR)$ such that $V,u_j=e_j$ for $j=1,2,ldots,m$. Define the matrix $K$ to be $$K:=sum_j=1^m,e_j,e_j^top=V,left(sum_j=1^m,u_j,u_j^topright),V^top=V,J,V^top,.$$
Setting $Y:=V^-1,X,V^top$, we see that $X^top,J,X=J$ if and only if $$Y^top,K,Y=K,.$$
Write
$$Y=beginbmatrixP&Q\R&Sendbmatrix,,$$
where $PintextMat_mtimes m(mathbbR)$, $QintextMat_mtimes (n-m)(mathbbR)$, $RintextMat_(n-m)times m(mathbbR)$, and $SintextMat_(n-m)times(n-m)(mathbbR)$. Also, $$K=beginbmatrixI_mtimes m&0_mtimes (n-m)\0_(n-m)times m&0_mtimes mendbmatrix,,$$
where $I_ktimes k$ is the $k$-by-$k$ identity matrix and $0_rtimes s$ is the $r$-by-$s$ zero matrix.
Thus, the condition $Y^top,K,Y=K$ means that
$$beginalign
beginbmatrixI_mtimes m&0_mtimes (n-m)\0_(n-m)times m&0_mtimes mendbmatrix
&=K=Y^top,K,Y
\
&=
beginbmatrix P^top& R^top\Q^top &S^topendbmatrixbeginbmatrixI_mtimes m&0_mtimes (n-m)\0_(n-m)times m&0_mtimes mendbmatrixbeginbmatrixP&Q\ R&Sendbmatrix
\
&=beginbmatrix P^top& R^top\Q^top &S^topendbmatrixbeginbmatrix P&Q\ 0_(n-m)times m&0_mtimes mendbmatrix=beginbmatrixP^top, P&P^top, Q\ Q^top, P &Q^top ,Qendbmatrix
endalign,.$$
That is, $Q^top, Q=0_(n-m)times (n-m)$. This shows that $Q=0_(n-m)times (n-m)$. Furthermore, we have
$$P^top,P =I_mtimes mtext or Pin O_m(mathbbR),.$$
There are no other conditions on $R$ and $S$.
Since $X=V,Y,left(V^-1right)^top$, we conclude that all solutions $X$ to $X^top,J,X=J$ take the form
$$V,beginbmatrixP&0_mtimes(n-m)\R&Sendbmatrix,left(V^-1right)^top,,$$
where $Pin O_m(mathbbR)$ is a real orthogonal matrix, $RintextMat_(n-m)times n(mathbbR)$, and $SintextMat_(n-m)times (n-m)(mathbbR)$.
The solution set is therefore a smooth real manifold, which is diffeomorphic to
$$O_m(mathbbR)times mathbbR^m(n-m)+(n-m)^2,.$$
Hence, the solution set is of dimension $$fracm(m-1)2+m(n-m)+(n-m)^2=n(n-m)+fracm(m-1)2geq fracn(n-1)2,.$$
There are therefore infinitely many nontrivial solutions when $ngeq2$.
answered Jul 30 at 21:43
Batominovski
22.8k22776
edited Jul 31 at 7:13
answered Jul 30 at 21:43
Batominovski
22.8k22776
answered Jul 30 at 21:43
Batominovski
22.8k22776
answered Jul 30 at 21:43
Batominovski
22.8k22776
22.8k22776
Then can we conclude that all the solution are the projection matrix onto span(u,v)?
– gimusi
Jul 30 at 21:50
This is great, and would generalize nicely to any number of rank-1 matrices. I am also curious what other solutions could exist.
– vibe
Jul 30 at 21:59
Interesting solution. Since in this case the $u_j$ are orthogonal, I guess you can choose $V = I$ to simplify the solution a little.
– vibe
Jul 31 at 17:26
Also in the case $m = 1$, we know that $X = u u^T / (u^T u)$ is a solution, but this would not necessarily have $0$ in the upper right block of the matrix?
– vibe
Jul 31 at 17:29
You can't always choose $V=I$ even when the $u_j$'s are orthogonal. Work on the small cases to see that $V=I$ doesn't usually work.
– Batominovski
Jul 31 at 17:32
add a comment |Â
Then can we conclude that all the solution are the projection matrix onto span(u,v)?
– gimusi
Jul 30 at 21:50
This is great, and would generalize nicely to any number of rank-1 matrices. I am also curious what other solutions could exist.
– vibe
Jul 30 at 21:59
Interesting solution. Since in this case the $u_j$ are orthogonal, I guess you can choose $V = I$ to simplify the solution a little.
– vibe
Jul 31 at 17:26
Also in the case $m = 1$, we know that $X = u u^T / (u^T u)$ is a solution, but this would not necessarily have $0$ in the upper right block of the matrix?
– vibe
Jul 31 at 17:29
You can't always choose $V=I$ even when the $u_j$'s are orthogonal. Work on the small cases to see that $V=I$ doesn't usually work.
– Batominovski
Jul 31 at 17:32
Then can we conclude that all the solution are the projection matrix onto span(u,v)?
– gimusi
Jul 30 at 21:50
Then can we conclude that all the solution are the projection matrix onto span(u,v)?
– gimusi
Jul 30 at 21:50
This is great, and would generalize nicely to any number of rank-1 matrices. I am also curious what other solutions could exist.
– vibe
Jul 30 at 21:59
This is great, and would generalize nicely to any number of rank-1 matrices. I am also curious what other solutions could exist.
– vibe
Jul 30 at 21:59
Interesting solution. Since in this case the $u_j$ are orthogonal, I guess you can choose $V = I$ to simplify the solution a little.
– vibe
Jul 31 at 17:26
Interesting solution. Since in this case the $u_j$ are orthogonal, I guess you can choose $V = I$ to simplify the solution a little.
– vibe
Jul 31 at 17:26
Also in the case $m = 1$, we know that $X = u u^T / (u^T u)$ is a solution, but this would not necessarily have $0$ in the upper right block of the matrix?
– vibe
Jul 31 at 17:29
Also in the case $m = 1$, we know that $X = u u^T / (u^T u)$ is a solution, but this would not necessarily have $0$ in the upper right block of the matrix?
– vibe
Jul 31 at 17:29
You can't always choose $V=I$ even when the $u_j$'s are orthogonal. Work on the small cases to see that $V=I$ doesn't usually work.
– Batominovski
Jul 31 at 17:32
You can't always choose $V=I$ even when the $u_j$'s are orthogonal. Work on the small cases to see that $V=I$ doesn't usually work.
– Batominovski
Jul 31 at 17:32
add a comment |Â
up vote
0
down vote
With reference to
$$X^T v v^T X = v v^T$$
assuming $X^T=fracvv^T$ that is a projection matrix onto $operatornamespan(v)$ we have
$$X^T v v^T X = fracvv^Tv v^Tfracvv^T=vv^T$$
For
$$X^T left( u u^T + v v^T right) X = u u^T + v v^T$$
we can assume $X^T$ as a projection matrix onto $operatornamespan(v,u)$.
answered Jul 30 at 21:36
gimusi
64.1k73480
add a comment |Â
up vote
0
down vote
With reference to
$$X^T v v^T X = v v^T$$
assuming $X^T=fracvv^T$ that is a projection matrix onto $operatornamespan(v)$ we have
$$X^T v v^T X = fracvv^Tv v^Tfracvv^T=vv^T$$
For
$$X^T left( u u^T + v v^T right) X = u u^T + v v^T$$
we can assume $X^T$ as a projection matrix onto $operatornamespan(v,u)$.
answered Jul 30 at 21:36
gimusi
64.1k73480
add a comment |Â
up vote
0
down vote
up vote
0
down vote
With reference to
$$X^T v v^T X = v v^T$$
assuming $X^T=fracvv^T$ that is a projection matrix onto $operatornamespan(v)$ we have
$$X^T v v^T X = fracvv^Tv v^Tfracvv^T=vv^T$$
For
$$X^T left( u u^T + v v^T right) X = u u^T + v v^T$$
we can assume $X^T$ as a projection matrix onto $operatornamespan(v,u)$.
answered Jul 30 at 21:36
gimusi
64.1k73480
With reference to
$$X^T v v^T X = v v^T$$
assuming $X^T=fracvv^T$ that is a projection matrix onto $operatornamespan(v)$ we have
$$X^T v v^T X = fracvv^Tv v^Tfracvv^T=vv^T$$
For
$$X^T left( u u^T + v v^T right) X = u u^T + v v^T$$
we can assume $X^T$ as a projection matrix onto $operatornamespan(v,u)$.
answered Jul 30 at 21:36
gimusi
64.1k73480
edited Jul 30 at 21:50
answered Jul 30 at 21:36
gimusi
64.1k73480
answered Jul 30 at 21:36
gimusi
64.1k73480
answered Jul 30 at 21:36
gimusi
64.1k73480
64.1k73480
add a comment |Â
add a comment |Â
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I don't understand how you obtain $$X^T v v^T X = v v^Timplies X=pm I$$
– gimusi
Jul 30 at 21:25
@gimusi The OP meant that $X=pm I$ are solutions to $X^top v v^top X=v v^top$.
– Batominovski
Jul 30 at 21:25
@Batominovski But it is not the only solution I guess, we can take also $X^T=$ a projection matrix onto $v$.
– gimusi
Jul 30 at 21:27
@gimusi The OP never claimed that whatever he found were the only solutions. Plus, if $X$ is a solution, then $X+N^top$ is a solution, where $N$ is a matrix with $vinker(N)$.
– Batominovski
Jul 30 at 21:29
@Batominovski: Since $u u^T + v v^T$ is rank 2 and therefore not invertible for $n > 2$, does this mean I can't write down a closed form expression for its pseudo inverse?
– vibe
Jul 30 at 21:29