Mixing
zeros of $x^*Ax$, a quadratic form
$begingroup$
The question hopefully says it all!
We have a Hermitian matrix $A=A^* in mathbb{C}^n$ and a quadratic form: $f(x)=x^*Ax,~xin mathbb{C}^n$
We want to find the solution of $f(x) = x^*Ax = 0$
When the matrix is positive semi definite (p.s.d.), the solution seems to be null-space of the matrix of $A$. This I found by diagonalising $A$. But suddenly I became helpless when $A$ has negative Eigen values too!
Please note that when the matrix is not p.s.d., the solution space contains the null space. (I.e. null space is always a solution)
linear-algebra matrices eigenvalues-eigenvectors quadratic-forms matrix-decomposition
asked Mar 20 '15 at 14:20
Loves ProbabilityLoves Probability
15319
$endgroup$
add a comment |
$begingroup$
The question hopefully says it all!
We have a Hermitian matrix $A=A^* in mathbb{C}^n$ and a quadratic form: $f(x)=x^*Ax,~xin mathbb{C}^n$
We want to find the solution of $f(x) = x^*Ax = 0$
When the matrix is positive semi definite (p.s.d.), the solution seems to be null-space of the matrix of $A$. This I found by diagonalising $A$. But suddenly I became helpless when $A$ has negative Eigen values too!
Please note that when the matrix is not p.s.d., the solution space contains the null space. (I.e. null space is always a solution)
linear-algebra matrices eigenvalues-eigenvectors quadratic-forms matrix-decomposition
asked Mar 20 '15 at 14:20
Loves ProbabilityLoves Probability
15319
$endgroup$
$begingroup$
Have you tried solving this in the case that $A$ is diagonal?
$endgroup$
– Omnomnomnom
Mar 20 '15 at 17:21
$begingroup$
In fact, it suffices to solve this in the case that $A$ has elements $pm 1$ and $0$.
$endgroup$
– Omnomnomnom
Mar 20 '15 at 17:23
$begingroup$
@omnomnomnom Thanks for the comment. That's exactly where my solution stage is. This tri element diagonal case gives me an equation with no further insights!
$endgroup$
– Loves Probability
Mar 20 '15 at 23:01
$begingroup$
For example, the solution for the matrix $$pmatrix{1\&1\&&-1}$$ is the cone $x^2+y^2=z^2$. Note that this is not a linear subspace.
$endgroup$
– Omnomnomnom
Mar 21 '15 at 5:07
add a comment |
$begingroup$
The question hopefully says it all!
We have a Hermitian matrix $A=A^* in mathbb{C}^n$ and a quadratic form: $f(x)=x^*Ax,~xin mathbb{C}^n$
We want to find the solution of $f(x) = x^*Ax = 0$
When the matrix is positive semi definite (p.s.d.), the solution seems to be null-space of the matrix of $A$. This I found by diagonalising $A$. But suddenly I became helpless when $A$ has negative Eigen values too!
Please note that when the matrix is not p.s.d., the solution space contains the null space. (I.e. null space is always a solution)
linear-algebra matrices eigenvalues-eigenvectors quadratic-forms matrix-decomposition
asked Mar 20 '15 at 14:20
Loves ProbabilityLoves Probability
15319
$endgroup$
The question hopefully says it all!
We have a Hermitian matrix $A=A^* in mathbb{C}^n$ and a quadratic form: $f(x)=x^*Ax,~xin mathbb{C}^n$
We want to find the solution of $f(x) = x^*Ax = 0$
When the matrix is positive semi definite (p.s.d.), the solution seems to be null-space of the matrix of $A$. This I found by diagonalising $A$. But suddenly I became helpless when $A$ has negative Eigen values too!
Please note that when the matrix is not p.s.d., the solution space contains the null space. (I.e. null space is always a solution)
linear-algebra matrices eigenvalues-eigenvectors quadratic-forms matrix-decomposition
linear-algebra matrices eigenvalues-eigenvectors quadratic-forms matrix-decomposition
asked Mar 20 '15 at 14:20
Loves ProbabilityLoves Probability
15319
asked Mar 20 '15 at 14:20
Loves ProbabilityLoves Probability
15319
edited Mar 20 '15 at 14:34
Loves Probability
asked Mar 20 '15 at 14:20
Loves ProbabilityLoves Probability
15319
asked Mar 20 '15 at 14:20
Loves ProbabilityLoves Probability
15319
asked Mar 20 '15 at 14:20
Loves ProbabilityLoves Probability
15319
15319
$begingroup$
Have you tried solving this in the case that $A$ is diagonal?
$endgroup$
– Omnomnomnom
Mar 20 '15 at 17:21
$begingroup$
In fact, it suffices to solve this in the case that $A$ has elements $pm 1$ and $0$.
$endgroup$
– Omnomnomnom
Mar 20 '15 at 17:23
$begingroup$
@omnomnomnom Thanks for the comment. That's exactly where my solution stage is. This tri element diagonal case gives me an equation with no further insights!
$endgroup$
– Loves Probability
Mar 20 '15 at 23:01
$begingroup$
For example, the solution for the matrix $$pmatrix{1\&1\&&-1}$$ is the cone $x^2+y^2=z^2$. Note that this is not a linear subspace.
$endgroup$
– Omnomnomnom
Mar 21 '15 at 5:07
add a comment |
$begingroup$
Have you tried solving this in the case that $A$ is diagonal?
$endgroup$
– Omnomnomnom
Mar 20 '15 at 17:21
$begingroup$
In fact, it suffices to solve this in the case that $A$ has elements $pm 1$ and $0$.
$endgroup$
– Omnomnomnom
Mar 20 '15 at 17:23
$begingroup$
@omnomnomnom Thanks for the comment. That's exactly where my solution stage is. This tri element diagonal case gives me an equation with no further insights!
$endgroup$
– Loves Probability
Mar 20 '15 at 23:01
$begingroup$
For example, the solution for the matrix $$pmatrix{1\&1\&&-1}$$ is the cone $x^2+y^2=z^2$. Note that this is not a linear subspace.
$endgroup$
– Omnomnomnom
Mar 21 '15 at 5:07
$begingroup$
Have you tried solving this in the case that $A$ is diagonal?
$endgroup$
– Omnomnomnom
Mar 20 '15 at 17:21
$begingroup$
Have you tried solving this in the case that $A$ is diagonal?
$endgroup$
– Omnomnomnom
Mar 20 '15 at 17:21
$begingroup$
In fact, it suffices to solve this in the case that $A$ has elements $pm 1$ and $0$.
$endgroup$
– Omnomnomnom
Mar 20 '15 at 17:23
$begingroup$
In fact, it suffices to solve this in the case that $A$ has elements $pm 1$ and $0$.
$endgroup$
– Omnomnomnom
Mar 20 '15 at 17:23
$begingroup$
@omnomnomnom Thanks for the comment. That's exactly where my solution stage is. This tri element diagonal case gives me an equation with no further insights!
$endgroup$
– Loves Probability
Mar 20 '15 at 23:01
$begingroup$
@omnomnomnom Thanks for the comment. That's exactly where my solution stage is. This tri element diagonal case gives me an equation with no further insights!
$endgroup$
– Loves Probability
Mar 20 '15 at 23:01
$begingroup$
For example, the solution for the matrix $$pmatrix{1\&1\&&-1}$$ is the cone $x^2+y^2=z^2$. Note that this is not a linear subspace.
$endgroup$
– Omnomnomnom
Mar 21 '15 at 5:07
$begingroup$
For example, the solution for the matrix $$pmatrix{1\&1\&&-1}$$ is the cone $x^2+y^2=z^2$. Note that this is not a linear subspace.
$endgroup$
– Omnomnomnom
Mar 21 '15 at 5:07
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Consider $Ntimes N$ matrices and $Ntimes 1$ vectors. Let $A=ULambda U^H$ be its eigen-decomposition. Then
begin{align}
x^HAx &=y^HLambda y && {text{where I define }y=Ux,~forall x } \
&=sum_{i=1}^{N}|y_i|^2lambda_i \
&= theta^Tlambda
end{align}
where $lambda$ is the vector with all eigenvalues and $theta$ is any vector whose entries are non-negative. Thus, for any $thetainleft(mathcal{N}(lambda)capmathbb{R}_{+}^Nright)$ where $mathcal{N}(lambda)$ is the null-space of $lambda$ vector (set of all vectors orthogonal to $lambda$) and the non-negative $N$-dimensional orthant (non-negative quadrant where all entries are non-negative), we have $theta^Tlambda=0$. Now consider the set $mathcal{D}$ of all diagonal matrices such that the diagonal entries lie on the unit circle in the complex plane. Then your solution set is
begin{align}
mathcal{S}_x,=,{U^HDsqrt{theta} ,mid,forall Dinmathcal{D}~,~forall thetainleft(mathcal{N}(lambda)capmathbb{R}_{+}^Nright) }
end{align}
where $sqrt{theta}$ is the entry-wise square root of $theta$.
$mathcal{D}$ is needed because it doesn't matter what the phase of each entry of $y$ is. Note that $y_{i}=D_{ii}sqrt{theta_i}$ and $|y_i|^2=|D_{ii}|^2theta_i=theta_i$. The difficult part is that $mathcal{S}_x$ is not a linear subspace and you won't have that nice properties. It is a highly non-linear transformation and I am not sure what intuitive sense you can derive out of it.
answered Mar 23 '15 at 13:53
dineshdileepdineshdileep
5,96111735
$endgroup$
$begingroup$
Thanks for the solution. This does give a good insight. But $mathcal{N}(lambda)bigcapmathbb{R}_{+}^{N}$ is not the solution of $y$ (but a solution of $ycdot^* y$), let alone desired $x$! (The question actually asks about $x$, as the zeros of the quadratic $x^*Ax$) Ofcourse, $y$ is a simple but non-linear mapping from $mathcal{N}(lambda)bigcapmathbb{R}_{+}^{N}$. But there is another unitary transformation to be applied after this non-linear transform. What does that all finally mean, is the mystery still!
$endgroup$
– Loves Probability
Mar 23 '15 at 15:22
$begingroup$
By the way, hope you remember me, Dileep! (I am Harish, who joined in ME(Telecom) along with you @IISc. I did my project under Prof. VS (2009-11 batch))
$endgroup$
– Loves Probability
Mar 23 '15 at 15:24
$begingroup$
You are right. It is far from stating the answer. I thought the reverse transformation is obvious. I have now explicitly stated it.
$endgroup$
– dineshdileep
Mar 24 '15 at 5:17
$begingroup$
Yes, I do know you harish. We were neighbors in hostel as well :):).
$endgroup$
– dineshdileep
Mar 24 '15 at 5:18
add a comment |
Your Answer
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Consider $Ntimes N$ matrices and $Ntimes 1$ vectors. Let $A=ULambda U^H$ be its eigen-decomposition. Then
begin{align}
x^HAx &=y^HLambda y && {text{where I define }y=Ux,~forall x } \
&=sum_{i=1}^{N}|y_i|^2lambda_i \
&= theta^Tlambda
end{align}
where $lambda$ is the vector with all eigenvalues and $theta$ is any vector whose entries are non-negative. Thus, for any $thetainleft(mathcal{N}(lambda)capmathbb{R}_{+}^Nright)$ where $mathcal{N}(lambda)$ is the null-space of $lambda$ vector (set of all vectors orthogonal to $lambda$) and the non-negative $N$-dimensional orthant (non-negative quadrant where all entries are non-negative), we have $theta^Tlambda=0$. Now consider the set $mathcal{D}$ of all diagonal matrices such that the diagonal entries lie on the unit circle in the complex plane. Then your solution set is
begin{align}
mathcal{S}_x,=,{U^HDsqrt{theta} ,mid,forall Dinmathcal{D}~,~forall thetainleft(mathcal{N}(lambda)capmathbb{R}_{+}^Nright) }
end{align}
where $sqrt{theta}$ is the entry-wise square root of $theta$.
$mathcal{D}$ is needed because it doesn't matter what the phase of each entry of $y$ is. Note that $y_{i}=D_{ii}sqrt{theta_i}$ and $|y_i|^2=|D_{ii}|^2theta_i=theta_i$. The difficult part is that $mathcal{S}_x$ is not a linear subspace and you won't have that nice properties. It is a highly non-linear transformation and I am not sure what intuitive sense you can derive out of it.
answered Mar 23 '15 at 13:53
dineshdileepdineshdileep
5,96111735
$endgroup$
$begingroup$
Thanks for the solution. This does give a good insight. But $mathcal{N}(lambda)bigcapmathbb{R}_{+}^{N}$ is not the solution of $y$ (but a solution of $ycdot^* y$), let alone desired $x$! (The question actually asks about $x$, as the zeros of the quadratic $x^*Ax$) Ofcourse, $y$ is a simple but non-linear mapping from $mathcal{N}(lambda)bigcapmathbb{R}_{+}^{N}$. But there is another unitary transformation to be applied after this non-linear transform. What does that all finally mean, is the mystery still!
$endgroup$
– Loves Probability
Mar 23 '15 at 15:22
$begingroup$
By the way, hope you remember me, Dileep! (I am Harish, who joined in ME(Telecom) along with you @IISc. I did my project under Prof. VS (2009-11 batch))
$endgroup$
– Loves Probability
Mar 23 '15 at 15:24
$begingroup$
You are right. It is far from stating the answer. I thought the reverse transformation is obvious. I have now explicitly stated it.
$endgroup$
– dineshdileep
Mar 24 '15 at 5:17
$begingroup$
Yes, I do know you harish. We were neighbors in hostel as well :):).
$endgroup$
– dineshdileep
Mar 24 '15 at 5:18
add a comment |
$begingroup$
Consider $Ntimes N$ matrices and $Ntimes 1$ vectors. Let $A=ULambda U^H$ be its eigen-decomposition. Then
begin{align}
x^HAx &=y^HLambda y && {text{where I define }y=Ux,~forall x } \
&=sum_{i=1}^{N}|y_i|^2lambda_i \
&= theta^Tlambda
end{align}
where $lambda$ is the vector with all eigenvalues and $theta$ is any vector whose entries are non-negative. Thus, for any $thetainleft(mathcal{N}(lambda)capmathbb{R}_{+}^Nright)$ where $mathcal{N}(lambda)$ is the null-space of $lambda$ vector (set of all vectors orthogonal to $lambda$) and the non-negative $N$-dimensional orthant (non-negative quadrant where all entries are non-negative), we have $theta^Tlambda=0$. Now consider the set $mathcal{D}$ of all diagonal matrices such that the diagonal entries lie on the unit circle in the complex plane. Then your solution set is
begin{align}
mathcal{S}_x,=,{U^HDsqrt{theta} ,mid,forall Dinmathcal{D}~,~forall thetainleft(mathcal{N}(lambda)capmathbb{R}_{+}^Nright) }
end{align}
where $sqrt{theta}$ is the entry-wise square root of $theta$.
$mathcal{D}$ is needed because it doesn't matter what the phase of each entry of $y$ is. Note that $y_{i}=D_{ii}sqrt{theta_i}$ and $|y_i|^2=|D_{ii}|^2theta_i=theta_i$. The difficult part is that $mathcal{S}_x$ is not a linear subspace and you won't have that nice properties. It is a highly non-linear transformation and I am not sure what intuitive sense you can derive out of it.
answered Mar 23 '15 at 13:53
dineshdileepdineshdileep
5,96111735
$endgroup$
$begingroup$
Thanks for the solution. This does give a good insight. But $mathcal{N}(lambda)bigcapmathbb{R}_{+}^{N}$ is not the solution of $y$ (but a solution of $ycdot^* y$), let alone desired $x$! (The question actually asks about $x$, as the zeros of the quadratic $x^*Ax$) Ofcourse, $y$ is a simple but non-linear mapping from $mathcal{N}(lambda)bigcapmathbb{R}_{+}^{N}$. But there is another unitary transformation to be applied after this non-linear transform. What does that all finally mean, is the mystery still!
$endgroup$
– Loves Probability
Mar 23 '15 at 15:22
$begingroup$
By the way, hope you remember me, Dileep! (I am Harish, who joined in ME(Telecom) along with you @IISc. I did my project under Prof. VS (2009-11 batch))
$endgroup$
– Loves Probability
Mar 23 '15 at 15:24
$begingroup$
You are right. It is far from stating the answer. I thought the reverse transformation is obvious. I have now explicitly stated it.
$endgroup$
– dineshdileep
Mar 24 '15 at 5:17
$begingroup$
Yes, I do know you harish. We were neighbors in hostel as well :):).
$endgroup$
– dineshdileep
Mar 24 '15 at 5:18
add a comment |
$begingroup$
Consider $Ntimes N$ matrices and $Ntimes 1$ vectors. Let $A=ULambda U^H$ be its eigen-decomposition. Then
begin{align}
x^HAx &=y^HLambda y && {text{where I define }y=Ux,~forall x } \
&=sum_{i=1}^{N}|y_i|^2lambda_i \
&= theta^Tlambda
end{align}
where $lambda$ is the vector with all eigenvalues and $theta$ is any vector whose entries are non-negative. Thus, for any $thetainleft(mathcal{N}(lambda)capmathbb{R}_{+}^Nright)$ where $mathcal{N}(lambda)$ is the null-space of $lambda$ vector (set of all vectors orthogonal to $lambda$) and the non-negative $N$-dimensional orthant (non-negative quadrant where all entries are non-negative), we have $theta^Tlambda=0$. Now consider the set $mathcal{D}$ of all diagonal matrices such that the diagonal entries lie on the unit circle in the complex plane. Then your solution set is
begin{align}
mathcal{S}_x,=,{U^HDsqrt{theta} ,mid,forall Dinmathcal{D}~,~forall thetainleft(mathcal{N}(lambda)capmathbb{R}_{+}^Nright) }
end{align}
where $sqrt{theta}$ is the entry-wise square root of $theta$.
$mathcal{D}$ is needed because it doesn't matter what the phase of each entry of $y$ is. Note that $y_{i}=D_{ii}sqrt{theta_i}$ and $|y_i|^2=|D_{ii}|^2theta_i=theta_i$. The difficult part is that $mathcal{S}_x$ is not a linear subspace and you won't have that nice properties. It is a highly non-linear transformation and I am not sure what intuitive sense you can derive out of it.
answered Mar 23 '15 at 13:53
dineshdileepdineshdileep
5,96111735
$endgroup$
Consider $Ntimes N$ matrices and $Ntimes 1$ vectors. Let $A=ULambda U^H$ be its eigen-decomposition. Then
begin{align}
x^HAx &=y^HLambda y && {text{where I define }y=Ux,~forall x } \
&=sum_{i=1}^{N}|y_i|^2lambda_i \
&= theta^Tlambda
end{align}
where $lambda$ is the vector with all eigenvalues and $theta$ is any vector whose entries are non-negative. Thus, for any $thetainleft(mathcal{N}(lambda)capmathbb{R}_{+}^Nright)$ where $mathcal{N}(lambda)$ is the null-space of $lambda$ vector (set of all vectors orthogonal to $lambda$) and the non-negative $N$-dimensional orthant (non-negative quadrant where all entries are non-negative), we have $theta^Tlambda=0$. Now consider the set $mathcal{D}$ of all diagonal matrices such that the diagonal entries lie on the unit circle in the complex plane. Then your solution set is
begin{align}
mathcal{S}_x,=,{U^HDsqrt{theta} ,mid,forall Dinmathcal{D}~,~forall thetainleft(mathcal{N}(lambda)capmathbb{R}_{+}^Nright) }
end{align}
where $sqrt{theta}$ is the entry-wise square root of $theta$.
$mathcal{D}$ is needed because it doesn't matter what the phase of each entry of $y$ is. Note that $y_{i}=D_{ii}sqrt{theta_i}$ and $|y_i|^2=|D_{ii}|^2theta_i=theta_i$. The difficult part is that $mathcal{S}_x$ is not a linear subspace and you won't have that nice properties. It is a highly non-linear transformation and I am not sure what intuitive sense you can derive out of it.
answered Mar 23 '15 at 13:53
dineshdileepdineshdileep
5,96111735
edited Mar 24 '15 at 5:14
answered Mar 23 '15 at 13:53
dineshdileepdineshdileep
5,96111735
answered Mar 23 '15 at 13:53
dineshdileepdineshdileep
5,96111735
answered Mar 23 '15 at 13:53
dineshdileepdineshdileep
5,96111735
5,96111735
$begingroup$
Thanks for the solution. This does give a good insight. But $mathcal{N}(lambda)bigcapmathbb{R}_{+}^{N}$ is not the solution of $y$ (but a solution of $ycdot^* y$), let alone desired $x$! (The question actually asks about $x$, as the zeros of the quadratic $x^*Ax$) Ofcourse, $y$ is a simple but non-linear mapping from $mathcal{N}(lambda)bigcapmathbb{R}_{+}^{N}$. But there is another unitary transformation to be applied after this non-linear transform. What does that all finally mean, is the mystery still!
$endgroup$
– Loves Probability
Mar 23 '15 at 15:22
$begingroup$
By the way, hope you remember me, Dileep! (I am Harish, who joined in ME(Telecom) along with you @IISc. I did my project under Prof. VS (2009-11 batch))
$endgroup$
– Loves Probability
Mar 23 '15 at 15:24
$begingroup$
You are right. It is far from stating the answer. I thought the reverse transformation is obvious. I have now explicitly stated it.
$endgroup$
– dineshdileep
Mar 24 '15 at 5:17
$begingroup$
Yes, I do know you harish. We were neighbors in hostel as well :):).
$endgroup$
– dineshdileep
Mar 24 '15 at 5:18
add a comment |
$begingroup$
Thanks for the solution. This does give a good insight. But $mathcal{N}(lambda)bigcapmathbb{R}_{+}^{N}$ is not the solution of $y$ (but a solution of $ycdot^* y$), let alone desired $x$! (The question actually asks about $x$, as the zeros of the quadratic $x^*Ax$) Ofcourse, $y$ is a simple but non-linear mapping from $mathcal{N}(lambda)bigcapmathbb{R}_{+}^{N}$. But there is another unitary transformation to be applied after this non-linear transform. What does that all finally mean, is the mystery still!
$endgroup$
– Loves Probability
Mar 23 '15 at 15:22
$begingroup$
By the way, hope you remember me, Dileep! (I am Harish, who joined in ME(Telecom) along with you @IISc. I did my project under Prof. VS (2009-11 batch))
$endgroup$
– Loves Probability
Mar 23 '15 at 15:24
$begingroup$
You are right. It is far from stating the answer. I thought the reverse transformation is obvious. I have now explicitly stated it.
$endgroup$
– dineshdileep
Mar 24 '15 at 5:17
$begingroup$
Yes, I do know you harish. We were neighbors in hostel as well :):).
$endgroup$
– dineshdileep
Mar 24 '15 at 5:18
$begingroup$
Thanks for the solution. This does give a good insight. But $mathcal{N}(lambda)bigcapmathbb{R}_{+}^{N}$ is not the solution of $y$ (but a solution of $ycdot^* y$), let alone desired $x$! (The question actually asks about $x$, as the zeros of the quadratic $x^*Ax$) Ofcourse, $y$ is a simple but non-linear mapping from $mathcal{N}(lambda)bigcapmathbb{R}_{+}^{N}$. But there is another unitary transformation to be applied after this non-linear transform. What does that all finally mean, is the mystery still!
$endgroup$
– Loves Probability
Mar 23 '15 at 15:22
$begingroup$
Thanks for the solution. This does give a good insight. But $mathcal{N}(lambda)bigcapmathbb{R}_{+}^{N}$ is not the solution of $y$ (but a solution of $ycdot^* y$), let alone desired $x$! (The question actually asks about $x$, as the zeros of the quadratic $x^*Ax$) Ofcourse, $y$ is a simple but non-linear mapping from $mathcal{N}(lambda)bigcapmathbb{R}_{+}^{N}$. But there is another unitary transformation to be applied after this non-linear transform. What does that all finally mean, is the mystery still!
$endgroup$
– Loves Probability
Mar 23 '15 at 15:22
$begingroup$
By the way, hope you remember me, Dileep! (I am Harish, who joined in ME(Telecom) along with you @IISc. I did my project under Prof. VS (2009-11 batch))
$endgroup$
– Loves Probability
Mar 23 '15 at 15:24
$begingroup$
By the way, hope you remember me, Dileep! (I am Harish, who joined in ME(Telecom) along with you @IISc. I did my project under Prof. VS (2009-11 batch))
$endgroup$
– Loves Probability
Mar 23 '15 at 15:24
$begingroup$
You are right. It is far from stating the answer. I thought the reverse transformation is obvious. I have now explicitly stated it.
$endgroup$
– dineshdileep
Mar 24 '15 at 5:17
$begingroup$
You are right. It is far from stating the answer. I thought the reverse transformation is obvious. I have now explicitly stated it.
$endgroup$
– dineshdileep
Mar 24 '15 at 5:17
$begingroup$
Yes, I do know you harish. We were neighbors in hostel as well :):).
$endgroup$
– dineshdileep
Mar 24 '15 at 5:18
$begingroup$
Yes, I do know you harish. We were neighbors in hostel as well :):).
$endgroup$
– dineshdileep
Mar 24 '15 at 5:18
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$begingroup$
Have you tried solving this in the case that $A$ is diagonal?
$endgroup$
– Omnomnomnom
Mar 20 '15 at 17:21
$begingroup$
In fact, it suffices to solve this in the case that $A$ has elements $pm 1$ and $0$.
$endgroup$
– Omnomnomnom
Mar 20 '15 at 17:23
$begingroup$
@omnomnomnom Thanks for the comment. That's exactly where my solution stage is. This tri element diagonal case gives me an equation with no further insights!
$endgroup$
– Loves Probability
Mar 20 '15 at 23:01
$begingroup$
For example, the solution for the matrix $$pmatrix{1\&1\&&-1}$$ is the cone $x^2+y^2=z^2$. Note that this is not a linear subspace.
$endgroup$
– Omnomnomnom
Mar 21 '15 at 5:07