Mixing
A matrix equation equivalence
$begingroup$
Let $Omega_{, mtimes m}$ be a real square positive definite symmetric matrix, $u_{mtimes 1}$ is a vector, $I_{mtimes m}$ is the identity matrix.
Let $x$ be a solution of a matrix equation
$$ u^T(Omega-x I)^{-1}u=1tag{1}$$
I have hypothesised that the solution can be found as an eigenvalue problem of a matrix
$$Lambda := Omega - uu^T$$
i.e. $x$ satisfies the equation
$$detleft(Lambda-x Iright)=0.tag{2}$$
Is this correct? Does the relation $(2)$ for $x$ indeed follow from $(1)$?
Motivation: I have managed to show the statement for $m=2$ and $m=3$ by diagonalisation of $Omega$ and then concluded it can hold for higher $m$. It goes as follows... Let $Omega=ODO^T$ be an orthonormal decomposition ($OO^T=O^TO=I$) with $D=operatorname{diag}(lambda_1,lambda_2,dots,lambda_m)$ with $lambda_1leqlambda_2leqdotsleqlambda_m$. Let me denote $u=Ov$, then $(1)$ takes form
$$v^T(D-x I)^{-1}v=1tag{3}$$
or
$$sum_{k=1}^m frac{v_k^2}{lambda_k-x}=1tag{4}.$$
For $m=2$ this becomes
$$v_1^2(lambda_2-x)+v_2^2(lambda_1-x)=lambda_1lambda_2 - x(lambda_1+lambda_2)+x^2,$$
which can be rearranged to
$$x^2 - x(lambda_1+lambda_2-(v_1^2+v_2^2))-lambda_1lambda_2left(frac{v_1^2}{lambda_1}+frac{v_1^2}{lambda_1}right)=0tag{5}.$$
The relation $(2)$ becomes under decomposition
$$detleft(D-vv^T-xIright)=0tag{6}$$
which is equal to
$$x^2-xoperatorname{Tr}left(D-vv^Tright)+detleft(D-vv^Tright)=0$$
which is, after some easy algebraic manipulations, equal to $(5)$. The similar proof holds for $m=3$. Is there, however, a general approach?
linear-algebra matrices
asked Jan 20 at 17:03
MachinatoMachinato
1,1481020
$endgroup$
add a comment |
$begingroup$
Let $Omega_{, mtimes m}$ be a real square positive definite symmetric matrix, $u_{mtimes 1}$ is a vector, $I_{mtimes m}$ is the identity matrix.
Let $x$ be a solution of a matrix equation
$$ u^T(Omega-x I)^{-1}u=1tag{1}$$
I have hypothesised that the solution can be found as an eigenvalue problem of a matrix
$$Lambda := Omega - uu^T$$
i.e. $x$ satisfies the equation
$$detleft(Lambda-x Iright)=0.tag{2}$$
Is this correct? Does the relation $(2)$ for $x$ indeed follow from $(1)$?
Motivation: I have managed to show the statement for $m=2$ and $m=3$ by diagonalisation of $Omega$ and then concluded it can hold for higher $m$. It goes as follows... Let $Omega=ODO^T$ be an orthonormal decomposition ($OO^T=O^TO=I$) with $D=operatorname{diag}(lambda_1,lambda_2,dots,lambda_m)$ with $lambda_1leqlambda_2leqdotsleqlambda_m$. Let me denote $u=Ov$, then $(1)$ takes form
$$v^T(D-x I)^{-1}v=1tag{3}$$
or
$$sum_{k=1}^m frac{v_k^2}{lambda_k-x}=1tag{4}.$$
For $m=2$ this becomes
$$v_1^2(lambda_2-x)+v_2^2(lambda_1-x)=lambda_1lambda_2 - x(lambda_1+lambda_2)+x^2,$$
which can be rearranged to
$$x^2 - x(lambda_1+lambda_2-(v_1^2+v_2^2))-lambda_1lambda_2left(frac{v_1^2}{lambda_1}+frac{v_1^2}{lambda_1}right)=0tag{5}.$$
The relation $(2)$ becomes under decomposition
$$detleft(D-vv^T-xIright)=0tag{6}$$
which is equal to
$$x^2-xoperatorname{Tr}left(D-vv^Tright)+detleft(D-vv^Tright)=0$$
which is, after some easy algebraic manipulations, equal to $(5)$. The similar proof holds for $m=3$. Is there, however, a general approach?
linear-algebra matrices
asked Jan 20 at 17:03
MachinatoMachinato
1,1481020
$endgroup$
add a comment |
$begingroup$
Let $Omega_{, mtimes m}$ be a real square positive definite symmetric matrix, $u_{mtimes 1}$ is a vector, $I_{mtimes m}$ is the identity matrix.
Let $x$ be a solution of a matrix equation
$$ u^T(Omega-x I)^{-1}u=1tag{1}$$
I have hypothesised that the solution can be found as an eigenvalue problem of a matrix
$$Lambda := Omega - uu^T$$
i.e. $x$ satisfies the equation
$$detleft(Lambda-x Iright)=0.tag{2}$$
Is this correct? Does the relation $(2)$ for $x$ indeed follow from $(1)$?
Motivation: I have managed to show the statement for $m=2$ and $m=3$ by diagonalisation of $Omega$ and then concluded it can hold for higher $m$. It goes as follows... Let $Omega=ODO^T$ be an orthonormal decomposition ($OO^T=O^TO=I$) with $D=operatorname{diag}(lambda_1,lambda_2,dots,lambda_m)$ with $lambda_1leqlambda_2leqdotsleqlambda_m$. Let me denote $u=Ov$, then $(1)$ takes form
$$v^T(D-x I)^{-1}v=1tag{3}$$
or
$$sum_{k=1}^m frac{v_k^2}{lambda_k-x}=1tag{4}.$$
For $m=2$ this becomes
$$v_1^2(lambda_2-x)+v_2^2(lambda_1-x)=lambda_1lambda_2 - x(lambda_1+lambda_2)+x^2,$$
which can be rearranged to
$$x^2 - x(lambda_1+lambda_2-(v_1^2+v_2^2))-lambda_1lambda_2left(frac{v_1^2}{lambda_1}+frac{v_1^2}{lambda_1}right)=0tag{5}.$$
The relation $(2)$ becomes under decomposition
$$detleft(D-vv^T-xIright)=0tag{6}$$
which is equal to
$$x^2-xoperatorname{Tr}left(D-vv^Tright)+detleft(D-vv^Tright)=0$$
which is, after some easy algebraic manipulations, equal to $(5)$. The similar proof holds for $m=3$. Is there, however, a general approach?
linear-algebra matrices
asked Jan 20 at 17:03
MachinatoMachinato
1,1481020
$endgroup$
Let $Omega_{, mtimes m}$ be a real square positive definite symmetric matrix, $u_{mtimes 1}$ is a vector, $I_{mtimes m}$ is the identity matrix.
Let $x$ be a solution of a matrix equation
$$ u^T(Omega-x I)^{-1}u=1tag{1}$$
I have hypothesised that the solution can be found as an eigenvalue problem of a matrix
$$Lambda := Omega - uu^T$$
i.e. $x$ satisfies the equation
$$detleft(Lambda-x Iright)=0.tag{2}$$
Is this correct? Does the relation $(2)$ for $x$ indeed follow from $(1)$?
Motivation: I have managed to show the statement for $m=2$ and $m=3$ by diagonalisation of $Omega$ and then concluded it can hold for higher $m$. It goes as follows... Let $Omega=ODO^T$ be an orthonormal decomposition ($OO^T=O^TO=I$) with $D=operatorname{diag}(lambda_1,lambda_2,dots,lambda_m)$ with $lambda_1leqlambda_2leqdotsleqlambda_m$. Let me denote $u=Ov$, then $(1)$ takes form
$$v^T(D-x I)^{-1}v=1tag{3}$$
or
$$sum_{k=1}^m frac{v_k^2}{lambda_k-x}=1tag{4}.$$
For $m=2$ this becomes
$$v_1^2(lambda_2-x)+v_2^2(lambda_1-x)=lambda_1lambda_2 - x(lambda_1+lambda_2)+x^2,$$
which can be rearranged to
$$x^2 - x(lambda_1+lambda_2-(v_1^2+v_2^2))-lambda_1lambda_2left(frac{v_1^2}{lambda_1}+frac{v_1^2}{lambda_1}right)=0tag{5}.$$
The relation $(2)$ becomes under decomposition
$$detleft(D-vv^T-xIright)=0tag{6}$$
which is equal to
$$x^2-xoperatorname{Tr}left(D-vv^Tright)+detleft(D-vv^Tright)=0$$
which is, after some easy algebraic manipulations, equal to $(5)$. The similar proof holds for $m=3$. Is there, however, a general approach?
linear-algebra matrices
linear-algebra matrices
asked Jan 20 at 17:03
MachinatoMachinato
1,1481020
asked Jan 20 at 17:03
MachinatoMachinato
1,1481020
edited Jan 20 at 19:33
Machinato
asked Jan 20 at 17:03
MachinatoMachinato
1,1481020
asked Jan 20 at 17:03
MachinatoMachinato
1,1481020
asked Jan 20 at 17:03
MachinatoMachinato
1,1481020
1,1481020
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Yes, it's correct since
$$begin{eqnarray}
(Lambda-xI)(Omega-xI)^{-1}u&=&(Omega-uu^T-xI)(Omega-xI)^{-1}u\
&=&(Omega-xI)(Omega-xI)^{-1}u-uu^T(Omega-xI)^{-1}u\
&=&u-ucdotleft(u^T(Omega-xI)^{-1}uright)=u-u=0.
end{eqnarray}$$ Hence $$
0ne (Omega-xI)^{-1}uin ker (Lambda-xI)
$$ and $det (Lambda-xI)=0$ follows.
Addendum
If $det(Lambda -xI)=0$ and $(Omega-xI)^{-1}$ exists, then $u^T(Omega-xI)^{-1}u=1$.
Proof: Let $vne 0$ be a vector such that
$$
Lambda v=xv,quad Omega v -(uu^T)v=xv.
$$ This gives
$$
(Omega-xI)v=(u^Tv)u
$$ and
$$
v=(u^Tv)(Omega-xI)^{-1}u.
$$ In particular, $u^Tvne 0$. By left-multiplying $u^T$ on both sides, it follows
$$
u^Tv = (u^Tv)cdot u^T(Omega-xI)^{-1}u.
$$ Since $u^Tvne 0$, we have
$$
1=u^T(Omega-xI)^{-1}u.
$$
answered Jan 20 at 19:45
SongSong
16.7k1741
$endgroup$
$begingroup$
That's great! Thank you very much for such a simple answer. However, I cannot see clerly how the other implications follows? I.e. from $det(Lambda-xI)=0$ to the original equation.
$endgroup$
– Machinato
Jan 20 at 20:09
1
$begingroup$
@Machinato I've added some part explaining that. I hope this will help.
$endgroup$
– Song
Jan 20 at 21:16
add a comment |
Your Answer
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1 Answer
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1 Answer
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active
oldest
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oldest
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oldest
votes
$begingroup$
Yes, it's correct since
$$begin{eqnarray}
(Lambda-xI)(Omega-xI)^{-1}u&=&(Omega-uu^T-xI)(Omega-xI)^{-1}u\
&=&(Omega-xI)(Omega-xI)^{-1}u-uu^T(Omega-xI)^{-1}u\
&=&u-ucdotleft(u^T(Omega-xI)^{-1}uright)=u-u=0.
end{eqnarray}$$ Hence $$
0ne (Omega-xI)^{-1}uin ker (Lambda-xI)
$$ and $det (Lambda-xI)=0$ follows.
Addendum
If $det(Lambda -xI)=0$ and $(Omega-xI)^{-1}$ exists, then $u^T(Omega-xI)^{-1}u=1$.
Proof: Let $vne 0$ be a vector such that
$$
Lambda v=xv,quad Omega v -(uu^T)v=xv.
$$ This gives
$$
(Omega-xI)v=(u^Tv)u
$$ and
$$
v=(u^Tv)(Omega-xI)^{-1}u.
$$ In particular, $u^Tvne 0$. By left-multiplying $u^T$ on both sides, it follows
$$
u^Tv = (u^Tv)cdot u^T(Omega-xI)^{-1}u.
$$ Since $u^Tvne 0$, we have
$$
1=u^T(Omega-xI)^{-1}u.
$$
answered Jan 20 at 19:45
SongSong
16.7k1741
$endgroup$
$begingroup$
That's great! Thank you very much for such a simple answer. However, I cannot see clerly how the other implications follows? I.e. from $det(Lambda-xI)=0$ to the original equation.
$endgroup$
– Machinato
Jan 20 at 20:09
1
$begingroup$
@Machinato I've added some part explaining that. I hope this will help.
$endgroup$
– Song
Jan 20 at 21:16
add a comment |
$begingroup$
Yes, it's correct since
$$begin{eqnarray}
(Lambda-xI)(Omega-xI)^{-1}u&=&(Omega-uu^T-xI)(Omega-xI)^{-1}u\
&=&(Omega-xI)(Omega-xI)^{-1}u-uu^T(Omega-xI)^{-1}u\
&=&u-ucdotleft(u^T(Omega-xI)^{-1}uright)=u-u=0.
end{eqnarray}$$ Hence $$
0ne (Omega-xI)^{-1}uin ker (Lambda-xI)
$$ and $det (Lambda-xI)=0$ follows.
Addendum
If $det(Lambda -xI)=0$ and $(Omega-xI)^{-1}$ exists, then $u^T(Omega-xI)^{-1}u=1$.
Proof: Let $vne 0$ be a vector such that
$$
Lambda v=xv,quad Omega v -(uu^T)v=xv.
$$ This gives
$$
(Omega-xI)v=(u^Tv)u
$$ and
$$
v=(u^Tv)(Omega-xI)^{-1}u.
$$ In particular, $u^Tvne 0$. By left-multiplying $u^T$ on both sides, it follows
$$
u^Tv = (u^Tv)cdot u^T(Omega-xI)^{-1}u.
$$ Since $u^Tvne 0$, we have
$$
1=u^T(Omega-xI)^{-1}u.
$$
answered Jan 20 at 19:45
SongSong
16.7k1741
$endgroup$
$begingroup$
That's great! Thank you very much for such a simple answer. However, I cannot see clerly how the other implications follows? I.e. from $det(Lambda-xI)=0$ to the original equation.
$endgroup$
– Machinato
Jan 20 at 20:09
1
$begingroup$
@Machinato I've added some part explaining that. I hope this will help.
$endgroup$
– Song
Jan 20 at 21:16
add a comment |
$begingroup$
Yes, it's correct since
$$begin{eqnarray}
(Lambda-xI)(Omega-xI)^{-1}u&=&(Omega-uu^T-xI)(Omega-xI)^{-1}u\
&=&(Omega-xI)(Omega-xI)^{-1}u-uu^T(Omega-xI)^{-1}u\
&=&u-ucdotleft(u^T(Omega-xI)^{-1}uright)=u-u=0.
end{eqnarray}$$ Hence $$
0ne (Omega-xI)^{-1}uin ker (Lambda-xI)
$$ and $det (Lambda-xI)=0$ follows.
Addendum
If $det(Lambda -xI)=0$ and $(Omega-xI)^{-1}$ exists, then $u^T(Omega-xI)^{-1}u=1$.
Proof: Let $vne 0$ be a vector such that
$$
Lambda v=xv,quad Omega v -(uu^T)v=xv.
$$ This gives
$$
(Omega-xI)v=(u^Tv)u
$$ and
$$
v=(u^Tv)(Omega-xI)^{-1}u.
$$ In particular, $u^Tvne 0$. By left-multiplying $u^T$ on both sides, it follows
$$
u^Tv = (u^Tv)cdot u^T(Omega-xI)^{-1}u.
$$ Since $u^Tvne 0$, we have
$$
1=u^T(Omega-xI)^{-1}u.
$$
answered Jan 20 at 19:45
SongSong
16.7k1741
$endgroup$
Yes, it's correct since
$$begin{eqnarray}
(Lambda-xI)(Omega-xI)^{-1}u&=&(Omega-uu^T-xI)(Omega-xI)^{-1}u\
&=&(Omega-xI)(Omega-xI)^{-1}u-uu^T(Omega-xI)^{-1}u\
&=&u-ucdotleft(u^T(Omega-xI)^{-1}uright)=u-u=0.
end{eqnarray}$$ Hence $$
0ne (Omega-xI)^{-1}uin ker (Lambda-xI)
$$ and $det (Lambda-xI)=0$ follows.
Addendum
If $det(Lambda -xI)=0$ and $(Omega-xI)^{-1}$ exists, then $u^T(Omega-xI)^{-1}u=1$.
Proof: Let $vne 0$ be a vector such that
$$
Lambda v=xv,quad Omega v -(uu^T)v=xv.
$$ This gives
$$
(Omega-xI)v=(u^Tv)u
$$ and
$$
v=(u^Tv)(Omega-xI)^{-1}u.
$$ In particular, $u^Tvne 0$. By left-multiplying $u^T$ on both sides, it follows
$$
u^Tv = (u^Tv)cdot u^T(Omega-xI)^{-1}u.
$$ Since $u^Tvne 0$, we have
$$
1=u^T(Omega-xI)^{-1}u.
$$
answered Jan 20 at 19:45
SongSong
16.7k1741
edited Jan 20 at 21:15
answered Jan 20 at 19:45
SongSong
16.7k1741
answered Jan 20 at 19:45
SongSong
16.7k1741
answered Jan 20 at 19:45
SongSong
16.7k1741
16.7k1741
$begingroup$
That's great! Thank you very much for such a simple answer. However, I cannot see clerly how the other implications follows? I.e. from $det(Lambda-xI)=0$ to the original equation.
$endgroup$
– Machinato
Jan 20 at 20:09
1
$begingroup$
@Machinato I've added some part explaining that. I hope this will help.
$endgroup$
– Song
Jan 20 at 21:16
add a comment |
$begingroup$
That's great! Thank you very much for such a simple answer. However, I cannot see clerly how the other implications follows? I.e. from $det(Lambda-xI)=0$ to the original equation.
$endgroup$
– Machinato
Jan 20 at 20:09
1
$begingroup$
@Machinato I've added some part explaining that. I hope this will help.
$endgroup$
– Song
Jan 20 at 21:16
$begingroup$
That's great! Thank you very much for such a simple answer. However, I cannot see clerly how the other implications follows? I.e. from $det(Lambda-xI)=0$ to the original equation.
$endgroup$
– Machinato
Jan 20 at 20:09
$begingroup$
That's great! Thank you very much for such a simple answer. However, I cannot see clerly how the other implications follows? I.e. from $det(Lambda-xI)=0$ to the original equation.
$endgroup$
– Machinato
Jan 20 at 20:09
1
1
$begingroup$
@Machinato I've added some part explaining that. I hope this will help.
$endgroup$
– Song
Jan 20 at 21:16
$begingroup$
@Machinato I've added some part explaining that. I hope this will help.
$endgroup$
– Song
Jan 20 at 21:16
add a comment |
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